Tuesday
May 26, 2015

Homework Help: Chemistry

Posted by Sophie on Sunday, May 4, 2014 at 8:27pm.

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.

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