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Question

A zinc electrode is placed in a 0.1M solution at 250C. assuming that the salt is 20% dissociated at this dilutions calculate the electrode reduction potential.E0(Zn2+|Zn)=0.76V

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Solution

Zn2++2eZn
α=20100=0.2,C=0.1M
concentration of Zn2+=Cα
=0.1×0.2
=0.02M
From nernst equation,
Ecell=E00.059nlog[1Zn2+]
Ecell=0.760.0592log[10.02]
Ecell=0.760.059×1.6982
Ecell=0.760.05
Ecell=0.81V

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