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)

A g, H g, C u, M g

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M g, C u, H g, A g

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A g, H g, C u

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C u, H g, A g

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Solution

The correct option is A $A g, H g, C u, M g$
Higher the oxidation potential, more easily it is oxidized and faster is the deposition of metal.
Decreasing order of $E_{o x i d a t i o n}^{⊖}$ is:
$E_{A g | A g^{⨁} (0.80)}^{⊖} > E_{2 H g | H g_{2}^{2 +}}^{⊖} (0.79) > E_{C u | C u^{2 +} (0.34)}^{⊖} > E_{M g | M g^{2 +} (- 2.37)}^{⊖}$
Hence, the sequence of deposition of metals at the cathode will be Ag, Hg, and Cu.

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Similar questions

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

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

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

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

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

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

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

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

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