Question

Calculate the cell potential for the following reaction as written at 25.00 °C, given that [Zn2 ] = 0.842 M and [Ni2 ] = 0.0100 M. Standard reduction potentials can be found here.

reaction:
Zn(s)+Ni^2+(aq)--->Zn^2+(aq)+Ni(s)

standard reduction Zn: -.76
standard reduction Ni: 0.26

Eo(cell)= 0.26-(-0.76)=1.02
E(cell)= Eo(cell) -RT/nF (ln(Zn/Ni))
E(cell)= 1.02 - ((8.314*298)/(2)(9.6485))(ln(.842/.01))
E(cell) =0.963

But the answer is wrong and I don't know why. Can someone explain it to me and where I did it wrong.

Answer 1

I worked it using the numbers you provided and came up with the same answer you did of 0.963 so there is nothing wrong with your math. The problem is that the reduction potential of Ni is not 0.26. It is -0.26.