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Metal rod A is dipped in 0.1 M solution ...

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

Text Solution

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