The passage of current liberates $H_{2}$ at cathode and $C l_{2}$ at anode. The solution is:

Copper chloride in water

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N a C l

in water

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H_{2} S O_{4}

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Water

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Solution

The correct option is B $N a C l$ in water
Since discharge potential of water is greater than that of sodium so water is reduced at cathode instead of $N a^{+}$
Cathode: $H_{2} O + e^{-} \to \frac{1}{2} H_{2} + O H^{-}$
Anode: $C l^{-} \to \frac{1}{2} C l_{2} + e^{-}$ .

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Similar questions

Q. The passage of current liberates

H_{2}

at cathode and

C l_{2}

at anode. The solution is_______________.

The passage of current through a solution of certain electrolyte results in the evolution of $H_{2}$ at cathode and $C l_{2}$ at anode. What could be the solution?

Q. Assertion :The electrolysis of NaCl solution gives

H_{2}

(g) at cathode and

C l_{2}

(g) at anode. Reason:

C l_{2}

has higher oxidation potential than

H_{2} O

Given are the reactions occuring during the electrolysis of sodium chloride solution:

$2 C l^{-} \to C l_{2} + 2 e^{-}$
$2 H_{2} O + 2 e^{-} \to H_{2} + 2 O H^{-}$

Which of the following is correct regarding liberation of electrolytic products?

Q. During electrolysis of water, and are liberated at anode and cathode respectively.

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The passage of current liberates H2 at cathode and Cl2 at anode. The solution is:

The correct option is B NaCl in waterSince discharge potential of water is greater than that of sodium so water is reduced at cathode instead of Na+Cathode: H2O+e−→12H2+OH−Anode: Cl−→12Cl2+e−.

The passage of current liberates $H_{2}$ at cathode and $C l_{2}$ at anode. The solution is:

The correct option is B $N a C l$ in water
Since discharge potential of water is greater than that of sodium so water is reduced at cathode instead of $N a^{+}$
Cathode: $H_{2} O + e^{-} \to \frac{1}{2} H_{2} + O H^{-}$
Anode: $C l^{-} \to \frac{1}{2} C l_{2} + e^{-}$ .