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Question

The E0cell for the given cell reaction is 0.32 V at 25C.
Fe(s)+Zn2+(aq)Zn(s)+Fe2+(aq)
What will be the value of log[Fe2+] at equilibrium when a piece of iron is placed in a 1 M of Zn2+ solution?

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Solution

We have the Nernst equation at equilibrium at 25C

E0=0.0591nlog K......(Eqn.1)

Since E0cell for the given reaction is negative, therefore, the reverse reaction is feasible for which E0cell will be +0.32 V.
Thus,

Zn(s)+Fe2+(aq)Fe(s)+Zn2+(aq);E0cell=+0.32 V

By Equation (1), we get

E0=0.0591nlog[Zn2+][Fe2+]

[Zn2+]=1 M

0.32=0.05912log1[Fe2+]

0.32=0.05912log[Fe2+]

log[Fe2+]=10.82910.83




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