On electrolysis of a solution of dilute $H_{2} S O_{4}$ between platinum electrodes, the gas evolved at the anode is:

S O_{2}

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S O_{3}

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O_{2}

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H_{2}

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Solution

The correct option is C $O_{2}$
Oxidation potential of $H_{2} O$ > Oxidation potential of $S O_{4}^{2 -}$
So oxidation of $H_{2} O$ occurs at anode and gives $O_{2} (g)$ .

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Similar questions

Q. On electrolyzing a solution of dilute

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

between platinum electrodes, the gas evolved at the anode is

In electrolysis of dilute H₂SO₄using platinum electrodes

[DPMT 1983; IIT 1983; Kurukshetra CET 2002; AFMC 2005]

Which of the following statements are correct?

a) The electrolysis of aqueous NaCl produces hydrogen gas at the cathode and chlorine gas at the anode .

b) The electrolysis of a $C u S O_{4}$ solution using Pt electrodes causes the liberation of $O_{2}$ at the anode and the deposition of copper at the cathode .

c) Oxygen and hydrogen are produced at the anode and cathode during the electrolysis of dilute aqueous solution of $H_{2} S O_{4}$ .

d) All electrolytic reactions are redox reactions .

Q. On electrolysis of dilute copper sulphate solution using platinum electrodes, the products obtained at cathode and anode respectively are:

Q. On electrolysis of dilute sulphuric acid using platinum electrodes, the product obtained at the anode will be:

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On electrolysis of a solution of dilute H2SO4 between platinum electrodes, the gas evolved at the anode is:

The correct option is C O2Oxidation potential of H2O > Oxidation potential of SO2−4So oxidation of H2O occurs at anode and gives O2(g).

On electrolysis of a solution of dilute $H_{2} S O_{4}$ between platinum electrodes, the gas evolved at the anode is:

The correct option is C $O_{2}$
Oxidation potential of $H_{2} O$ > Oxidation potential of $S O_{4}^{2 -}$
So oxidation of $H_{2} O$ occurs at anode and gives $O_{2} (g)$ .