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A zinc rod is dipped in 0.1 M ZnSO(4) so...

A zinc rod is dipped in 0.1 M `ZnSO_(4)` solution. The salt is 95% dissociated of this dilution at 298 K. Calculate electrode potential.
`(E_(Zn^(2+)//Zn)=-0.76 V)`.

Text Solution

Verified by Experts

Salt is 95% ionised
`[A^(2+)] = (95)/(100) xx 0.1 = 0.095M`
The electrode reaction is
`A^(2+) + 2e^(-) rarr A(s)`
Applying Nernst equation
`E_(A^(2+)//A) = E_(A^(2+)//A)^(@) - (0.0591)/(n) "log " (1)/([A^(2+)])`
`=-0.76V - (0.0591)/(2) "log" (1)/(0.095)`
`=-0.76V - 0.0295 "log" (1000)/(95)`
`=-0.76V -0.0295 [log 1000-log 95]`
`=-0.76V- 0.029 [3-1.977]`
`=-0.79021V`
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