## Practice Makes Perfect

**Solve Equations Using the Division and Multiplication Properties of Equality**

In the following exercises, solve each equation using the Division and Multiplication Properties of Equality and check the solution.

Exercise (PageIndex{1})

(8x=56)

**Answer**(x=7)

Exercise (PageIndex{2})

(7 p=63)

Exercise (PageIndex{3})

(-5 c=55)

**Answer**(c=-11)

Exercise (PageIndex{4})

(-9 x=-27)

Exercise (PageIndex{5})

(-809=15 y)

**Answer**(y = -frac{809}{15})

Exercise (PageIndex{6})

(-731=19 y)

Exercise (PageIndex{7})

(-37 p=-541)

**Answer**(p=-frac{541}{37})

Exercise (PageIndex{8})

(-19 m=-586)

Exercise (PageIndex{9})

(0.25 z=3.25)

**Answer**z= 13

Exercise (PageIndex{10})

(0.75 a=11.25)

Exercise (PageIndex{11})

(-13x=0)

**Answer**(x=0)

Exercise (PageIndex{12})

(24x=0)

Exercise (PageIndex{13})

(frac{x}{4} = 35)

**Answer**(x=140)

Exercise (PageIndex{14})

(frac{z}{2}=54)

Exercise (PageIndex{15})

(-20=frac{q}{-5})

**Answer**(q=100)

Exercise (PageIndex{16})

(frac{c}{-3}=-12)

Exercise (PageIndex{17})

(frac{y}{9}=-16)

**Answer**(y=-144)

Exercise (PageIndex{18})

(frac{q}{6}=-38)

Exercise (PageIndex{19})

(frac{m}{-12}=45)

**Answer**(m=-540)

Exercise (PageIndex{20})

(-24=frac{p}{-20})

Exercise (PageIndex{21})

(-y=6)

**Answer**(y=-6)

Exercise (PageIndex{22})

(-u=15)

Exercise (PageIndex{23})

(-v=-72)

**Answer**(v=72)

Exercise (PageIndex{24})

(-x=-39)

Exercise (PageIndex{25})

(frac{2}{3} y=48)

**Answer**(y=72)

Exercise (PageIndex{26})

(frac{3}{5} r=75)

Exercise (PageIndex{27})

(-frac{5}{8} w=40)

**Answer**(w=-64)

Exercise (PageIndex{28})

(24=-frac{3}{4} x)

Exercise (PageIndex{29})

(-frac{2}{5}=frac{1}{10} a)

**Answer**(a=-4)

Exercise (PageIndex{30})

(-frac{1}{3} q=-frac{5}{6})

Exercise (PageIndex{31})

(-frac{7}{10} x=-frac{14}{3})

**Answer**(x=frac{20}{3})

Exercise (PageIndex{32})

(frac{3}{8} y=-frac{1}{4})

Exercise (PageIndex{33})

(frac{7}{12}=-frac{3}{4} p)

**Answer**(p=-frac{7}{9})

Exercise (PageIndex{34})

(frac{11}{18}=-frac{5}{6} q)

Exercise (PageIndex{35})

(-frac{5}{18}=-frac{10}{9} u)

**Answer**(u=frac{1}{4})

Exercise (PageIndex{36})

(-frac{7}{20}=-frac{7}{4} v)

**Solve Equations That Require Simplification**

In the following exercises, solve each equation requiring simplification.

Exercise (PageIndex{37})

(100-16=4 p-10 p-p)

**Answer**(p=-12)

Exercise (PageIndex{38})

(-18-7=5 t-9 t-6 t)

Exercise (PageIndex{39})

(frac{7}{8} n-frac{3}{4} n=9+2)

**Answer**(n=88)

Exercise (PageIndex{40})

(frac{5}{12} q+frac{1}{2} q=25-3)

Exercise (PageIndex{41})

(0.25 d+0.10 d=6-0.75)

**Answer**d=15

Exercise (PageIndex{42})

(0.05 p-0.01 p=2+0.24)

Exercise (PageIndex{43})

(-10(q-4)-57=93)

**Answer**(q=-11)

Exercise (PageIndex{44})

(-12(d-5)-29=43)

Exercise (PageIndex{45})

(-10(x+4)-19=85)

**Answer**(x=-frac{72}{5})

Exercise (PageIndex{46})

(-15(z+9)-11=75)

**Mixed Practice**

In the following exercises, solve each equation.

Exercise (PageIndex{47})

(frac{9}{10} x=90)

**Answer**(x=100)

Exercise (PageIndex{48})

(frac{5}{12} y=60)

Exercise (PageIndex{49})

(y+46=55)

**Answer**(y=9)

Exercise (PageIndex{50})

(x+33=41)

Exercise (PageIndex{51})

(frac{w}{-2}=99)

**Answer**(w=-198)

Exercise (PageIndex{52})

(frac{s}{-3}=-60)

Exercise (PageIndex{53})

(27=6 a)

**Answer**(a=frac{9}{2})

Exercise (PageIndex{54})

(-a=7)

Exercise (PageIndex{55})

(-x=2)

**Answer**(x=-2)

Exercise (PageIndex{56})

(z-16=-59)

Exercise (PageIndex{57})

(m-41=-14)

**Answer**(m=27)

Exercise (PageIndex{58})

(0.04 r=52.60)

Exercise (PageIndex{59})

(63.90=0.03 p)

**Answer**(p=2130)

Exercise (PageIndex{60})

(-15 x=-120)

Exercise (PageIndex{61})

(84=-12 z)

**Answer**(y=-7)

Exercise (PageIndex{62})

(19.36=x-0.2 x)

Exercise (PageIndex{63})

(c-0.3 c=35.70)

**Answer**(c=51)

Exercise (PageIndex{64})

(-y=-9)

Exercise (PageIndex{65})

(-x=-8)

**Answer**(x=8)

**Translate to an Equation and Solve**

In the following exercises, translate to an equation and then solve.

Exercise (PageIndex{66})

187 is the product of (-17) and (m)

Exercise (PageIndex{67})

133 is the product of (-19) and (n)

**Answer**(133=-19 n ; n=-7)

Exercise (PageIndex{68})

(-184) is the product of 23 and (p)

Exercise (PageIndex{69})

(-152) is the product of 8 and (q)

**Answer**(-152=8 q ; q=-19)

Exercise (PageIndex{70})

(u) divided by 7 is equal to (-49)

Exercise (PageIndex{71})

(r) divided by 12 is equal to (-48)

**Answer**(frac{r}{12}=-48 ; r=-576)

Exercise (PageIndex{72})

(h) divided by (-13) is equal to (-65)

Exercise (PageIndex{73})

(j) divided by (-20) is equal to (-80)

**Answer**(frac{j}{-20}=-80 ; j=1,600)

Exercise (PageIndex{74})

The quotient (c) and (-19) is (38 .)

Exercise (PageIndex{75})

The quotient of (b) and (-6) is 18

**Answer**(frac{b}{-6}=18 ; b=-108)

Exercise (PageIndex{76})

The quotient of (h) and 26 is (-52)

Exercise (PageIndex{77})

The quotient (k) and 22 is (-66)

**Answer**(frac{k}{22}=-66 ; k=-1,452)

Exercise (PageIndex{78})

Five-sixths of (y) is 15

Exercise (PageIndex{79})

Three-tenths of (x) is 15

**Answer**(frac{3}{10} x=15 ; x=50)

Exercise (PageIndex{80})

Four-thirds of (w) is 36

Exercise (PageIndex{81})

Five-halves of (v) is 50

**Answer**(frac{5}{2} v=50 ; v=20)

Exercise (PageIndex{82})

The sum of nine-tenths and (g) is two-thirds.

