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 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, find the sequence of deposition of metals on the cathode.

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Solution

Electrolysis will occur if $E_{e x t} / E_{g a l v a n i c}$ . The value of $E_{g a l v a n i c}$ (in which metal is oxidized) is $E_{H_{2} O | O_{2} | P t} - E_{M^{n +} / M}$
So $E_{e x t}$ will be reached in order ${A g}^{+} / A g, {H g}_{2}^{2 +} / H g$ and ${C u}^{2 +} / C u$ . Hence the sequence of deposition of metals on the cathode will be $A g, H g$ and $C u$ .

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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}

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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