Electrochemistry
A galvanic cell is an electrochemical cell that converts the chemical energy of a spontaneous redox reaction into electrical energy whereas, an electrolytic cell is a device which uses electrical energy to carry out the non-spontaneous chemical reactions. We can construct innumerable number of galvanic cells on the pattern of Daniell cell (Cu-Zn cell) by taking combinations of different half-cells. Each half-cell consists of a metallic electrode dipped into an electrolyte. The two half-cells are connected by a metallic wire through a voltmeter and a switch externally. The electrolytes of the two half-cells are connected internally through a salt bridge. In a galvanic cell, the half-cell in which oxidation takes place is called anode and it has a negative potential with respect to the solution and in an accepted convention we keep the anode on the left and the cathode on the right side. The potential difference between the two electrodes of a galvanic cell is called the cell potential and is measured in volts. The cell potential is the difference between the electrode potentials (reduction potentials) of the cathode and anode. It is called the cell electromotive force (emf) of the cell when no current is drawn through the cell.
If the concentration of all the reacting species is unity then find the standard Gibbs free energy (in kJ mol−1) for the given redox reaction:
Ni(s) + 2Ag+(aq) → Ni2+(aq) + 2Ag(s)
Given \( E^{\circ}_{\mathrm{Ni^{2+}/Ni}} \) = −0.25 V, \( E^{\circ}_{\mathrm{Ag^{+}/Ag}} \) = +0.80 V; F = 96500 C/mol
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → −202.65
The standard Gibbs energy of a cell reaction is related to its standard emf: the electrical work the cell can do is the charge passed (nF) times the voltage (E°), and this equals the fall in Gibbs energy:
\( \displaystyle \Delta_rG^{\circ}=-nFE^{\circ}_{\text{cell}} \)
A positive E° gives a negative ΔrG°, i.e. a spontaneous reaction.
In Ni(s) + 2Ag+(aq) → Ni2+(aq) + 2Ag(s), each Ni atom gives 2 electrons, so n = 2. Nickel is oxidised (anode) and silver ion is reduced (cathode):
\( \displaystyle E^{\circ}_{\text{cell}}=E^{\circ}_{\text{cathode}}-E^{\circ}_{\text{anode}}=0.80-(-0.25)=1.05\ \mathrm{V} \)
\( \displaystyle \Delta_rG^{\circ}=-2\times96500\times1.05 \)
\( \displaystyle \Delta_rG^{\circ}=-202650\ \mathrm{J\,mol^{-1}} \)
\( \displaystyle \Delta_rG^{\circ}=-202.65\ \mathrm{kJ\,mol^{-1}} \)
The question asks for the value in kJ mol−1, so the answer is −202.65.
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 – …
