In the reaction, $2 {CuSO}_{4} + 4 KI ⟶ {Cu}_{2} I_{2} + I_{2} + 2 K_{2} {SO}_{4}$ the equivalent weight of Cu will be:
(Atomic weight of Cu = 63.5).

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15.88

31.75

127

63.5

answer is B.

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Detailed Solution

$\begin{array}{l} 2 {CuSO}_{4} + 4 KI \to {Cu}_{2} I_{2} + I_{2} + 2 K_{2} {SO}_{4} \\ {Cu}^{2 +} + 1 e^{-} \to {Cu}^{+} \\ E_{Cu} = ? V.F. = 1 \\ E_{Cu} = \frac{Atomic wht.}{V.F.} = \frac{63.5}{1} = 63.5 \end{array}$

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