A 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 potential) are.
$A g^{+} / A g = + 0.80, H g_{2}^{2 +} / 2 H g = + 0.79$
$C u^{2 +} / C u = + 0.34, M g^{2 +} / M g = - 2.37$
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 B Ag, Hg, Cu
A cation having highest reduction potential will be reduced find and so on. However $M g^{2 +}$ in aqueous solution will not be reduced $⎛ ⎜ ⎝ E_{M g^{2} / M g}^{o} < E_{H_{2} O / \frac{1}{2} H_{2} / O H^{-}} ⎞ ⎟ ⎠$ .

Instead water would be reduced in preference.

So option C is correct.

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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},

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

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