1

Electrochemistry Practice Problems

F = 96,500 coulombs • mole–1 R = 8.314 joules • mole–1 • K–1

1. Calculate E° and K for the following reactions:

A) Mg (s) + Cl2 (g) ↔ Mg2+ + 2 Cl– Mg(s) ↔ Mg2+ + 2e- +2.360 volts Cl2(g) + 2e- ↔ 2Cl- +1.360 volts Mg (s) + Cl2 (g) ↔ Mg2+ + 2 Cl- +3.720 volts

126)298)(314.8)(3.2()72.3)(500,96)(2(

1010 ==⎥⎦

⎤⎢⎣

⎡

K B) 5 MnO2 (s) + 4 H+ ↔ 3 Mn2+ + 2 MnO4

– + 2H2O 5 MnO2 (s) + 10 H2O ↔ 5 MnO4

– + 20 H+ + 15e- -1.692 volts 3 MnO4

– + 24 H+ + 15e- ↔ 3 Mn2+ + 12H2O +1.507 volts

5 MnO2 (s) + 4 H+ ↔ 3 Mn2+ + 2 MnO4– + 2H2O -0.185 volts

47)298)(314.8)(3.2()185.0)(500,96)(15(

1010 −⎥⎦

⎤⎢⎣

⎡ −

==K

KRTnFEG ln0 −=−=∆ ⎟⎠⎞

⎜⎝⎛

= RTnFE

K 3.2

0

10

⎟⎠⎞

⎜⎝⎛−=⎟

⎠⎞

⎜⎝⎛−=

reactantsproductslog

n0.059E

reactantsproductsln

nFRTEE 10

00cell

2

C) I2(s) + 5Br2(aq) + 6H2O ↔ 2IO3– + 10 Br– + 12H+

5Br2(aq) + 10e- ↔ 10 Br– +1.098 volts

I2(s) + 6H2O ↔ 2IO3– + 12H+ + 10e- -1.210 volts I2(s) + 5Br2(aq) + 6H2O ↔ 2IO3– + 10 Br– + 12H+ -0.112 volts

19)298)(314.8)(3.2()112.0)(500,96)(10(

1010 −⎥⎦

⎤⎢⎣

⎡ −

==K 2. Pt(s) ⏐ Fe3+(0.1 M), Fe2+(0.01M) ⏐⏐ Cr2O72-(0.01M), Cr3+(0.2M), H+(1.0M) ⏐Pt(s)

A. Write an oxidation half reaction for the left electrode and a reduction half reaction for the right electrode.

Fe2+ ↔ Fe3+ + e- -0.771 volts Cr2O7

2- + 14H+ + 6e- ↔ 2Cr3+ + 7H2O +1.36 volts B. Write a balanced reaction for the cell reaction. 6Fe2+ + Cr2O7

2- + 14H+ + ↔ 6Fe3+ + 2Cr3+ + 7H2O C. Calculate Eo for the cell. In which direction is the reaction spontaneous; left to right or right to

left? Eo = -0.771 + 1.36 = +0.589 volts D. Calculate the cell potential, taking into account the concentrations listed.

[ ] [ ][ ] [ ][ ] ⎥

⎥⎦

⎤

⎢⎢⎣

⎡−=

+−+

++

14272

62

23630 log059.0

HOCrFe

CrFen

EEcell

[ ] [ ][ ] [ ][ ]

voltsEEcell 524.0101.001.0

2.01.0log6059.0

146

260 +=⎥

⎦

⎤⎢⎣

⎡−=

3

3, A) Draw a diagram of the glass pH sensing electrode. Label the parts of your drawing carefully.

B) Write an equation which describes voltage as a function of pH for this electrode.

)(059.0 pHCEcell −=

C) Describe the cause of "alkaline error", and what effect it has on pH measurements. When [H+] is very low and [Na+] is high, the electrode responds to Na+ and the apparent pH is lower than the true pH. This is called alkaline error.

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4, A) Draw a diagram of the silver/silver chloride reference electrode. Label the parts of your drawing carefully.

B) What is meant by a liquid junction potential?

Whenever dissimilar electrolyte solutions are in contact, a voltage difference called the junction potential develops at their interface. This small voltage (usually a few millivolts) is found at each end of a salt bridge connecting two half-cells. The junction potential puts a fundamental limitation on the accuracy of direct potentiometric measurements, because we usually do not know the contribution of the junction to the measured voltage. To see why the junction potential occurs, consider a solution of NaCl in contact with distilled water. Na+ and Cl− ions begin to diffuse from the NaCl solution into the water. However, Cl− ion has a greater mobility than Na+. That is, Cl− diffuses faster than Na+. As a result, a region rich in Cl−, with excess negative charge, develops at the front. Behind it is a positively charged region depleted of Cl−. The result is an electric potential difference at the junction of the NaCl and H2O phases.

C) What salt shows the least tendency to cause a liquid junction potential, and why?

Potassium chloride, KCl, because the mobilities and diffusion coefficients of potassium ion and chloride ion are nearly identical.

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5. Using data from the table of reduction potentials, calculate the solubility product constant (Ksp) for lead sulfate.

Pb(s) ↔ Pb2+ + 2e- +0.126 volts PbSO4(s) +2e- ↔ Pb(s) + SO4

- -0.355 volts PbSO4(s) ↔ Pb2+ + SO4

2- -0.229 volts

spcell Kn

EE log059.00 0 −==

spcell KE log2059.0229.00 −−==

763.7log −=spK

81073.1 −≈ xKsp

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6. Calculate the cell potential for the following cell:

Pt (s) ⏐ H2 (1 atm) ⏐ acetate buffer (pH = 3.50) ⏐⏐ KCl (1.00 M) ⏐AgCl (s) ⏐ Ag (s) H2(g) ↔ 2H+ + 2e- 0.000 volts 2AgCl(s) + 2e- ↔ 2Ag(s) + 2Cl- +0.222 volts H2(g) + 2AgCl(s) ↔ 2H+ + 2Ag(s) + 2Cl- +0.222 volts [ ] [ ]

[ ]2

22

log2059.0222.0

HClHEcell

−+

−=

[ ] [ ]

[ ]111016.3log

2059.0222.0

224−−=

xEcell

voltsEcell 428.0206.0222.0 +=+=