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Calculate the current (in mA) required to deposite 0.195g of platinum metal in 5.0 hours from a solution of ${[P t C l_{6}]}^{2 -}$
(Atomic mass Pt = 195)

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310

21.44

5.36

answer is C.

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Detailed Solution

$\frac{W}{E} = \frac{I \times t}{96500}; \frac{0.195}{195} \times 4 = \frac{5 \times 60 \times 60 \times I}{96500}$

$I = 21.44 m A$

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