Values of \(E^o\) reduction are given below:
\(Sn^{2+}|Sn = -0.14V\), \(Zn^{2+}|Zn = -0.76 V\), \(Cr^{3+}|Cr = -0.74V\), \(Mg^{2+}|Mg = -2.36 V\)
The ion which will get deposited at cathode in an electrochemical cell would be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(Sn^{2+}\)
At the cathode of an electrochemical cell, reduction takes place. The ion with the highest (least negative) standard reduction potential has the greatest tendency to gain electrons, so it is the one reduced and deposited as metal.
Given values, highest first:
\( \ce{Sn^{2+}} \)|Sn (−0.14 V) > \( \ce{Cr^{3+}} \)|Cr (−0.74 V) > \( \ce{Zn^{2+}} \)|Zn (−0.76 V) > \( \ce{Mg^{2+}} \)|Mg (−2.36 V)
\( \ce{Sn^{2+}} \) has the highest E°, so it is reduced most readily at the cathode:
\( \displaystyle \ce{Sn^2+(aq) + 2e- -> Sn(s)} \)
Why the other options are incorrect
Option (2) \(Zn^{2+}\) — Incorrect. E° = −0.76 V is lower than that of \( \ce{Sn^{2+}} \), so \( \ce{Zn^{2+}} \) is a weaker oxidising agent and is not reduced before \( \ce{Sn^{2+}} \).
Option (3) \(Cr^{3+}\) — Incorrect. E° = −0.74 V, again lower than −0.14 V; \( \ce{Cr^{3+}} \) is less easily reduced.
Option (4) \(Mg^{2+}\) — Incorrect. E° = −2.36 V is the most negative; \( \ce{Mg^{2+}} \) is the hardest to reduce of the four (magnesium metal is a strong reducing agent).
NCERT (2026-27 edition):
Measurement of Electrode Potential – …
