An electrochemical cell is made by connecting Nickel and Chromium electrodes. With the help of information given below, the correct value of standard cell potential, would be:
\(Ni^{2+} + 2e^- \longrightarrow Ni(s)\, \ E^o = -0.25V\)
\(Cr^{3+} + 3e^- \longrightarrow Cr(s)\, \ E^o = -0.74 V\)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → +0.49 V
In a galvanic cell the electrode with the higher reduction potential acts as the cathode (right) and the other as the anode (left). The emf is
\( \displaystyle E^\circ_{cell} = E^\circ_{right} - E^\circ_{left} = E^\circ_{cathode} - E^\circ_{anode} \)
\( \ce{Ni^{2+}} \)/Ni has E° = −0.25 V and \( \ce{Cr^{3+}} \)/Cr has E° = −0.74 V. Nickel is higher, so nickel is the cathode and chromium the anode:
\( \displaystyle \text{Anode } (\times 2):\ \ce{Cr(s) -> Cr^3+(aq) + 3e-} \)
\( \displaystyle \text{Cathode } (\times 3):\ \ce{Ni^2+(aq) + 2e- -> Ni(s)} \)
\( \displaystyle \ce{2Cr(s) + 3Ni^2+(aq) -> 2Cr^3+(aq) + 3Ni(s)} \)
\( \displaystyle E^\circ_{cell} = (-0.25\ \text{V}) - (-0.74\ \text{V}) = +0.49\ \text{V} \)
E° is an intensive property, so multiplying the half-reactions by 2 and 3 does not change the electrode potentials.
NCERT (2026-27 edition):
Galvanic Cells – “… 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. It is now an accepted convention that we keep the anode on the left and the cathode on the right while representing the galvanic cell. A galvanic cell is generally represented by putting a vertical line between metal and electrolyte solution and putting a double vertical line between the two electrolytes connected by a salt bridge. Under this convention the emf of the cell is positive and is given by the potential of the half-cell on the right hand side minus the potential of the half-cell on the left hand side i.e.,
Ecell = Eright – Eleft”