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Electrochemistry

Question
CBSEENCH12006020

Write the Nernst equation and emf of the following cells at 298 k.
Mg(s) | Mg2+ (0.001 M) || Cu2+ (0.0001 M) | Cu(s)

Solution

For an electrochemical cell reaction
aA + bB → cC + dD
The Nernst’s equation for cell reaction is
Ecell = E0cell-0.059n log Cc DdAa Bb
The values of a, b, c, d and n can be obtained from the balanced cell reaction. (i) Anode reaction:
Mg(s)  Mg2+(aq) + 2e-
Cathode reaction:
Cu2+(aq) + 2e-  Cu(s)
Overall cell reaction:
          Mg(s) + Cu2+(aq)  Mg2+(aq) + Cu(s)
Here,  n= 2,  E0cell = E0cathode - E0anode = 0.34 V - (-2.37) V = 2.71 V
The Nernst equation for Ecell at 298 can be written as
Ecell =E0cell         = -0.059nlog10-310-4          = 2.71- 0.0295 = 2.68 V