# The EMF of a cell

Gold Member

## 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

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mjc123
Homework Helper
Why have you got log(0.1)^3 ? Write out the Nernst equation in full for this cell.

Gold Member
Ok, so we have E°=-1.75, RT/F=0.059, Q=[Ni2+]^3/[Au3+]^2=(0.1)^3/(1)^2=(0.1)^3
replacing in E=E°-2.303(RT/nF)logQ we get the equation I wrote in"Attempt at a solution"

mjc123
Homework Helper
Sorry, I misread the question, I thought [Au3+] was also 0.1M.
I think you've got the signs wrong. E for the cell is ER - EL by convention
Ecell = E(Au3+/Au) - E(Ni2+/Ni)
= E0(Au3+/Au) - E0(Ni2+/Ni) + RT/nF*log([Au3+]2/[Ni2+]3)

Gold Member
Sorry, I misread the question, I thought [Au3+] was also 0.1M.
I think you've got the signs wrong. E for the cell is ER - EL by convention
Ecell = E(Au3+/Au) - E(Ni2+/Ni)
= E0(Au3+/Au) - E0(Ni2+/Ni) + RT/nF*log([Au3+]2/[Ni2+]3)
The magnitude should remain the same (note that reversing the terms inside log reverses its sign)

Gold Member
Alright, I realized my mistake. I got confused in sign conventions. Thank you for your help :D

mjc123
Homework Helper