The standard electrode potentials of the Ag+/Ag and Zn2+/Zn electrodes at 298 K are +0.80 V and −0.76 V respectively. What is the standard Gibbs energy of the following reaction at 298 K? (F = 96500 C mol−1)
\( \ce{Zn(s) + 2Ag+(aq) -> Zn^2+(aq) + 2Ag(s)} \)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → −301.1 kJ mol−1
The standard Gibbs energy of a cell reaction is found from its standard emf: ΔrG° = −nFE°cell, where n is the number of electrons transferred in the reaction as written. Silver ions are reduced, so silver is the cathode and zinc the anode; each zinc atom gives two electrons, so n = 2.
\( \displaystyle E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.80\ \mathrm{V} - (-0.76\ \mathrm{V}) = 1.56\ \mathrm{V} \)
\( \displaystyle \Delta_r G^\circ = -nFE^\circ_{cell} = -2 \times 96500\ \mathrm{C\,mol^{-1}} \times 1.56\ \mathrm{V} \)
\( \displaystyle \Delta_r G^\circ = -301080\ \mathrm{J\,mol^{-1}} = -301.1\ \mathrm{kJ\,mol^{-1}} \)
The negative value shows that the reaction is spontaneous.
NCERT (2026-27 edition):
2.3.2 Electrochemical Cell and Gibbs Energy of the Reaction – “…If the concentration of all the reacting species is unity, then E(cell) = E°(cell) and we have
ΔrG° = – nFE°(cell) (2.16).
Thus, from the measurement of E°(cell) we can obtain an important thermodynamic quantity, ΔrG°, standard Gibbs energy of the reaction.”