Question

Calculate the cell potential for the following reaction as written at 25.00 °C, given that [Cr2 ] = 0.759 M and [Ni2 ] = 0.0120 M. Standard reduction potentials can be found here.

Cr + Ni^2+ <----> Cr^2+ + Ni

Answer 1

Cr (s) + Ni²⁺ (aq) <-------> Cr²⁺ (aq) + Ni (s)

Here Cr has been oxidised from 0 to +2 ,so this reaction is taking place at anode and Ni²⁺ has been reduced to 0 from +2 ,so this reaction is taking place at cathode .

Also, Cr²⁺(aq) + 2 e^- ---->  Cr(s) , E⁰ = -0.91 V

and Ni²⁺(aq) + 2 e^- ----> Ni(s) , E⁰ = -0.25 V

E⁰cell = E⁰_(cathode) - E⁰_(anode)

=> E⁰_cell = -0.25 - (-0.91) = 0.65 V

This would have been the e.m.f. of cell if the solution were at 1 M each but here [Cr²⁺] = 0.759 M and [Ni²⁺] = 0.0120M

By using Nernst equation we will have---

E_cell = E⁰_cell - RT/nF ln [Cr²⁺]/[Ni²⁺]

where R = 8.314 J/mol. K
F = 96500 C
T = 25 + 273 = 298 K
n = number of electrons involved in electrode reaction = 2
E⁰_cell = 0.65 V

Now,

E_cell= 0.65 - (8.314 x 298) /(2 x 96500) ln 0.759/0.0120

=> E_cell= 0.65 - (2477.572 / 193000) ln 63.25

=> E_cell = 0.65 - 0.0128 x 4.147

=> E_cell = 0.65 - 0.0530816

=> E_cell = 0.59 V