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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) |
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Answer» SOLUTION :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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