Exercise (PageIndex{83})

The sum of two-fifths and (f) is one-half.

**Answer**(frac{2}{5}+f=frac{1}{2} ; f=frac{1}{10})

Exercise (PageIndex{84})

The difference of (p) and one-sixth is two-thirds.

Exercise (PageIndex{85})

The difference of (q) and one-eighth is three-fourths.

**Answer**(q-frac{1}{8}=frac{3}{4} ; q=frac{7}{8})

**Translate and Solve Applications**

In the following exercises, translate into an equation and solve.

Exercise (PageIndex{86})

**Kindergarten** Connie’s kindergarten class has 24 children. She wants them to get into 4 equal groups. How many children will she put in each group?

Exercise (PageIndex{87})

**Balloons** Ramona bought 18 balloons for a party. She wants to make 3 equal bunches. How many balloons did she use in each bunch?

**Answer**6 balloons

Exercise (PageIndex{88})

**Tickets** Mollie paid $36.25 for 5 movie tickets. What was the price of each ticket?

Exercise (PageIndex{89})

**Shopping** Serena paid $12.96 for a pack of 12 pairs of sport socks. What was the price of pair of sport socks?

**Answer**$1.08

Exercise (PageIndex{90})

**Sewing** Nancy used 14 yards of fabric to make flags for one-third of the drill team. How much fabric, would Nancy need to make flags for the whole team?

Exercise (PageIndex{91})

**MPG** John’s SUV gets 18 miles per gallon (mpg). This is half as many mpg as his wife’s hybrid car. How many miles per gallon does the hybrid car get?

**Answer**36 mpg

Exercise (PageIndex{92})

**Height** Aiden is 27 inches tall. He is (frac{3}{8}) as tall as his father. How tall is his father?

Exercise (PageIndex{92})

**Real estate** Bea earned ($ 11,700) commission for selling a house, calculated as (frac{6}{100}) of the selling price. What was the selling

price of the house?

**Answer**$195,000

## Everyday Math

Exercise (PageIndex{93})

**Commissions** Every week Perry gets paid ($150) plus 12% of his total sales amount. Solve the equation

(840=150+0.12(a-1250)) for (a) to find the total amount Perry must sell in order to be paid ($ 840) one week.

Exercise (PageIndex{94})

**Stamps** Travis bought $9.45 worth of 49-cent stamps and 21-cent stamps. The number of 21-cent stamps was 5 less than the number of 49-cent stamps. Solve the equation 0.49s+0.21(s−5)=9.45 for *s*, to find the number of 49-cent stamps Travis bought.

**Answer**15 49-cent stamps

## Writing Exercises

Exercise (PageIndex{95})

Frida started to solve the equation −3x=36 by adding 3 to both sides. Explain why Frida’s method will not solve the equation.

Exercise (PageIndex{96})

Emiliano thinks (x=40) is the solution to the equation (frac{1}{2} x=80 .) Explain why he is wrong.

**Answer**Answers will vary.

## Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

## Exercises

Identify each half-reaction below as either oxidation or reduction.

Identify each half-reaction below as either oxidation or reduction.

Assuming each pair of half-reactions below takes place in an acidic solution, write a balanced equation for the overall reaction.

Balance the equations below assuming they occur in an acidic solution.

(a) H 2 O 2 + Sn 2+ ⟶ H 2 O + Sn 4+ H 2 O 2 + Sn 2+ ⟶ H 2 O + Sn 4+

(b) PbO 2 + Hg ⟶ Hg 2 2+ + Pb 2+ PbO 2 + Hg ⟶ Hg 2 2+ + Pb 2+

(c) Al + Cr 2 O 7 2− ⟶ Al 3+ + Cr 3+ Al + Cr 2 O 7 2− ⟶ Al 3+ + Cr 3+

Identify the oxidant and reductant of each reaction of the previous exercise.

Balance the equations below assuming they occur in a basic solution.

(a) SO 3 2− ( a q ) + Cu(OH) 2 ( s ) ⟶ SO 4 2− ( a q ) + Cu(OH) ( s ) SO 3 2− ( a q ) + Cu(OH) 2 ( s ) ⟶ SO 4 2− ( a q ) + Cu(OH) ( s )

(b) O 2 ( g ) + Mn(OH) 2 ( s ) ⟶ MnO 2 ( s ) O 2 ( g ) + Mn(OH) 2 ( s ) ⟶ MnO 2 ( s )

(c) NO 3 − ( a q ) + H 2 ( g ) ⟶ NO ( g ) NO 3 − ( a q ) + H 2 ( g ) ⟶ NO ( g )

(d) Al ( s ) + CrO 4 2− ( a q ) ⟶ Al(OH) 3 ( s ) + Cr(OH) 4 − ( a q ) Al ( s ) + CrO 4 2− ( a q ) ⟶ Al(OH) 3 ( s ) + Cr(OH) 4 − ( a q )

Identify the oxidant and reductant of each reaction of the previous exercise.

Why don’t hydroxide ions appear in equations for half-reactions occurring in acidic solution?

Why don’t hydrogen ions appear in equations for half-reactions occurring in basic solution?

Why must the charge balance in oxidation-reduction reactions?

### 16.2 Galvanic Cells

Write cell schematics for the following cell reactions, using platinum as an inert electrode as needed.

(a) Mg ( s ) + Ni 2+ ( a q ) ⟶ Mg 2+ ( a q ) + Ni ( s ) Mg ( s ) + Ni 2+ ( a q ) ⟶ Mg 2+ ( a q ) + Ni ( s )

(b) 2 Ag + ( a q ) + Cu ( s ) ⟶ Cu 2+ ( a q ) + 2Ag ( s ) 2 Ag + ( a q ) + Cu ( s ) ⟶ Cu 2+ ( a q ) + 2Ag ( s )

(c) Mn ( s ) + Sn(NO 3 ) 2 ( a q ) ⟶ Mn(NO 3 ) 2 ( a q ) + Sn ( s ) Mn ( s ) + Sn(NO 3 ) 2 ( a q ) ⟶ Mn(NO 3 ) 2 ( a q ) + Sn ( s )

(d) 3 CuNO 3 ( a q ) + Au(NO 3 ) 3 ( a q ) ⟶ 3Cu(NO 3 ) 2 ( a q ) + Au ( s ) 3 CuNO 3 ( a q ) + Au(NO 3 ) 3 ( a q ) ⟶ 3Cu(NO 3 ) 2 ( a q ) + Au ( s )

Assuming the schematics below represent galvanic cells as written, identify the half-cell reactions occurring in each.

(a) Mg ( s ) │ Mg 2+ ( a q ) ║ Cu 2+ ( a q ) │ Cu ( s ) Mg ( s ) │ Mg 2+ ( a q ) ║ Cu 2+ ( a q ) │ Cu ( s )

(b) Ni ( s ) │ Ni 2+ ( a q ) ║ Ag + ( a q ) │ Ag ( s ) Ni ( s ) │ Ni 2+ ( a q ) ║ Ag + ( a q ) │ Ag ( s )

Write a balanced equation for the cell reaction of each cell in the previous exercise.

Balance each reaction below, and write a cell schematic representing the reaction as it would occur in a galvanic cell.

