Question

What is the voltage of an electrolytic cell with nickel and copper electrodes?

Answer 1

The voltage is the cell's electromotive force derived from the reduction potentials of each cell, here we have,

`Ni^(2+)(aq) + 2e^(-) to Ni(s)` where `E = -0.23V`
`Cu^(2+)(aq) + 2e^(-) to Cu(s)` where `E = 0.34V`

Quantitatively, this is saying copper ions are more easily reduced than nickel ions. Hence, the cell's reaction would be,

`Cu^(2+)(aq) + Ni(s) rightleftharpoons Ni^(2+)(aq) + Cu(s)`, where

`E = E_("red") - E_("ox")`
`therefore E^° = 0.57V`