Question

Calculate the standard electrode potential of the Ni²⁺/Ni electrode, if the cell potential of the cell is 0.59 V.

Ni | Ni²⁺ (0.01 M) || Cu²⁺ (0.1 M) | Cu

Given: \[\ce{E^{\circ}_{Cu{^{2+}/Cu}}}\] = +0.34 V

Answer 1

Given: Ni | Ni²⁺ (0.01 M) || Cu²⁺ (0.1 M) | Cu = 0.59

\[\ce{E^{\circ}_{Cu{^{2+}/Cu}}}\] = +0.34 V

Anode (oxidation): \[\ce{Ni -> Ni^2+ + 2e-}\]

Cathode (reduction): \[\ce{Cu^2+ + 2e- -> Cu}\]

By using Nernst equation

The cell potential at 298 K is

`E_"cell" = E_"cell"^circ - 0.0591/n log Q`

= `E_"cell"^circ - 0.0591/2 log  0.01/0.1`

= `E_"cell"^circ - 0.0591/2 log (0.1)`

= `E_"cell"^circ - 0.0591/2 (-1)`

`E_"cell" = E_"cell"^circ + 0.02955`

`E_"cell"^circ` = 0.590 − 0.02955

`E_"cell"^circ` = 0.566045

`E_"cell"^circ = E_"cathode"^circ - E_"anode"^circ`

`0.566045 = 0.34 - E_(text(Ni)^(2+)//text(Ni))^circ`

`E_(text(Ni)^(2+)//text(Ni))^circ = 0.34 - 0.56045` 

= −0.22045 V

∴ The standard electrode potential of the Ni²⁺/Ni electrode is −0.22045 V.