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

Dec 28, 2017

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

$N {i}^{2 +} \left(a q\right) + 2 {e}^{-} \to N i \left(s\right)$ where $E = - 0.23 V$
$C {u}^{2 +} \left(a q\right) + 2 {e}^{-} \to C u \left(s\right)$ where $E = 0.34 V$

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

$C {u}^{2 +} \left(a q\right) + N i \left(s\right) r i g h t \le f t h a r p \infty n s N {i}^{2 +} \left(a q\right) + C u \left(s\right)$, where

$E = {E}_{\text{red") - E_("ox}}$
therefore E^° = 0.57V