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Electrochemistry

Question
CBSEENCH12006126

A Cell is prepared by dipping a copper rod in 0.01 M copper sulphate solution, and zinc rod in 0.02 M ZnSO4 solution. The standard reduction potentials of copper and zinc are + 0.34 V and – 0.76 V respectively.
(a) What will be the cell reaction?
(b) How will the cell be represented?
(c) What will be the emf. of the cell?

Solution

(a) As E°Zn2+/Zn < E°Cu2+/Cu . Zinc electrode acts as anode and copper electrode acts as a cathode. The cell reaction is
Zn(s) + Cu2+(aq) (0.01 M) → Zn2+(aq) (0.02 M) + Cu(s)
(b) The cell is represented as
Zn(s) | Zn2+(aq) (0.02 M) || Cu2+(aq) (0.01 M) + Cu(s)
(c) The emf of the cell can be calculated by using Nernst equation
E= E°-0.059nlogZn2+Cu2+E  = +1.10 - 0.0592log 0.020.01E =  +1.10 - 0.0592×0.301E = +1.10 - 0.0089  or E = + 1.091 V