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Electrochemistry

Question
CBSEENCH12006000

Zinc rod is dipped in 0.1 M solution of ZnSO4. The salt is 95% dissociated at this dilution at 298 K. Calculate the electrode potential. [Given that: E°Zn2+/Zn = – 0.76 V.]

Solution

Concentration of Zn2+(aq)
             = 0.1 + 95100=0.095
           Zn2+(aq) + 2e-  Zn
According to Nernst equation,
            E = E°+0.0591nlogZn2+(aq)[Zn]
             = -0.76 + 0.0591nlog0.0951= 0.76+0.02953 × (-1.0223)= -0.76 - 0.03021 = -0.79 V.