Question

At 298 K write Nernst equation for the following cell – Ni $\vert$ Ni$^{2+}$ (0.01 M) $\Vert$ Cu$^{2+}$ (0.1 M) $\vert$ Cu. If the emf of the above cell ($E_{cell}$) is 0.59 V then calculate the standard emf of the cell ($E^\circ_{cell}$)

Hint:

Remember: $E_{cell}$ decreases with increasing concentration of products and increases with increasing concentration of reactants.

Answer 1

Step 1: Write the cell reaction.
\[ Ni (s) + Cu^{2+} (aq) \;\longrightarrow\; Ni^{2+} (aq) + Cu (s) \] 

Step 2: Nernst equation.
General Nernst equation: \[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log Q \] where, $n = 2$ (electrons transferred) 
$Q = \dfrac{[Ni^{2+}]}{[Cu^{2+}]} = \dfrac{0.01}{0.1} = 0.1$ 

Step 3: Substitution.
\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{2} \log (0.1) \] \[ E_{cell} = E^\circ_{cell} - 0.02955 \times (-1) \] \[ E_{cell} = E^\circ_{cell} + 0.02955 \] 

Step 4: Calculate $E^\circ_{cell$.}
Given: $E_{cell} = 0.59 \, V$ \[ 0.59 = E^\circ_{cell} + 0.02955 \] \[ E^\circ_{cell} = 0.59 - 0.02955 \] \[ E^\circ_{cell} \approx 0.56 \, V \] 

Conclusion:
The standard emf of the cell is: \[ \boxed{0.56 \, V} \]