The standard reduction potential for \(Sn^{4+}/Sn^{2+}\) is \(+ 0.15 V\) and for \(Cr^{3+}/Cr\) is \(- 0.74 V\). These two half-cells, coupled in their standard states, are connected to make a cell. The galvanic cell can be correctly represented by:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(Cr(s)|Cr^{3+}(aq)||Sn^{4+} (aq) | Sn^{2+}(aq)\)
In a galvanic cell the couple with the higher (more positive) reduction potential is the cathode (reduction) and the other is the anode (oxidation). The cell is written anode on the left, cathode on the right, with a double line for the salt bridge; in each half-cell the species is written in the order in which it reacts.
\( \ce{Sn^{4+}} \)/\( \ce{Sn^{2+}} \), E° = +0.15 V (higher) — cathode: \( \ce{Sn^4+(aq) + 2e- -> Sn^2+(aq)} \)
\( \ce{Cr^{3+}} \)/Cr, E° = −0.74 V (lower) — anode: \( \ce{Cr(s) -> Cr^3+(aq) + 3e-} \)
Cell reaction (electrons balanced, 6 e−):
\( \ce{2Cr(s) + 3Sn^4+(aq) -> 2Cr^3+(aq) + 3Sn^2+(aq)} \)
\( E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.15 - (-0.74) = +0.89\ V \)
The cell is Cr(s)|Cr3+(aq)||Sn4+(aq)|Sn2+(aq).
Why the other options are incorrect
Option (1) \(Sn^{2+}(aq) |Sn^{4+} (aq)||Cr(s)|Cr^{3+}(aq)\) — Incorrect. It puts the tin couple as the anode (\( \ce{Sn^{2+}} \) oxidised) and chromium on the right as an oxidation; this cell would have E° = −0.89 V and is not spontaneous.
Option (2) \(Sn^{4+}(aq)|Sn^{2+}(aq) ||Cr^{3+} (aq)|Cr(s)\) — Incorrect. It shows \( \ce{Cr^{3+}} \) being reduced at the cathode; the couple with the lower E° cannot be the cathode.
Option (4) \(Cr(s)|Cr^{3+}(aq)||Sn^{2+}(aq)|Sn^{4+}(aq)\) — Incorrect. The anode is right, but writing \( \ce{Sn^{2+}} \)|\( \ce{Sn^{4+}} \) on the cathode side shows \( \ce{Sn^{2+}} \) being oxidised to \( \ce{Sn^{4+}} \); at the cathode \( \ce{Sn^{4+}} \) is reduced.
NCERT (2026-27 edition):
Galvanic Cells – …
