How long (in hours) must a current of $5$ ampere be maintained to electroplate $60 g$ of calcium from molten $C a C l_{2}$ ?

27 h o u r s

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8.3 h o u r s

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11 h o u r s

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16 h o u r s

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Solution

The correct option is D $16 h o u r s$
The reduction reaction:-
$C a^{2 +} + 2 e \to C a$
(i) Mole ratio $= \frac{m o l e s o f C a f o r m e d}{m o l e s o f e l e c t r o n s} = \frac{1}{2}$
(ii) Moles of $C a = \frac{w e i g h t}{m o l a r m a s s} = \frac{60}{40} = 1.5$
(iii) Moles of $C a$ formed $= \frac{Q (C)}{96500 C m o l^{- 1} e^{-}} \times m o l e r a t i o$
$1.5 m o l = \frac{Q (C)}{96500 C m o l^{- 1} e^{-}} \times \frac{1}{2}$
$∴ Q (C) = 1.5 \times \times 96500 \times 2 = 289500 C$

Coulombs = ampere X seconds
Seconds = 289X10^5 / 5 = 57900 seconds
57900 s X (1 hr / 3600 s) = 16 hours

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