Calculate the cell potential for the following reaction as written at 25.00 °C, given that [Zn2+ ] = 0.850 M and [Ni2+ ] = 0.0130 M. Standard reduction potentials can be found here. Zn(s) + Ni2+(aq)Zn2+(aq) + Ni(s)
In the reaction, Zn is oxidising and Ni is reducing.
Reduction potential of Zn is -0.762 V
Oxidation potential of Zn will be +0.762 V.
Reduction potential of Ni is -0.236 V
Eo = Eo(Zn) + Eo(Ni)
= 0.762 - 0.236 = 0.526 V
[Zn2+ ] = 0.850 M
[Ni2+ ] = 0.0130 M
n = number of electrons = 2
At temperature = 25 oC, EMF of cell = Eo - 0.06/n log [Zn2+]/[Ni2+]
= 0.526 - 0.06/2 log[0.850/0.0130]
= 0.526 - 0.03 log [65.38]
= 0.526 - 0.03*1.815
= 0.526 - 0.054
= 0.472 V
cell potential = 0.472 V
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