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Standard X

Chemistry

Electrolysis and Electrolytes

Electrolysis ...

Question

Electrolysis of an aqueous solution of $N a O H$ between platinum electrodes yields :

_{2}

at cathode only

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_{2}

at anode only

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_{2}

at cathode, O

_{2}

at anode

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_{2}

at anode, O

_{2}

at cathode

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Solution

The correct option is C H $_{2}$ at cathode, O $_{2}$ at anode
Reduction potential of $H^{+}$ $>$ Reduction potential of $N a^{+}$
$∴ H^{+}$ gets reduced at cathode
$2 H^{+} + 2 e^{-} \to H_{2}$

At, anode $O H^{-}$ gets oxidised to $O_{2}$

$2 O H^{-} \to H_{2} O + \frac{1}{2} O_{2} + 2 e^{-}$

Option C is correct.

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(i) - - -

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(i i i) - --- -

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Q. Assertion: Starting from aq. copper [II] sulphate solution, equations for the reactions at the cathode and anode during electrolysis of aq.

C u S O_{4}

using active copper electrodes is;
at anode:

C u^{+ 2} + 2 e^{-} \to C u

at cathode:

C u \to C u^{+ 2} + 2 e^{-}

Reason: Oxidation always take place at anode and reduction at cathode.

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 .

Aqueous $C u S O_{4}$ is electrolyzed using $A g$ as anode and $C u$ as cathode, then product of electrolysis at anode and cathode are, respectively

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Electrolysis of an aqueous solution of NaOH between platinum electrodes yields :

The correct option is C H2 at cathode, O2 at anodeReduction potential of H+ > Reduction potential of Na+∴H+gets reduced at cathode2H++2e−→H2At, anode OH− gets oxidised to O22OH−→H2O+12O2+2e−Option C is correct.

Electrolysis of an aqueous solution of $N a O H$ between platinum electrodes yields :