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Standard X

Chemistry

Application of Electrolysis

On electrolyz...

Question

On electrolyzing a solution of dilute $H_{2} S O_{4}$ between platinum electrodes, the gas evolved at the anode and cathode respectively are :

S O_{2}

and

O_{2}

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S O_{3}

and

H_{2}

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O_{2}

and

H_{2}

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H_{2}

and

O_{2}

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Solution

The correct option is C $O_{2}$ and $H_{2}$
When $H_{2} S O_{4}$ is electrolysed, there are two different anion presents in solution $O H^{-}, S O_{4}^{2 -}$ .
But as oxidation potential of hydroxide ion is higher so it will oxidise first.

Reaction:
$2 H^{+} + 2 e \to H_{2}$ (Cathode)
$2 O H^{-} \to H_{2} O + (1 / 2) O_{2} + 2 e$ (Anode)

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Similar questions

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Q. In a fuel cell,

H_{2}

and

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67.2 litres

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react in 15 minutes and the entire current is used for electrode deposition of

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from

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solution, how many grams of

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Anode :

H_{2} + 2 O H^{-} \to 2 H_{2} O + 2 e^{-}

Cathode :

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

Q. In a fuel cell,

H_{2}

and

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react to produce electricity. In the process,

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gas is oxidized at the anode and

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67.2 litres

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react in 15 minutes and the entire current is used for electrode deposition of

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from

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solution, how many grams of

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will be deposited and what is the average current produced?
Anode :

H_{2} + 2 O H^{-} \to 2 H_{2} O + 2 e^{-}

Cathode :

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

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On electrolyzing a solution of dilute H2SO4 between platinum electrodes, the gas evolved at the anode and cathode respectively are :

The correct option is C O2 and H2When H2SO4 is electrolysed, there are two different anion presents in solution OH−,SO2−4.But as oxidation potential of hydroxide ion is higher so it will oxidise first.Reaction:2H++2e→H2 (Cathode)2OH−→H2O+(1/2)O2+2e (Anode)

On electrolyzing a solution of dilute $H_{2} S O_{4}$ between platinum electrodes, the gas evolved at the anode and cathode respectively are :