Electrolysis of $N a C l (a q)$ is carried out using inert electrodes, choose the correct option:

When concentrated $N a C l$ is used, oxygen gas is liberated at anode.

When dilute $N a C l$ is used, chlorine gas is liberated at anode.

When concentrated $N a C l$ is used, chlorine gas is liberated at anode.

When dilute $N a C l$ is used, oxygen gas is liberated at anode.

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Solution

Electrolysis of concentrated $N a C l (a q)$ carried out using inert electrodes:

Since ,
$E_{H_{2} O / H_{2}}^{0} > E_{N a^{+} / N a}^{0}$

Reaction at cathode:
$2 H_{2} O (l) + 2 e^{-} ⇌ H_{2} (g) + 2 O H^{-} (a q)$

Reaction at anode:
Even if $E_{C l^{-} / C l_{2}}^{0} < E_{H_{2} O / O_{2}}^{0}$
Kinetically rate of oxidation of water is very slow. To increase its rate, the greater potential difference is applied called as overpotential.Because of this overpotential, oxidation of $C l^{-}$ ions becomes more feasible
The reaction that occurs at anode is :
$2 C l^{-} ⇌ C l_{2} (g) + 2 e^{-}$
So, chroine gas is involved at anode.

Electrolysis of dilute $N a C l (a q)$ carried out using inert electrodes:

Since ,
$E_{H_{2} O / H_{2}}^{0} > E_{N a^{+} / N a}^{0}$

Reaction at cathode:
$2 H_{2} O (l) + 2 e^{-} ⇌ H_{2} (g) + 2 O H^{-} (a q)$

Reaction at anode: If concentration of $C l^{-}$ is low, oxidation of $H_{2} O$ to give $O_{2}$ is favoured.

$2 H_{2} O (l) ⇌ 4 H^{+} (a q) + O_{2} (g) + 4 e^{-}$

When dilute $N a C l$ is used, oxygen gas is liberated at anode.
Hence (c) is correct.

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Similar questions

During the electrolysis of NaCl, sodium is obtained at the ________whereas chlorine gas is liberated at the ________.

I. Gas X is liberated at the anode during the electrolysis of molten NaCl.

II. Gas Y is liberated at the cathode during the electrolysis of NaCl solution.

Identify gases X and Y.

E^{o}

O_{2}

C l_{2}

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