(a) Al ( s ) + Zr 4+ ( a q ) ⟶ Al 3+ ( a q ) + Zr ( s ) Al ( s ) + Zr 4+ ( a q ) ⟶ Al 3+ ( a q ) + Zr ( s )

(b) Ag + ( a q ) + NO ( g ) ⟶ Ag ( s ) + NO 3 − ( a q ) ( acidic solution ) Ag + ( a q ) + NO ( g ) ⟶ Ag ( s ) + NO 3 − ( a q ) ( acidic solution )

(c) SiO 3 2− ( a q ) + Mg ( s ) ⟶ Si ( s ) + Mg ( OH ) 2 ( s ) (basic solution) SiO 3 2− ( a q ) + Mg ( s ) ⟶ Si ( s ) + Mg ( OH ) 2 ( s ) (basic solution)

(d) ClO 3 − ( a q ) + MnO 2 ( s ) ⟶ Cl − ( a q ) + MnO 4 − ( a q ) (basic solution) ClO 3 − ( a q ) + MnO 2 ( s ) ⟶ Cl − ( a q ) + MnO 4 − ( a q ) (basic solution)

Identify the oxidant and reductant in each reaction of the previous exercise.

From the information provided, use cell notation to describe the following systems:

(a) In one half-cell, a solution of Pt(NO_{3})_{2} forms Pt metal, while in the other half-cell, Cu metal goes into a Cu(NO_{3})_{2} solution with all solute concentrations 1 *M*.

(b) The cathode consists of a gold electrode in a 0.55 *M* Au(NO_{3})_{3} solution and the anode is a magnesium electrode in 0.75 *M* Mg(NO_{3})_{2} solution.

(c) One half-cell consists of a silver electrode in a 1 *M* AgNO_{3} solution, and in the other half-cell, a copper electrode in 1 *M* Cu(NO_{3})_{2} is oxidized.

Why is a salt bridge necessary in galvanic cells like the one in Figure 16.3?

An active (metal) electrode was found to gain mass as the oxidation-reduction reaction was allowed to proceed. Was the electrode an anode or a cathode? Explain.

An active (metal) electrode was found to lose mass as the oxidation-reduction reaction was allowed to proceed. Was the electrode an anode or a cathode? Explain.

The masses of three electrodes (A, B, and C), each from three different galvanic cells, were measured before and after the cells were allowed to pass current for a while. The mass of electrode A increased, that of electrode B was unchanged, and that of electrode C decreased. Identify each electrode as active or inert, and note (if possible) whether it functioned as anode or cathode.

### 16.3 Electrode and Cell Potentials

Calculate the standard cell potential for each reaction below, and note whether the reaction is spontaneous under standard state conditions.

(a) Mg ( s ) + Ni 2+ ( a q ) ⟶ Mg 2+ ( a q ) + Ni ( s ) Mg ( s ) + Ni 2+ ( a q ) ⟶ Mg 2+ ( a q ) + Ni ( s )

(b) 2 Ag + ( a q ) + Cu ( s ) ⟶ Cu 2+ ( a q ) + 2Ag ( s ) 2 Ag + ( a q ) + Cu ( s ) ⟶ Cu 2+ ( a q ) + 2Ag ( s )

(c) Mn ( s ) + Sn(NO 3 ) 2 ( a q ) ⟶ Mn(NO 3 ) 2 ( a q ) + Sn ( s ) Mn ( s ) + Sn(NO 3 ) 2 ( a q ) ⟶ Mn(NO 3 ) 2 ( a q ) + Sn ( s )

(d) 3 Fe(NO 3 ) 2 ( a q ) + Au(NO 3 ) 3 ( a q ) ⟶ 3Fe(NO 3 ) 3 ( a q ) + Au ( s ) 3 Fe(NO 3 ) 2 ( a q ) + Au(NO 3 ) 3 ( a q ) ⟶ 3Fe(NO 3 ) 3 ( a q ) + Au ( s )

Calculate the standard cell potential for each reaction below, and note whether the reaction is spontaneous under standard state conditions.

(a) Mn ( s ) + Ni 2+ ( a q ) ⟶ Mn 2+ ( a q ) + Ni ( s ) Mn ( s ) + Ni 2+ ( a q ) ⟶ Mn 2+ ( a q ) + Ni ( s )

(b) 3 Cu 2+ ( a q ) + 2Al ( s ) ⟶ 2Al 3+ ( a q ) + 3Cu ( s ) 3 Cu 2+ ( a q ) + 2Al ( s ) ⟶ 2Al 3+ ( a q ) + 3Cu ( s )

(c) Na ( s ) + LiNO 3 ( a q ) ⟶ NaNO 3 ( a q ) + Li ( s ) Na ( s ) + LiNO 3 ( a q ) ⟶ NaNO 3 ( a q ) + Li ( s )

(d) Ca(NO 3 ) 2 ( a q ) + Ba ( s ) ⟶ Ba(NO 3 ) 2 ( a q ) + Ca ( s ) Ca(NO 3 ) 2 ( a q ) + Ba ( s ) ⟶ Ba(NO 3 ) 2 ( a q ) + Ca ( s )

Write the balanced cell reaction for the cell schematic below, calculate the standard cell potential, and note whether the reaction is spontaneous under standard state conditions.

Cu ( s ) │ Cu 2+ ( a q ) ║ Au 3+ ( a q ) │ Au ( s ) Cu ( s ) │ Cu 2+ ( a q ) ║ Au 3+ ( a q ) │ Au ( s )

Determine the cell reaction and standard cell potential at 25 °C for a cell made from a cathode half-cell consisting of a silver electrode in 1 *M* silver nitrate solution and an anode half-cell consisting of a zinc electrode in 1 *M* zinc nitrate. Is the reaction spontaneous at standard conditions?

Determine the cell reaction and standard cell potential at 25 °C for a cell made from an anode half-cell containing a cadmium electrode in 1 *M* cadmium nitrate and an anode half-cell consisting of an aluminum electrode in 1 *M* aluminum nitrate solution. Is the reaction spontaneous at standard conditions?

Write the balanced cell reaction for the cell schematic below, calculate the standard cell potential, and note whether the reaction is spontaneous under standard state conditions.

Pt ( s ) │ H 2 ( g ) │ H + ( a q ) ║ Br 2 ( a q ) , Br − ( a q ) │ Pt ( s ) Pt ( s ) │ H 2 ( g ) │ H + ( a q ) ║ Br 2 ( a q ) , Br − ( a q ) │ Pt ( s )

### 16.4 Potential, Free Energy, and Equilibrium

For each pair of standard cell potential and electron stoichiometry values below, calculate a corresponding standard free energy change (kJ).

For each pair of standard free energy change and electron stoichiometry values below, calculate a corresponding standard cell potential.

Determine the standard cell potential and the cell potential under the stated conditions for the electrochemical reactions described here. State whether each is spontaneous or nonspontaneous under each set of conditions at 298.15 K.

(a) Hg ( l ) + S 2− ( a q , 0.10 M ) + 2 Ag + ( a q , 0.25 M ) ⟶ 2 Ag ( s ) + HgS ( s ) Hg ( l ) + S 2− ( a q , 0.10 M ) + 2 Ag + ( a q , 0.25 M ) ⟶ 2 Ag ( s ) + HgS ( s )

(b) The cell made from an anode half-cell consisting of an aluminum electrode in 0.015 *M* aluminum nitrate solution and a cathode half-cell consisting of a nickel electrode in 0.25 *M* nickel(II) nitrate solution.

