What is \(\Delta G^\circ\) for the given reaction?
\(Zn(s) + Cu^{2+}(aq) \longrightarrow Zn^{2+}(aq) + Cu(s)\)
(Given: \(E^\circ_{Zn^{2+}/Zn} = -0.76\ V\) & \(E^\circ_{Cu^{2+}/Cu} = +0.34\ V\))
\(F = 96500\ C\ mol^{-1}\)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\Delta G^\circ = -212.3\ kJ/mol\)
The Gibbs energy of a cell reaction is related to the cell emf by
\( \displaystyle \Delta_r G^\circ = -nFE^\circ_{\text{cell}} \)
where n is the number of electrons transferred. In this reaction zinc is oxidised (anode) and \( \ce{Cu^{2+}} \) is reduced (cathode):
\( \displaystyle \text{Anode: } \ce{Zn(s) -> Zn^2+(aq) + 2e-} \qquad \text{Cathode: } \ce{Cu^2+(aq) + 2e- -> Cu(s)} \)
So n = 2, and
\( \displaystyle E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.34 - (-0.76) = 1.10\ \mathrm{V} \)
\( \displaystyle \Delta_r G^\circ = -2 \times 96500\ \mathrm{C\,mol^{-1}} \times 1.10\ \mathrm{V} = -212300\ \mathrm{J\,mol^{-1}} = -212.3\ \mathrm{kJ\,mol^{-1}} \)
The negative value shows that the reaction (the Daniell cell) is spontaneous.
Shortcut: A positive E°cell always gives a negative ΔG°, which rules out Options (2) and (4).
Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27 edition):
Electrochemical Cell and Gibbs Energy of the Reaction – “Example 2.3: The standard electrode potential for Daniell cell is 1.1V. Calculate the standard Gibbs energy for the reaction:”
\( \ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)} \)