Which of the following does not evolve oxygen at anodes when electrolysis is carried?

dil,

H_{2} {S O}_{4}

with Pt electrode

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Fused NaOH with Pt electrode

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Acidic water with Pt electrode

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dil,

H_{2} {S O}_{4}

with Cu electrode

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Solution

The correct option is A dil, $H_{2} {S O}_{4}$ with Cu electrode
In the electrolysis of dil. $H_{2} S O_{4}$ using Cu electrodes, the reaction at anode occurs as:
Cu $\to C u^{2}$ $^{+}$ + $e^{-}$
The electrode is not inert an so participates in the reaction. In all other reaction, inert electrode i.e. Pt is used. Thus oxygen is evolved in these reactions.
Therefore D is the correct solution.

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Similar questions

H_{2}

H_{2} {S O}_{4}

_{2}

H_{2} {S O}_{4}

A) Electrolysis of aq. $N a_{2} S O_{4}$ using Pt electrodes	1) 1 Faraday
B) The charge carried by $6.023 \times 10^{23}$ electrons is	2) 3 Faradays
C) The amount of electricity required to deposit 27 grams of Aluminium at cathode from molten $A l_{2} O_{3}$ is	3) $H_{2 (g)} /$ Pt
D) A gas in contact with an inert electrode	4) $O_{2}$ at anode $H_{2}$ at cathode

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