(c) The cell comprised of a half-cell in which aqueous bromine (1.0 *M*) is being oxidized to bromide ion (0.11 *M*) and a half-cell in which Al 3+ (0.023 *M*) is being reduced to aluminum metal.

Determine Δ*G* and Δ*G*° for each of the reactions in the previous problem.

Use the data in Appendix L to calculate equilibrium constants for the following reactions. Assume 298.15 K if no temperature is given.

(a) AgCl ( s ) ⇌ Ag + ( a q ) + Cl − ( a q ) AgCl ( s ) ⇌ Ag + ( a q ) + Cl − ( a q )

(b) CdS ( s ) ⇌ Cd 2+ ( a q ) + S 2− ( a q ) at 377 K CdS ( s ) ⇌ Cd 2+ ( a q ) + S 2− ( a q ) at 377 K

(c) Hg 2+ ( a q ) + 4 Br − ( a q ) ⇌ [ HgBr 4 ] 2− ( a q ) Hg 2+ ( a q ) + 4 Br − ( a q ) ⇌ [ HgBr 4 ] 2− ( a q )

(d) H 2 O ( l ) ⇌ H + ( a q ) + OH − ( a q ) at 25 °C H 2 O ( l ) ⇌ H + ( a q ) + OH − ( a q ) at 25 °C

### 16.5 Batteries and Fuel Cells

Consider a battery made from one half-cell that consists of a copper electrode in 1 *M* CuSO_{4} solution and another half-cell that consists of a lead electrode in 1 *M* Pb(NO_{3})_{2} solution.

(a) What is the standard cell potential for the battery?

(b) What are the reactions at the anode, cathode, and the overall reaction?

(c) Most devices designed to use dry-cell batteries can operate between 1.0 and 1.5 V. Could this cell be used to make a battery that could replace a dry-cell battery? Why or why not.

(d) Suppose sulfuric acid is added to the half-cell with the lead electrode and some PbSO_{4}(*s*) forms. Would the cell potential increase, decrease, or remain the same?

Consider a battery with the overall reaction: Cu ( s ) + 2 Ag + ( a q ) ⟶ 2Ag ( s ) + Cu 2+ ( a q ) . Cu ( s ) + 2 Ag + ( a q ) ⟶ 2Ag ( s ) + Cu 2+ ( a q ) .

(a) What is the reaction at the anode and cathode?

(b) A battery is “dead” when its cell potential is zero. What is the value of *Q* when this battery is dead?

(c) If a particular dead battery was found to have [Cu 2+ ] = 0.11 *M*, what was the concentration of silver ion?

Why do batteries go dead, but fuel cells do not?

Use the Nernst equation to explain the drop in voltage observed for some batteries as they discharge.

Using the information thus far in this chapter, explain why battery-powered electronics perform poorly in low temperatures.

### 16.6 Corrosion

Which member of each pair of metals is more likely to corrode (oxidize)?

Consider the following metals: Ag, Au, Mg, Ni, and Zn. Which of these metals could be used as a sacrificial anode in the cathodic protection of an underground steel storage tank? Steel is an alloy composed mostly of iron, so use −0.447 V as the standard reduction potential for steel.

If a sample of iron and a sample of zinc come into contact, the zinc corrodes but the iron does not. If a sample of iron comes into contact with a sample of copper, the iron corrodes but the copper does not. Explain this phenomenon.

Suppose you have three different metals, A, B, and C. When metals A and B come into contact, B corrodes and A does not corrode. When metals A and C come into contact, A corrodes and C does not corrode. Based on this information, which metal corrodes and which metal does not corrode when B and C come into contact?

Why would a sacrificial anode made of lithium metal be a bad choice

### 16.7 Electrolysis

If a 2.5 A current flows through a circuit for 35 minutes, how many coulombs of charge moved through the circuit?

For the scenario in the previous question, how many electrons moved through the circuit?

Write the half-reactions and cell reaction occurring during electrolysis of each molten salt below.

What mass of each product is produced in each of the electrolytic cells of the previous problem if a total charge of 3.33 × × 10 5 C passes through each cell?

How long would it take to reduce 1 mole of each of the following ions using the current indicated?

A current of 2.345 A passes through the cell shown in Figure 16.19 for 45 minutes. What is the volume of the hydrogen collected at room temperature if the pressure is exactly 1 atm? (Hint: Is hydrogen the only gas present above the water?)

An irregularly shaped metal part made from a particular alloy was galvanized with zinc using a Zn(NO_{3})_{2} solution. When a current of 2.599 A was used, it took exactly 1 hour to deposit a 0.01123-mm layer of zinc on the part. What was the total surface area of the part? The density of zinc is 7.140 g/cm 3 .

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## Q11 /VC

All reactants and products are in their standard states, and use data from the standard electrode reduction potentials table to predict whether the reaction is spontaneous in the forward reaction:

- (mathrm
(aq) + 2Ag (s)>) - (mathrm <2Al (s) + 3Zn^<2+>(aq) ightarrow 2Al^ <3+>(aq) + 3Zn (s)>)
- (mathrm
(aq) + Ag^+ (aq) ightarrow Fe^ <3+>(aq) + Ag (s)>) - (mathrm <2Fe^<2+>(aq) + I_2 (s) ightarrow 2Fe^ <3+>(aq) + 2I^- (aq)>)

### S11 /VC

Write cell reactions for the electrochemical cells diagrammed, and use data from the table of standard electrode potentials to calculate (mathrm<>

>) for each reaction. | - (mathrm
(aq)||Zn (s) |Zn^ <2+>(aq)>) - (mathrm
) - (mathrm
(aq)||Mn^ <2+>(aq)|Mn(s)>) - (mathrm
(aq), Fe^ <3+>(aq)||Pb^<2+>(aq)|Pb(s)>)

Reduction: (mathrm

(aq) + 2e^- ightarrow Zn(s) hspace <68 pt>quad E^circ = -0.7621:V>)

Net Reaction: (mathrm(aq) ightarrow Fe^ <2+>(aq) + Zn(s) quad E^circ_ = -0.3321: V>) | Reduction: (mathrm

(aq) + 2e^- ightarrow Mn(s) hspace<71 pt>quad E^circ = -1.182:V>)

Net Reaction: (mathrm(aq) ightarrow Co^ <2+>(aq) + Mn(s) quad E^circ_ = -0.905:V>) | Reduction: (mathrm

(aq) + 2e^- ightarrow Pb(s) hspace<76 pt>quad E^circ= -0.125:V>)

Net Reaction: (mathrm <2Fe^<2+>+ Pb^<2+>(aq) ightarrow 2Fe^ <3+>(aq) + Pb(s) quad E^circ_= -0.896:V>) | For each of the following reactions below, draw a voltaic cell. In your drawing include the anode, cathode, and show the flow of electrons. Balance the equation and calculate the (mathrm<>

>). | Reduction: (mathrm

(aq) + 2e^- ightarrow Hg (s) hspace<43 pt>quad E^circ=0.86:V>) Net: (mathrm<2Rb(s) + Hg^<2+>(aq) ightarrow 2Rb^+(aq) + Hg (s) quad E^circ_

