If time required when a current of $9.65 A$ is passed through $80 mL$ of $0.1 M NaCl$ solution in order to make its $p H = 12$ is $x sec$ and time required when a current of $965 A$ is passed through an aqueous solution of $C u S O_{4}$ to produce $0.28 L$ of oxygen gas at STP is $y sec$ . Then the value of $\frac{x}{y}$ would be

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Solution

$p H = 12 p O H = 2$
$∴ [O H^{-}] = 10^{- 2}$
$Equivalents of [O H^{-}] = \frac{9.65 \times x}{96500} \Rightarrow 10^{- 2} \times 80 \times 10^{- 3} = \frac{9.65 \times x}{96500} \Rightarrow 8 \times 10^{- 4} = \frac{9.65 \times x}{96500}$
$\Rightarrow x = 8$ sec
Equivalents of oxygen $= \frac{0.28}{5.6} = 0.05$
$∴ 0.05 = \frac{965 \times y}{96500}$
$\Rightarrow y = 5$
So, $\frac{x}{y} = \frac{8}{5} = 1.6$

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If time required when a current of 9.65 A is passed through 80 mL of 0.1 M NaCl solution in order to make its pH=12 is x sec and time required when a current of 965 A is passed through an aqueous solution of CuSO4 to produce 0.28 L of oxygen gas at STP is y sec. Then the value of xy would be

pH=12pOH=2 ∴[OH−]=10−2 Equivalents of [OH−]=9.65×x96500⇒10−2×80×10−3=9.65×x96500⇒8×10−4=9.65×x96500 ⇒x=8 sec Equivalents of oxygen =0.285.6=0.05 ∴0.05=965×y96500 ⇒y=5 So, xy=85=1.6