Q.Determine the standard emf of the cell and standard free energy change of the cell reaction $Zn\,|\,Zn^{2+}\,\|\,Ni^{2+}\,|\,Ni$. The standard reduction potentials of $Zn^{2+}|Zn$ and $Ni^{2+}|Ni$ half cells are $-0.76$ V and $-0.25$ V respectively.

With Zn as the anode (oxidation) and Ni as the cathode (reduction), the standard cell potential works out to +0.51 V, giving a standard free energy change of −98.43 kJ/mol via ΔG° = −nFE°.

Cell and half-reactions: In $Zn\,|\,Zn^{2+}\,\|\,Ni^{2+}\,|\,Ni$ (standard cell notation: anode on left, cathode on right):

• Anode (oxidation): $Zn \rightarrow Zn^{2+} + 2e^-$, standard reduction potential $E^\circ(Zn^{2+}/Zn) = -0.76\ V$
• Cathode (reduction): $Ni^{2+} + 2e^- \rightarrow Ni$, standard reduction potential $E^\circ(Ni^{2+}/Ni) = -0.25\ V$
• Overall: $Zn + Ni^{2+} \rightarrow Zn^{2+} + Ni$

Standard EMF of the cell:

$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = E^\circ(Ni^{2+}/Ni) - E^\circ(Zn^{2+}/Zn)$

$E^\circ_{cell} = (-0.25\ V) - (-0.76\ V) = -0.25 + 0.76 = +0.51\ V$

The positive value confirms the cell reaction is spontaneous as written (Zn reduces Ni$^{2+}$), consistent with Zn being more easily oxidised (more negative $E^\circ$) than Ni.

Standard free energy change:

Number of electrons transferred in the balanced overall reaction, $n=2$ (both half-reactions involve 2 electrons); Faraday constant $F = 96500\ C\,mol^{-1}$. …