=3.79:V>) | What are the (mathrm

) for the reactions below? Reduction: (mathrm

(aq) + 2e^- ightarrow Cr (s) hspace<30.5 pt>quad E^circ=-0.90:V>) Net: (mathrm

(aq) ightarrow Ti^<2+>(aq) + Cr (s) quad E^circ_ =0.73:V>) | Reduction: (mathrm<(Cu^<2+>(aq) + e^- ightarrow Cu^+(aq)) imes2 hspace<39 pt>quad E^circ=0.159:V>)

Find (mathrm

), (mathrm ), (ce ), and given that the reactants and products are in their standard state, if they go to completion or not for the following reaction: Oxidation: (mathrm<(2SO_4^<2->(aq) ightarrow S_2O_8^<2->(aq) + 2e^-) imes5 hspace<127 pt>quad <-E>^circ=-2.01:V>)

Since (ce

) is small ((mathrm<<1>)) the reaction will not go to completion. Calculate the theoretical cell voltage for the reaction between copper and zinc given that the overall reaction is:

(mathrm(aq) ightarrow Zn^<2+>(aq) + Cu(s)>) Oxidation: (mathrm

(aq) + 2e^- hspace<1 pt>quad E^circ= 0.340: V>)

Reduction: (mathrm(aq) + 2e^- ightarrow Zn(s) quad E^circ= -0.763: V>)

(mathrm<>= -0.763:V - 0.340:V = -1.103:V>) | A voltaic cell has an (mathrm<>

>) value of 1.536 V. What is the concentration of (ce | ) in the cell? Using the Nernst equation, find (mathrm<>

>) for the following cells: | Oxidation: (mathrm<2Al(s) ightarrow 2Al^<3+>(aq) + 6e^- hspace <115 pt>quad E^circ_

= 1.676: V>) |

Reduction: (mathrm<3Fe^<3+>(aq) + 6e^- ightarrow 3Fe(s) hspace <110 pt>quad E^circ_= -0.440: V>) |

Net Reaction: (mathrm<2Al(s) + 3Fe^<2+>(1.1:M) ightarrow 2Al^<3+>(0.30:M) + 3Fe(s) quad E^circ_= 1.236: V>) | Oxidation: (mathrm

(aq) + 2e^- hspace<131 pt>quad <-E>^circ_ =0.277: V>) |

Reduction: (mathrm<2Fe^<3+>(aq) + 2e^- ightarrow 2Fe^<2+>(aq) hspace<112 pt>quad E^circ_=0.771>) |

Net Reaction: (mathrm(1.0:M) ightarrow Co^<2+>(0.6:M) +2Fe^<2+>(0.9:M) quad E^circ_ =1.048: V>) | If (ce<[Cu^2+]>) is maintained at 1.0 M, what is the minimum (ce<[Ag+]>) for which the reaction from 19.2, given below, is spontaneous in the forward direction?

Remember - in order for the reaction to be spontaneous, (mathrm

). Write the equation as a cell diagram:

Solve for (mathrm<>

>) of the following voltaic cell | Reduction: (mathrm

(aq) + 2e^- ightarrow Cu(s) quad E^circ= +0.340>) Then use the Nernst Equation to solve for (mathrm<>

>) | By referring back to figure 19-20, explain what would happen at each individual circumstance

- zinc is wrapped around the head and tip of the iron nail
- a hole is poked at the center of an iron nail
- the nail is completely covered with copper

- Through cathodic protection, zinc would get oxidized first. The zinc would protect the nail from oxidation.
- Oxidation would occur even more because there would now be another head and tip. With more strained regions, the air would be able to oxidize the nail more.
- The entire nail will be oxidized since copper isn't a sacrificial anode. It won't protect it from corrosion.

Calculate the amount in grams of metal that is deposited at the cathode by running a current of 3.15 A for 78 min in an electrolysis reaction for an aqueous solution containing a) (ce

) b) (ce ) c) (ce ) d) (ce ) - (mathrm<0.153:mol: electrons imes dfrac<1:mol: Zn><2:mol: e^->= 0.0764:mol: Zn imes dfrac<65.41:g><1:mol>= 5.00:g: Zn>)
- (mathrm<0.153:mol: electrons imes dfrac<1:mol: Sn><2:mol: e^->= 0.0764:mol: Sn imes dfrac<118.7:g><1:mol>= 9.07:g: Sn>)
- (mathrm<0.153:mol: electrons imes dfrac<1:mol: Fe><1:mol: e^->= 0.153:mol: Fe imes dfrac<55.85:g><1:mol>= 8.55:g: Fe>)
- (mathrm<0.153:mol: electrons imes dfrac<1:mol: Ni><2:mol: e^->= 0.0764:mol: Ni imes dfrac<58.69:g><1:mol>= 4.48:g: Ni>)

Assuming all reactants and products are currently in their standard states, determine which of the following reactions occur spontaneously and which can occur only through the use of electrolysis. Also, for those that require electrolysis, determine what the minimum voltage required is.

- (mathrm
(aq) ightarrow Cu^<2+>(aq) + Zn(s)>) - (mathrm <2Al + 3Cu^<2+> ightarrow 3Cu + 2Al^<3+>>)
- (mathrm
) - (mathrm <2Fe^<3+>+ 2Cl^- ightarrow 2Fe^ <2+>+ Cl_2(g)>)

Since voltage is negative, it requires electrolysis with an applied voltage of greater than 1.103 V

This is a spontaneous reaction under standard conditions because it has positive V.

This is a spontaneous reaction under standard conditions because it has positive V.

Since the V is negative, it requires electrolysis with an applied V of greater than 0.587 V.

In the two galvanic cells created the following variations occur:

[mathrm

(aq) + Ag^+(aq) + H_2O ightarrow LO^<2+>(aq) + 2H^+ + Ag(s) \E^circ= 0.439: V>] Use other values as well as the data presented here to determine (mathrm

) for the half reaction:

## Exercise 3: Mixed logit model

A sample of residential electricity customers were asked a series of choice experiments. In each experiment, four hypothetical electricity suppliers were described. The person was asked which of the four suppliers he/she would choose. As many as 12 experiments were presented to each person. Some people stopped before answering all 12. There are 361 people in the sample, and a total of 4308 experiments. In the experiments, the characteristics of each supplier were stated. The price of the supplier was either :

- a fixed price at a stated cents per kWh, with the price varying over suppliers and experiments.
- a time-of-day (TOD) rate under which the price is 11 cents per kWh from 8am to 8pm and 5 cents per kWh from 8pm to 8am. These TOD prices did not vary over suppliers or experiments: whenever the supplier was said to offer TOD, the prices were stated as above.
- a seasonal rate under which the price is 10 cents per kWh in the summer, 8 cents per kWh in the winter, and 6 cents per kWh in the spring and fall. Like TOD rates, these prices did not vary. Note that the price is for the electricity only, not transmission and distribution, which is supplied by the local regulated utility.

The length of contract that the supplier offered was also stated, in years (such as 1 year or 5 years.) During this contract period, the supplier guaranteed the prices and the buyer would have to pay a penalty if he/she switched to another supplier. The supplier could offer no contract in which case either side could stop the agreement at any time. This is recorded as a contract length of 0.

Some suppliers were also described as being a local company or a "well-known" company. If the supplier was not local or well-known, then nothing was said about them in this regard .

- Run a mixed logit model without intercepts and a normal distribution for the 6 parameters of the model, using 100 draws, halton sequences and taking into account the panel data structure.

- Using the estimated mean coefficients, determine the amount that a customer with average coefficients for price and length is willing to pay for an extra year of contract length.

