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

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 C $A g, H g, C u$
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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Similar questions

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

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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