Question

A voltaic cell utilizes the following reaction and operates at 298 K: 3Ce4+(aq) + Cr(s) →3Ce3+(aq)...

A voltaic cell utilizes the following reaction and operates at 298 K: 3Ce4+(aq) + Cr(s) →3Ce3+(aq) + Cr3+(aq).

1. What is the emf of this cell when [Ce4+]= 2.0 M , [Ce3+]= 1.8×10−2 M , and [Cr3+]= 1.7×10−2 M ?

2. What is the emf of the cell when [Ce4+]= 0.45 M ,[Ce3+]= 0.79 M , and [Cr3+]= 1.3 M ?

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Homework Answers

Answer #1

1. Ecell = Ecathode - Eanode

             = 1.61 - (-0.74)

             = 2.35 V

Nernst equation,

E = Ecell - 0.0592/n logQ

Q = [Ce3+]^3[Cr3+]/[Ce4+]^3

    = 2.35 - 0.0592/3 log(1.8 x 10^-2)^3(1.7 x 10^-2)/(2)^3

    = 2.51 V

2. Feeding the given concentration values we get,

E = Ecell - 0.0592/n logQ

Q = [Ce3+]^3[Cr3+]/[Ce4+]^3

    = 2.35 - 0.0592/3 log(0.79)^3(1.3)/(0.45)^3

    = 2.33 V

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