The mean coefficient of length is -0.20. The consumer with this average coefficient dislikes having a longer contract. So this person is willing to pay to reduce the length of the contract. The mean price coefficient is -0.97. A customer with these coefficients is willing to pay 0.20/0.97=0.21, or one-fifth a cent per kWh extra to have a contract that is one year shorter.

- Determine the share of the population who are estimated to dislike long term contracts (ie have a negative coefficient for the length.) egin
[5]

The coefficient of length is normally distributed with mean -0.20 and standard deviation 0.35. The share of people with coefficients below zero is the cumulative probability of a standardized normal deviate evaluated at 0.20 / 0.3 5=0. 57. Looking 0.57 up in a table of the standard normal distribution, we find that the share below 0.57 is 0.72. About seventy percent of the population are estimated to dislike long-term contracts.

- The price coefficient is assumed to be normally distributed in these runs. This assumption means that some people are assumed to have positive price coefficients, since the normal distribution has support on both sides of zero. Using your estimates from exercise 1, determine the share of customers with positive price coefficients. As you can see, this is pretty small share and can probably be ignored. However, in some situations, a normal distribution for the price coefficient will give a fairly large share with the wrong sign. Revise the program to make the price coefficient fixed rather than random. A fixed price coefficient also makes it easier to calculate the distribution of willingness to pay (wtp) for each non-price attribute. If the price coefficient is fixed, the distribtion of wtp for an attribute has the same distribution as the attribute's coefficient, simply scaled by the price coefficient. However, when the price coefficient is random, the distribution of wtp is the ratio of two distributions, which is harder to work with.

The price coefficient is distributed normal with mean -0.97 and standard deviation 0.20. The cumulative standard normal distribution evaluated at 0.97 / 0.20 = 4.85 is more than 0.999, which means that more than 99.9% of the population are estimated to have negative price coefficients. Essentially no one is estimated to have a positive price coefficient.

- You think that everyone must like using a known company rather than an unknown one, and yet the normal distribution implies that some people dislike using a known company. Revise the program to give the coefficient of a uniform distribution between zero and b,where b is estimated (do this with the price coefficient fixed).

The price coefficient is uniformly distributed with parameters 1.541 and 1.585.

A lognormal is specified as (exp(b+se)) where (e) is a standard normal deviate. The parameters of the lognormal are (b) and (s) . The mean of the lognormal is (exp(b+0.5s^2)) and the standard deviation is the mean times (sqrt<(exp(s^2))-1>) .

## Added-variable plots

*added variable*plots (aka:*partial residual*plots) are very useful, because they show the**unique**partial relationship of Y to each X, controlling (or adjusting) for all other predictors. In these plots “adjusting” means that prestige | others is actually the residual for prestige in the regression from the other variable.

The

function is avPlots() . Try this. Several observations stand out in both the influence plot and in the partial residual plots, and were identified in the plots by their labels (id.n= ). Try to think why these might be unusual in terms of the context of this data.

## Advanced Quantitative Methods

This question involves the use of multiple linear regression on the Auto data set.

- Compute the matrix of correlations between the variables using the function cor() . You will need to exclude the name variable, which is qualitative.

- Use the lm() function to perform a multiple linear regression with mpg as the response and all other variables except name as the predictors. Use the summary() function to print the results. Comment on the output. For instance:

**Yes, there is a relatioship between the predictors and the response by testing the null hypothesis of whether all the regression coefficients are zero. The F-statistic is far from 1 (with a small p-value), indicating evidence against the null hypothesis.****Looking at the p-values associated with each predictor’s t-statistic, we see that displacement, weight, year, and origin have a statistically significant relationship, while cylinders, horsepower, and acceleration do not.****The regression coefficient for year, 0.7507727 , suggests that for every one year, mpg increases by the coefficient. In other words, cars become more fuel efficient every year by almost 1 mpg / year.**- Use the plot() function to produce diagnostic plots of the linear regression fit. Comment on any problems you see with the fit. Do the residual plots suggest any unusually large outliers? Does the leverage plot identify any observations with unusually high leverage?

**Seems to be non-linear pattern, linear model not the best fit. From the leverage plot, point 14 appears to have high leverage, although not a high magnitude residual.****There are possible outliers as seen in the plot of studentized residuals because there are data with a value greater than 3.**- Use the * and : symbols to fit linear regression models with interaction effects. Do any interactions appear to be statistically significant?

**Interaction between displacement and weight is statistically signifcant, while the interactiion between cylinders and displacement is not.**### 2.2.2 Exercise

This question should be answered using the Carseats dataset from the ISLR package.

- Provide an interpretation of each coefficient in the model. Be careful, some of the variables in the model are qualitative!

**Price: suggests a relationship between price and sales given the low p-value of the t-statistic. The coefficient states a negative relationship between Price and Sales: as Price increases, Sales decreases.****UrbanYes: The linear regression suggests that there is not enough evidence for arelationship between the location of the store and the number of sales based.****USYes: Suggests there is a relationship between whether the store is in the US or not and the amount of sales. A positive relationship between USYes and Sales: if the store is in the US, the sales will increase by approximately 1201 units.**- Write out the model in equation form, being careful to handle the qualitative variables properly.

**Sales = 13.04 + -0.05 Price + -0.02 UrbanYes + 1.20 USYes****Price and USYes, based on the p-values, F-statistic, and p-value of the F-statistic.**- On the basis of your response to the previous question, fit a smaller model that only uses the predictors for which there is evidence of association with the outcome.

**Based on the RSE and R^2 of the linear regressions, they both fit the data similarly, with linear regression from (e) fitting the data slightly better.****All studentized residuals appear to be bounded by -3 to 3, so no potential outliers are suggested from the linear regression.****There are a few observations that greatly exceed ((p+1)/n) ( 0.0075567 ) on the leverage-statistic plot that suggest that the corresponding points have high leverage.**### 2.2.3 Exercise

In this exercise you will create some simulated data and will fit simple linear regression models to it. Make sure to use set.seed() prior to starting part (a) to ensure consistent results.

- Using the rnorm() function, create a vector, x , containing 100 observations drawn from a (N(0,1)) distribution. This represents a feature, X .

- Using the rnorm() function, create a vector, eps , containing 100 observations drawn from a (N(0,0.25)) distribution i.e. a normal distribution with mean zero and variance 0.25.

- Using x and eps , generate a vector y according to the model [Y = -1 + 0.5X + epsilon] What is the length of the vector y ? What are the values of (eta_0) and (eta_1) in this linear model?

**y is of length 100. (eta_0) is -1, (eta_1) is 0.5.****A linear relationship between x and y with a positive slope, with a variance as is to be expected.**- Fit a least squares linear model to predict y using x . Comment on the model obtained. How do (hat<eta>_0) and (hat<eta>_1) compare to (eta_0) and (eta_1) ?

**The linear regression fits a model close to the true value of the coefficients as was constructed. The model has a large F-statistic with a near-zero p-value so the null hypothesis can be rejected.**- Display the least squares line on the scatterplot obtained in (d). Draw the population regression line on the plot, in a different color. Use the legend() command to create an appropriate legend.

- Now fit a polynomial regression model that predicts (y) using (x) and (x^2) . Is there evidence that the quadratic term improves the model fit? Explain your answer.

