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Electrolysis and Electrolytes

On passing 0....

Question

On passing 0.5 mole of electrons through
$C u S o_{4}$ and $H g_{2} (N O_{3})_{2}$ solutions in series
using inert electrodes :

A

0.5 mol of Cu deposited

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B

0.5 mol of Hg deposited

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C

0.125 mol of

O_{2}

produced

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D

0.5 mol of

O_{2}

produced

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Solution

The correct options are
B 0.5 mol of Hg deposited
D 0.125 mol of $O_{2}$ produced
To deposit one mole of mercury, one mole of electrons is required.
${H g}^{+} + e^{-} \to H g$
Hence 0.5 mole of electron will deposit 0.5 mole of mercury.
To deposit one mole of oxygen, four moles of electrons is required.
${4 O H}^{-} \to 2 H_{2} {O + O}_{2} + 4 e^{-}$
Hence 0.5 mole of electrons will deposit 0.125 mole of oxygen.

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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
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(IIT-JEE, 1984)

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