Solutions-electrochemistry Practice Problems 1069

Electrochemistry Practice Problems F = 96,500 coulombs • mole–1

R = 8.314 joules • mole–1 • K–1

0

∆G = − nFE = − RT ln K E cell = E 0 −

K = 10

0 ⎛ nFE ⎞ ⎜ ⎟ ⎝ 2.3 RT ⎠

RT ⎛ products ⎞ 0.059 products ⎞ 0 ln⎜ log10 ⎛⎜ ⎟=E − ⎟ nF ⎝ reactants ⎠ n ⎝ reactants ⎠

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

K = 10 B)

⎡ ( 2 )( 96 , 500 )( 3.72 ) ⎤ ⎢ ( 2.3)(8.314 )( 298 ) ⎥ ⎣ ⎦

= 10126

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

K = 10

⎡ (15)( 96,500 )( −0.185 ) ⎤ ⎢ ( 2.3)(8.314 )( 298 ) ⎥ ⎣ ⎦

= 10 −47

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

C)

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

K = 10

⎡ (10 )( 96 ,500 )( −0.112 ) ⎤ ⎢ ( 2.3)(8.314 )( 298 ) ⎥ ⎣ ⎦

= 10 −19

Pt(s) ⏐ Fe3+(0.1 M), Fe2+(0.01M) ⏐⏐ Cr2O72-(0.01M), Cr3+(0.2M), H+(1.0M) ⏐Pt(s)

2.

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 Cr2O72- + 14H+ + 6e- ↔ 2Cr3+ + 7H2O

+1.36 volts

B. Write a balanced reaction for the cell reaction. 6Fe2+ + Cr2O72- + 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 the concentrations listed.

E cell

Ecell

[

][

] ][ ]

3+ 6 3+ 2 ⎡ 0 . 059 Fe Cr = E0 − log ⎢ 6 n ⎢⎣ Fe 2+ Cr2 O7 2− H +

[

][

⎤ ⎥ 14 ⎥⎦

0.059 ⎡ [0.1]6 [0.2]2 ⎤ =E − log ⎢ = +0.524volts 6 14 ⎥ 6 ⎣[0.01] [0.01][1] ⎦ 0

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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.

Ecell = C − 0.059( pH )

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 , 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 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 PbSO4(s) ↔ Pb2+ + SO42-

Ecell = 0 = E 0 −

-

-0.355 volts

-0.229 volts

0.059 log K sp n

Ecell = 0 = −0.229 −

0.059 log K sp 2

log K sp = −7.763

K sp ≈ 1.73 x10−8

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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

2

Ecell

0.059 H + Cl − = 0.222 − log [H 2 ] 2

Ecell

0.059 3.16 x10− 4 [1]2 log = 0.222 − [1] 2

[

Ecell = 0.222 + 0.206 = +0.428volts

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]

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