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Representation of Cell

Electrolysis ...

Question

Electrolysis of aqueous $C u S O_{4}$ with inert electrodes gives :

C u

at cathode, anode gets dissolved

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C u

at cathode,

O_{2}

at anode

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O_{2}

at anode,

H_{2}

at cathode

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O_{2}

at anode, cathode gets dissolved

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Solution

The correct option is B $C u$ at cathode, $O_{2}$ at anode
Electrolyte $C u S O_{4}$ dissociates as $C u^{2 +}$ and $S O_{4}^{2 -}$ along with $H +$ and $O H^{-}$ ions in the aqueous solution.
At cathode (reduction):
As $C u^{2 +}$ has higher reduction potential than $H^{+}$ , it gets preferentially reduced.
$2 C u^{2 +} + 4 e^{-} \to 2 C u$
At anode (oxidation):
As $O H^{-}$ has higher oxidation potential than $S O_{4}^{2 -}$ , it gets preferentially oxidised.
$4 O H^{-} \to 2 H_{2} O + O_{2} + 4 e^{-}$
Hence, $C u$ at cathode and $O_{2}$ at anode are produced by electrolysis of aqueous $C u S O_{4}$ with inert electrodes.

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Similar questions

Aqueous $C u S O_{4}$ is electrolyzed using $A g$ as anode and $C u$ as cathode, then product of electrolysis at anode and cathode are, respectively

A g N O_{3}

(i) - - -

(i i) - -- -

(i i i) - --- -

(i v) - -- -

C u S O_{4}

C u^{+ 2} + 2 e^{-} \to C u

C u \to C u^{+ 2} + 2 e^{-}

C u S O_{4}

N a O H, 4 g

O_{2}

H_{2}

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