Question

A solution containing 1 mol per litre of each $Cu(N{O}_3{)}_2,AgN{O}_3,$ and $H{g}_2(N{O}_3{)}_2$ is being electrolyzed by using inert electrodes. The values of standard electrode potentials in volts (reduction potential) are
$Ag|A{g}^{⨁}=+0.80,2Hg|H{g}_2^{2+}=+0.79$
$Cu|C{u}_{2+}=+0.34,Mg|M{g}^{2+}=-2.37$
with increasing voltage, the sequence of deposition of metals at the cathode will be
(IIT-JEE, 1984)

• A. Ag, Hg, Cu, Mg
• B. Mg, Cu, Hg, Ag
• C. Ag, Hg, Cu
• D. Cu, Hg, Ag

Answer 1

The correct option is A Ag, Hg, Cu, Mg

Higher the oxidation potential, more easily it is oxidized and faster is the deposition of metal.
Decreasing order of $E_{oxidation}^{\ominus }$ is:
$E_{Ag|A{g}^{⨁}\left(0.80\right)}^{\ominus }>E_{2Hg|H{g}_2^{2+}}^{\ominus }\left(0.79\right)>E_{Cu|C{u}^{2+}\left(0.34\right)}^{\ominus }>E_{Mg|M{g}^{2+}(-2.37)}^{\ominus }$
Hence, the sequence of deposition of metals at the cathode will be Ag, Hg, and Cu.