**There is evidence that model fit has increased over the training data given the slight increase in (R^2) . Although, the p-value of the t-statistic suggests that there isn’t a relationship between y and (x^2) .**- Repeat (a)-(f) after modifying the data generation process in such a way that there is less noise in the data. The model should remain the same. You can do this by decreasing the variance of the normal distribution used to generate the error term (epsilon) in (b). Describe your results.

**As expected, the error observed in the values of (R^2) decreases considerably.**- Repeat (a)-(f) after modifying the data generation process in such a way that there is more noise in the data. The model should remain the same. You can do this by increasing the variance of the normal distribution used to generate the error term (epsilon) in (b). Describe your results.

**As expected, the error observed in (R^2) and (RSE) increases considerably.**- What are the confidence intervals for (eta_0) and (eta_1) based on the original data set, the noisier data set, and the less noisy data set? Comment on your results.

**All intervals seem to be centered on approximately 0.5, with the second fit’s interval being narrower than the first fit’s interval and the last fit’s interval being wider than the first fit’s interval.**### 2.2.4 Optional Exercise

Follow the instructions on regsim documentation page to install the package. If you don’t already have devtools package installed, then install that first.

## Model with Categorical Variables or Factors

Sometimes, we may be also interested in using categorical variables as predictors. According to the information posted in the website of National Heart Lung and Blood Institute (http://www.nhlbi.nih.gov/health/public/heart/obesity/lose_wt/risk.htm), individuals with body mass index (BMI) greater than or equal to 25 are classified as overweight or obesity. In our dataset, the variable adiposity is equivalent to BMI.

Create a categorical variable

which takes value of "overweight or obesity" if adiposity >= 25 and "normal or underweight" otherwise.*bmi*,With the variable

you generated from the previous exercise, we go ahead to model our data.*bmi*age + fatfreeweight + neck + factor(bmi), data = fatdata)

age + fatfreeweight + neck + factor(bmi),

-13.4222 -3.0969 -0.2637 2.7280 13.3875

Estimate Std. Error t value Pr(>|t|)

(Intercept) -21.31224 6.32852 -3.368 0.000879 ***

age 0.01698 0.02887 0.588 0.556890

fatfreeweight -0.23488 0.02691 -8.727 3.97e-16 ***

neck 1.83080 0.22152 8.265 8.63e-15 ***

factor(bmi)overweight or obesity 7.31761 0.82282 8.893 < 2e-16 ***

Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 5.146 on 247 degrees of freedom

Multiple R-squared: 0.5662, Adjusted R-squared: 0.5591

F-statistic: 80.59 on 4 and 247 DF, p-value: < 2.2e-16

Note that although factor

*bmi*has two levels, the result only shows one level: "overweight or obesity", which is called the**"treatment effect"**. In other words, the level "normal or underweight" is considered as baseline or reference group and the estimate of factor(bmi)*overweight or obesity**7.3176*is the effect difference of these two levels on percent body fat.**Multiple Linear Regression in R (R Tutorial 5.3) MarinStatsLectures**[Contents]Content �. Some Rights Reserved.

Date last modified: January 6, 2016.

Boston University School of Public Health

## 20.9 Quantitative Aspects of Electrolysis

The stoichiometry of a half-reaction tells us how many electrons are needed to achieve an electrolytic process. For example, the reduction of Na + to Na is a one-electron process:

Thus, 1 mol of electrons will plate out 1 mol of Na metal, 2 mol of electrons will plate out 2 mol of Na metal, and so forth. Similarly, 2 mol of electrons are required to produce 1 mol of copper from Cu 2 + , and 3 mol of electrons are required to produce 1 mol of aluminum from Al 3 + :

For any half-reaction, the amount of a substance that is reduced or oxidized in an electrolytic cell is directly proportional to the number of electrons passed into the cell.

The quantity of charge passing through an electrical circuit, such as that in an electrolytic cell, is generally measured in coulombs. As noted in Section 20.5, the charge on 1 mol of electrons is 96,500 C (1 faraday):

In terms of more familiar electrical units, a coulomb is the quantity of charge passing a point in a circuit in 1 s when the current is 1 ampere (A). (Conversely, current is the rate of flow of electricity. An ampere (often referred to merely as an amp) is the current associated with the flow of 1 C past a point each second.) Therefore, the number of coulombs passing through a cell can be obtained by multiplying the amperage and the elapsed time in seconds:

Figure 20.22 shows how the quantities of substances produced or consumed in electrolysis are related to the quantity of electrical charge that is used. The same relationships can also be applied to voltaic cells.

FIGURE 20.22 The steps relating the quantity of electrical charge used in electrolysis to the amounts of substances oxidized or reduced.

### Sample Exercise 20.15

Calculate the mass of aluminum produced in 1.00 hr by the electrolysis of molten AlCl 3 if the electrical current is 10.0 A.

SOLUTION First , we use Equation 20.20 to calculate the coulombs of electrical charge that are passed into the electrolytic cell:

Second , we calculate the number of moles of electrons (the number of faradays of electrical charge) that pass into the cell:

Third , we relate the number of moles of electrons to the number of moles of aluminum being formed. The half-reaction for the reduction of Al 3 + is

Thus, 3 mol of electrons (3 F of electrical charge) are required to form 1 mol of Al:

Finally , we convert moles to grams

Because each step involves a multiplication by a new factor, the steps can be combined into a single sequence of factors:

### Practice Exercise

The half-reaction for formation of magnesium metal upon electrolysis of molten MgCl 2 is Mg 2 + + 2e - Mg. Calculate the mass of magnesium formed upon passage of a current of 60.0 A for a period of 4.00 × 10 3 s. Answer: 30.2 g of Mg

## Electrical Work

This is a good point at which to consider the relationship between electrochemical processes and work. We have already seen that a positive value of E is associated with a negative value for the free-energy change and thus, with a spontaneous process. We also know that for any spontaneous process, G is a measure of the maximum useful work, w max , that can be extracted from the process: G = w max . (For more information, see Section 19.5) Because G = -nFE, the maximum useful electrical work obtainable from a voltaic cell is

A word about sign conventions is in order here. Remember that work done by the system on its surroundings is indicated by a negative sign for w (For more information, see Section 5.2) Therefore, a negative value for w max corresponds to a spontaneous process, for which E is positive.

Notice that the maximum work obtainable is proportional to the cell potential E . We can think of E as a kind of pressure, a measure of the driving force for the process. Recall from Section 20.4 that the units of E are J/C w max is also proportional to the number of coulombs that flow, measured by nF . When we put these quantities together, we can see how the units cancel to leave us with units of energy:

In an electrolytic cell we use an external source of energy to bring about a nonspontaneous electrochemical process. In this case, G for the cell process is positive and E cell is negative. To force the process to occur, we must apply an external potential, E ext , which must be larger in magnitude than E cell : E ext >- E cell . For example, if a nonspontaneous process has E cell = ǂ.9 V, then the external potential E ext must be greater than 0.9 V in order to cause the process to occur.

When an external potential E ext is applied to a cell, the surroundings are doing work on the system. The amount of work performed is given by

Notice that, unlike Equation 20.21, there is no minus sign in Equation 20.22. The work calculated in Equation 20.22 will be a positive number because the surroundings are doing work on the system.

Electrical work is usually expressed in energy units of watts times time. The watt is a unit of electrical power, that is, the rate of energy expenditure:

Therefore, a watt-second is a joule. The unit employed by electric utilities is the kilowatt-hour (kWh), which works out to be 3.6 × 10 6 J:

Using these considerations, we can calculate the maximum work obtainable from the voltaic cells and the work required to bring about desired electrolysis reactions.

