Question

Consider the cell described below at 273 K:

Sn | Sn²⁺ (1.45 M) || Fe³⁺ (2.33 M) | Fe

Given the standard reduction potentials found on the sheet attached to the exam, calculate the cell potential after the reaction has operated long enough for the [Fe³⁺] to have changed by 1.242 M

Answer 1

The reactions:

Sn2+ + 2 e− ⇌ Sn(s) −0.13

Fe3+ + 3 e− ⇌ Fe(s) −0.04

Clearly, Sn is oxidizing

Sn(s) ⇌Sn2+ + 2 e− E = +0.13

Fe3+ + 3 e− ⇌ Fe(s) −0.04

Balance

3Sn(s) ⇌ 3Sn2+ + 6e− E = +0.13

2Fe3+ + 6e− ⇌ 2Fe(s) E =−0.04

E° = Ered + Eox = 0.13 - 0.04 = 0.09 V

now

E = E° -0.0592/n * log(Q)

n = 2x3 = 6 electrons

E = E° -0.0592/n * log(Q)

E = 0.09 -0.0592/6 * log(Q)

Q = [Sn+2]^3 / [Fe+3]^2

initially

[Sn+2] = 1.45

[Fe+3] = 2.33

after reactions

[Sn+2] = 1.45 + 3x

[Fe+3] = 2.33 - 2x

note that we need Fe+3 = 1.242 so

[Fe+3] = 2.33 - 2x = 1.242

x = (1.242-2.33)/(-2) = 0.544

so

[Sn+2] = 1.45 + 3x = 1.45+3*0.544 = 3.082

substitute

E = 0.09  -0.0592/6 * log((3.082^3) / (1.242^2))

E = 0.077387V