- #1

Krushnaraj Pandya

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## Homework Statement

The EMF of the cell : Ni(s) | Ni2+ (0.1M) || Au 3+ (1.0M) | Au (s) is

[ given E°(Ni2+/Ni)= -0.25V; E° (Au3+ / Au )= +1.50V ]

## Homework Equations

E°cell=E°(cathode)-E°(anode) For SRP.

Nernst equation

## The Attempt at a Solution

SOP is given, therefore converting to SRP we take minus sign of both. Now Ni to Ni2+ is at anode and Au3+ to Au is at cathode. Therefore E(cell)= -1.5-0.25= -1.75V

Writing balanced reaction 3Ni+2Au3+ gives 3Ni2+ + 2Au, no. of electrons transferred=6.

Putting into nernst equation, EMF should be -1.75-(0.059/6)log(0.1)^3 which gives the wrong answer. Where am I wrong, I'd appreciate some help