Two platinum electrodes were immersed in a solution of CuSO $_{4}$ and current was passed through the solution. After sometime it was found that colour of CuSO $_{4}$ disappeared with the evolution of gas at the electrode. The colourless solution contains:

platinum sulphate

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copper sulphate

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copper hydroxide

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sulphuric acid

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Solution

The correct option is B sulphuric acid
$S O_{4}^{2 -}$ oxidises less readily as compared to $O H^{-}$
$∴$ at anode $O_{2}$ is released and
$S O_{4}^{2 -}$ ions remain is solution
whereas the copper ion having higher reduction potential then $H^{+}$ will get reduced more easily then $H^{+}$
and at cathode $C u^{+} + e^{-} \to C u$ (metal).
Sulphate ion and hydrogen ion combine to form sulphuric acid.
$S O_{4}^{2 -}$ + 2 $H^{+}$ $\to$ $H_{2} S O_{4}$

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Two platinum electrodes were immersed in a solution of CuSO4 and current was passed through the solution. After sometime it was found that colour of CuSO4 disappeared with the evolution of gas at the electrode. The colourless solution contains:

Two platinum electrodes were immersed in a solution of CuSO $_{4}$ and current was passed through the solution. After sometime it was found that colour of CuSO $_{4}$ disappeared with the evolution of gas at the electrode. The colourless solution contains: