calculate cell potential

Question

Calculate the cell potential for the galvanic cell in which the given reaction occurs at 25 °C, given that [Sn2+]=0.0680 M[Sn2+]=0.0680 M, [Co3+]=0.0406 M[Co3+]=0.0406 M, [Sn4+]=0.00237 M[Sn4+]=0.00237 M, and [Co2+]=0.01195 M[Co2+]=0.01195 M. Standard reduction potentials can be found in this table.

Sn2+(aq)+2Co3+(aq)↽−−⇀ Sn4+(aq)+2Co2+(aq)

Answer 1

Co³⁺ + 1e- ==> Co²⁺ x2

Sn²⁺ ==> Sn⁴⁺ + 2e-

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2Co³⁺ + Sn²⁺ ==> 2Co²⁺ + Sn⁴⁺ balanced redox equation

Nernst equation:

Ecell = Eºcell - RT/nF ln Q and at 25ºC and converting to log₁₀ we have...

Ecell = Eºcell - 0.0592/n log Q

Ecell = ?

Eºcell = 1.769

n = 2 mols electrons

Q = [Co²⁺]²[Sn⁴⁺] / [Co³⁺]²[Sn²⁺] = (0.01195)²(0.00237) / (0.0406)²(0.068) = 3.02x10^-3

Solving for Ecell...

Ecell = 1.769 - (0.0592/2) log 3.02x10^-3

Ecell = 1.769 - (0.0296 x -2.52) = 1.769 + 0.0746

Ecell = 1.8436 V

(be sure to check all of the math)

Answer 2

In the table of standard reduction potentials,

Sn4+(aq) +2e– → Sn2+(aq)

+0.151

and

Co3+(aq) + e– → Co2+(aq)

+1.92

From the overall parent reaction, we know that Sn2+ is oxidized to Sn4+ meaning that the standard reduction potential for Sn4+ --> Sn2+ will be flipped for our purposes since in our reaction it is oxidized from 2+ --> 4+.

So, -(0.151)

+

+1.92 =

1.769

Cell potential for galvanic cell supposed to be positive since a positive E or cell potential indicates spontaneity.