### Sample Exercise 20.16

Calculate the number of kilowatt-hours of electricity required to produce 1.0 × 10 3 kg of aluminum by electrolysis of Al 3 + if the applied emf is 4.50 V.

SOLUTION We need to employ Equation 20.22 to calculate w for an applied potential of 4.50 V. First we need to calculate nF , the number of coulombs required.

We can now employ Equation 20.22 to calculate w . In doing so, we must apply the unit conversion factor of Equation 20.23:

This quantity of energy does not include the energy used to mine, transport, and process the aluminum ore, and to keep the electrolysis bath molten during electrolysis. A typical electrolytic cell used to reduce aluminum ore to aluminum metal is only 40 percent efficient, 60 percent of the electrical energy being dissipated as heat. It therefore requires on the order of 33 kWh of electricity to produce 1 kg of aluminum. The aluminum industry consumes about 2 percent of the electrical energy generated in the United States. Because this is used mainly for reduction of aluminum, recycling this metal saves large quantities of energy.

### Practice Exercise

Calculate the number of kilowatt-hours of electricity required to produce 1.00 kg of Mg from electrolysis of molten MgCl 2 if the applied emf is 5.00 V. Assume that the process is 100 percent efficient. Answer: 11.0 kWh

### Sample Exercise 20.17

A 12-V lead-storage battery contains 410 g of lead in its anode plates and a stoichiometrically equivalent amount of PbO 2 in the cathode plates. (a) What is the maximum number of coulombs of charge it can deliver without being recharged? (b) For how many hours could the battery deliver a current of 1.0 A assuming that the current does not fall during discharge? (c) What is the maximum electrical work that the battery can do in kilowatt-hours?

SOLUTION (a) The lead anode undergoes a two-electron oxidation:

Thus, 2 F 1 mol Pb. Using this relationship, we have

Although the emf of a cell is independent of the masses of the solid reactants involved in the cell, the total electrical charge the cell can deliver depends on these quantities. The size and surface area further affect the rate at which electrical charge can be delivered.

(b) We calculate the number of hours of operation at a current level of 1.0 A by recalling that a coulomb corresponds to a current of 1 A flowing for 1 s:

This battery might be described as a 110 A-hr battery.

(c) The maximum work is given by the product - nFE, Equation 20.21. The product nF is the amount of charge, 3.82 × 10 5 C, calculated earlier in this problem:

(The negative sign indicates that the system does work on its surroundings.)

### Practice Exercise

A "deep discharge" lead-acid battery for marine use is advertised as having an 80-A-hr capacity. What amount of Pb would be oxidized if this battery were discharged so as to consume 80 percent of its capacity? Answer: 247 g of Pb

## Exercises

- Determine the dimensions of this dataset. How many genes, SNPs and samples are included?
- Principle components of the expression data.
- Compute the principle components.
- Create a plot of the variances for the first 10 PCs.
- How much of the total variance is explained by the first 10 PCs?

- Model the expression measured by probe 3710685 as a function of SNP rs4077515 and the first 10 PCs.
- Create a plot of gene expression by genotype with the effect of the PCs removed.
- How does this compare to the simple linear regression model for this SNP/gene pair.

## Exercise 2: Nested logit model

The data set HC from mlogit contains data in R format on the choice of heating and central cooling system for 250 single-family, newly built houses in California.

- Gas central heat with cooling gcc ,
- Electric central resistence heat with cooling ecc ,
- Electric room resistence heat with cooling erc ,
- Electric heat pump, which provides cooling also hpc ,
- Gas central heat without cooling gc ,
- Electric central resistence heat without cooling ec ,
- Electric room resistence heat without cooling er .

Heat pumps necessarily provide both heating and cooling such that heat pump without cooling is not an alternative.

- depvar gives the name of the chosen alternative,
- ich.alt are the installation cost for the heating portion of the system,
- icca is the installation cost for cooling
- och.alt are the operating cost for the heating portion of the system
- occa is the operating cost for cooling
- income is the annual income of the household

Note that the full installation cost of alternative gcc is ich.gcc+icca , and similarly for the operating cost and for the other alternatives with cooling.

- Run a nested logit model on the data for two nests and one log-sum coefficient that applies to both nests. Note that the model is specified to have the cooling alternatives ( gcc>, ecc>, erc>, hpc>) in one nest and the non-cooling alternatives ( gc>, ec>, `er>) in another nest.

- The estimated log-sum coefficient is (0.59) . What does this estimate tell you about the degree of correlation in unobserved factors over alternatives within each nest?

The correlation is approximately (1-0.59=0.41) . It's a moderate correlation.

- Test the hypothesis that the log-sum coefficient is 1.0 (the value that it takes for a standard logit model.) Can the hypothesis that the true model is standard logit be rejected?

We can use a t-test of the hypothesis that the log-sum coefficient equal to 1. The t-statistic is :

The critical value of t for 95% confidence is 1.96. So we can reject the hypothesis at 95% confidence.

We can also use a likelihood ratio test because the multinomial logit is a special case of the nested model.

Note that the hypothesis is rejected at 95% confidence, but not at 99% confidence.

- Re-estimate the model with the room alternatives in one nest and the central alternatives in another nest. (Note that a heat pump is a central system.)

- What does the estimate imply about the substitution patterns across alternatives? Do you think the estimate is plausible?

The log-sum coefficient is over 1. This implies that there is more substitution across nests than within nests. I don't think this is very reasonable, but people can differ on their concepts of what's reasonable.

We cannot reject the hypothesis at standard confidence levels.

- How does the value of the log-likelihood function compare for this model relative to the model in exercise 1, where the cooling alternatives are in one nest and the heating alternatives in the other nest.

The (ln L) is worse (more negative.) All in all, this seems like a less appropriate nesting structure.

- Rerun the model that has the cooling alternatives in one nest and the non-cooling alternatives in the other nest (like for exercise 1), with a separate log-sum coefficient for each nest.

- Which nest is estimated to have the higher correlation in unobserved factors? Can you think of a real-world reason for this nest to have a higher correlation?

The correlation in the cooling nest is around 1-0.60 = 0.4 and that for the non-cooling nest is around 1-0.45 = 0.55. So the correlation is higher in the non-cooling nest. Perhaps more variation in comfort when there is no cooling. This variation in comfort is the same for all the non-cooling alternatives.

- Are the two log-sum coefficients significantly different from each other? That is, can you reject the hypothesis that the model in exercise 1 is the true model?

We can use a likelihood ratio tests with models nl and nl3 .

The restricted model is the one from exercise 1 that has one log-sum coefficient. The unrestricted model is the one we just estimated. The test statistics is 0.6299. The critical value of chi-squared with 1 degree of freedom is 3.8 at the 95% confidence level. We therefore cannot reject the hypothesis that the two nests have the same log-sum coefficient.

- Rewrite the code to allow three nests. For simplicity, estimate only one log-sum coefficient which is applied to all three nests. Estimate a model with alternatives gcc , ecc and erc in a nest, hpc in a nest alone, and alternatives gc , ec and er in a nest. Does this model seem better or worse than the model in exercise 1, which puts alternative hpc in the same nest as alternatives gcc , ecc and erc ?

The (ln L) for this model is (-180.26) , which is lower (more negative) than for the model with two nests, which got (-178.12) .

## Watch the video: Form 5 Mathematics KSSM Chapter 2 - Matrices. Self Practice (December 2021).