Calculate the cell potential for the reaction as written at 25.00 °C , given that [Cr2+]=0.860 M and [Sn2+]=0.0120 M . Use the standard reduction potentials in this table.

Answer 1

You didn't supply standard reduction potentials so I've used the ones in my table. You also didn't provide the reaction as written so I must assume it is a functioning galvanic cell with a positive cell potential.

Cr²⁺ + 2e- ==> Cr(s) Eº = -0.91

Sn²⁺ + 2e- ==> Sn(s) Eº = -0.14

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Sn²⁺ + Cr(s) ==> Sn(s) + Cr²⁺ Eºcell = 0.77 V

Nernst equation @ 25ºC: Ecell = Eºcell - 0.0592/n log Q where Q = [Cr2+/[Sn2+]

Q = 0.860 / 0.0120 = 71.67

log Q = 1.86

Substituting into the Nernst equation...

Ecell = 0.77 - 0.0592/2 (1.86) = 0.77 - 0.055

Ecell = 0.72 V