Question

Calculate the \[\ce{E^{\circ}_{cell}}\] of the following if the cell potential (E_cell) is 0.59 V.

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

Answer 1

Given: Cell potential (E_cell) = 0.59 V

The concentrations of the ions are:

Ni²⁺ = 0.1 M

Cu²⁺ = 0.01 M

The reaction involves the half-reactions for:

\[\ce{Ni^{2+} + 2e- -> Ni}\] and \[\ce{Cu^{2+} + 2e- -> Cu}\]

Using the Nernst Equation:

E_cell = \[\ce{E^{\circ}_{cell} - \frac{0.059}{n} log \frac{[Products]}{[Reactants]}}\]

E_cell = \[\ce{E^{\circ}_{cell} - \frac{0.059}{n} log \frac{[Cu^{2+}]_{cathode}}{[Ni^{2+}]_{anode}}}\]

0.59 = \[\ce{E^{\circ}_{cell} - \frac{0.059}{2} log \frac{0.01}{0.1}}\]

0.59 = \[\ce{E^{\circ}_{cell} - \frac{0.059}{2} \times (-1)}\]

0.59 = \[\ce{E^{\circ}_{cell} + 0.0295}\]

\[\ce{E^{\circ}_{cell}}\] = 0.59 − 0.0295

\[\ce{E^{\circ}_{cell}}\] = 0.5605 V

Hence, the standard cell potential, \[\ce{E^{\circ}_{cell}}\] is 0.56 V.