Calculate the current (in mA) required to deposit $0.195 g$ of platinum metal in $5.0$ hours from a solution of $[P t C l_{6}]^{2 -}$ .
(Atomic weight: $C r = 52.0$ )

310

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31

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21.44

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5.36

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Solution

The correct option is C $21.44$
$p t {C l}_{6}^{2 -} + 4 e^{-} \to P t + 6 {C l}^{(-)}$
moles of $p t = \frac{g i v e n m a s s}{m o l a r m a s s} = \frac{0.195}{195} = 0.001 m o l$
4 mol. $F$ give 1 mole of pt.
$∴$ for 0.001 mol pt require 0.004 mol of $e^{-}$
$∴$ no. of Faraday $= \frac{c h a r g e}{96500}$

$0.004 = \frac{1 \times 5 \times 60 \times 60}{96500} = 21.4 m A$

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Similar questions

Calculate the current (in mA) required to deposit 0.195g of platinum metal in 5 hours from a solution of [PtCl₆]^2- : (Atomic mass of Pt = 195)

(a) 310 (b) 31 (c) 21.44 (d) 5.36

P t C l_{6}^{2 -}

P t = 195)

[P t C l_{6}]^{2 -}

C r_{2} (S O_{4})_{3}

C r

C r (N O_{3})_{3}

0.54 g

A g

C r (N O_{3})_{3}

C r = 52.0 u; A g = 108 u

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