An aqueous solution containing one mole per litre of each $C u (N O_{3})_{2}, A g N O_{3}, H g_{2} (N O_{3})_{2}$ and $M g (N O_{3})_{2}$ , is being electrolysed by using inert electrodes. The values of standard electrode potentials in volts (reduction potentials) are $A g / A g^{+} = + 0.80, 2 H g / H g_{2}^{2 +} = + 0.79, C u / C u^{2 +} = + 0.34, M g / M g^{2 +} = - 2.37$
With increasing voltage, the sequence of deposition of metals on the cathode will be

Ag, Hg, Cu, Mg

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Mg, Cu, Hg, Ag

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Ag, Hg, Cu

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Cu, Hg, Ag

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Solution

The correct option is C Ag, Hg, Cu
A cation that has the highest reduction potential will be reduced first and so on.
However, $M g^{2 +}$ in aqueous solution will not be reduced $(E_{M g^{2 +} / M g}^{\circ} < E_{_{H_{2} O} / \frac{1}{2} H_{2} + O H^{-}}) .$ Instead water would be reduced in preference.

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A solution containing 1 mol per litre of each $C u (N O_{3})_{2}, A g N O_{3},$ 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
$A g | A g^{⨁} = + 0.80, 2 H g | H g_{2}^{2 +} = + 0.79$
$C u | C u^{2 +} = + 0.34, M g | M g^{2 +} = - 2.37$
with increasing voltage, the sequence of deposition of metals at the cathode will be
(IIT-JEE, 1984)

Q. A solution containing one mole per litre each of

C u (N O_{3})_{2}, A g N O_{3}, H g_{2} (N O_{3})_{2}

and

M g (N O_{3})_{2}

is being electrolysed by using inert electrodes. The values of standard electrode potentials in volt (reduction potentials) are:

A g / A g^{+} = 0.80,

2 H g / H g_{2}^{2 +} = 0. 79

C u / C u^{2 +} = + 0.34

and

M g / M g^{2 +} = - 2.37

With increasing voltage, the sequence of deposition of metals on the cathode will be?

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The correct option is C Ag, Hg, CuA cation that has the highest reduction potential will be reduced first and so on. However, Mg2+ in aqueous solution will not be reduced (E∘Mg2+/Mg<EH2O/12H2+OH−). Instead water would be reduced in preference.