During electrolysis of 1 litre of aq.CuSO₄, the P^H of the solution is changed from 5.0 to 4.3. Then the number of faradays consumed in the process is (x × 10^–y), here, (x+y) =

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answer is 9.

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Detailed Solution

$(H_{2} O + {CuSO}_{4}) \overset{electrolysis / 2 F}{\to} Cu + 1 / 2 O_{2} ↑ + H_{2} {SO}_{4}$
$2 F \overset{passed}{⟶} 2 mole of H^{\oplus}$
Decrease in [H⁺] = $5 \times 10^{- 5} - 10^{- 5} = 4 \times 10^{- 5}$

current passed = $4 \times 10^{- 5}$ F

x + y = 4 + 5 = 9

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