The gas evolved at the anode when $K_{2} S O_{4} (a q)$ is electrolyzed between Pt electrodes is:

O_{2}

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H_{2}

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S O_{2}

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S O_{3}

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Solution

The correct option is D $O_{2}$
Here,
Electrolysis of $H_{2} O$ takes place:

$2 H_{2} O \to O_{2} + 4 H^{\oplus} + 4 e^{-}$

and $O_{2}$ evolved.

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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

Q. Assertion: When dilute

H_{2} S O_{4}

is electrolyzed between Pt electrodes, the gas liberated at the anode will be

O_{2} (g)

.
Reason: Oxidation potential of

H_{2} O > O x i d a t i o n p o t e n t i a l o f S O_{4}^{2 -}

The reaction taking place at the anode when an aqueous solution of $C u S O_{4}$ is electrolyzed using inert $P t$ electrodes:

When the aqueous solution of $L i_{2} S O_{4}$ electrolyzed the products formed at cathode and anode respectively are

When aqueous $A g N O_{3}$ is electrolyzed using Pt electrodes, if 3.2 g of $O_{2}$ is liberated at the anode, the Amount of Ag that will be deposited at the cathode in grams is (at wt. of Ag = 108)

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The gas evolved at the anode when K2SO4(aq) is electrolyzed between Pt electrodes is:

The correct option is D O2Here,Electrolysis of H2O takes place:2H2O→O2+4H⊕+4e−and O2 evolved.

The gas evolved at the anode when $K_{2} S O_{4} (a q)$ is electrolyzed between Pt electrodes is:

The correct option is D $O_{2}$
Here,
Electrolysis of $H_{2} O$ takes place:

$2 H_{2} O \to O_{2} + 4 H^{\oplus} + 4 e^{-}$

and $O_{2}$ evolved.