$100 m L$ of a buffer of 1 M $N H_{3} (a q)$ and $1 M$ $N H_{4}^{\oplus} (a q)$ are placed in two voltaic cells separately. A current of $1.5 A$ is passed through both cells for $20 m i n$ . If the electrolysis of water only takes place :
$2 H_{2} O + O_{2} + 4 e^{-} \to, 4 ⊖ O H (R H S)$
$2 H_{2} O \to H^{\oplus} + O_{2} + 4 e^{-} (L H S)$ , then:

p H

L H S

half cell will increase.

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p H

R H S

half cell will increase.

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p H

of both half cells will increase.

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p H

of both half cells will decrease.

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Solution

The correct option is D $p H$ of both half cells will decrease.
At $L H S$ there is formation of $H^{\oplus}$ ion and in RHS $⊖ O H$ ion.
$L H S$ : $N H_{4} O H + H^{\oplus} ⇌ N H_{4}^{\oplus} + H_{2} O$
$R H S$ : $N H_{4}^{\oplus} + ⊖ O H ⇌ N H_{4} O H$
At $L H S$ , concentration of $N H_{4} O H$ decreases and that of $N H_{4}^{\oplus}$ increases and reverse at $R H S$ .
$p O H = p K_{b} + l o g \frac{[N H_{4}^{\oplus}]}{[N H_{4} O H]}$
Hence, $p O H$ at $L H S$ increases, $p H$ decreases and that of $R H S$ increases.

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