Question

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Consider the cell described below at 251 K: Pb | Pb2+ (1.47 M) || Fe3+ (2.11...

Consider the cell described below at 251 K:

Pb | Pb2+ (1.47 M) || Fe3+ (2.11 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 1.078 M

Solutions

Expert Solution

Cathode (Reduction)

Fe3+ + 3e-              Fe        E° = - 0.036 V

Anode (Oxidation)

Pb              Pb2+ + 2e-        E° = + 0.13 V

Net reaction

2Fe3+ + 3Pb            3Pb2+ + 2Fe     E°cell = + 0.094 V

Constructing the ICF table,

2Fe3+

3Pb

3Pb2+

2Fe

I

2.11 M

1.47 M

C

-2x

+3x

F

2.11 - 2x

1.47 + 3x

Here, 2x = 1.078,

Therefore, x = 0.539

Therefore,

2Fe3+

3Pb

3Pb2+

2Fe

I

2.11 M

1.47 M

C

-1.078 M

+1.617 M

F

1.032 M

3.087 M

Ecell = Eocell – (RT/nF)ln[Pb2+]3/[Fe3+]2

Ecell = 0.094 – (8.314 x 251/6 x 96500)ln[3.087]3/[1.032]2

Ecell = 0.0820 V

queen_honey_blossom answered 2 months ago
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