Question

How much current is required to deposit 0.195g of elemental Pt from a solution containing $[PtC{l}_6{]}^{2-}$ ion with a time period of 2 hrs? (Atomic Mass of Pt = 195)

• A. 0.054A
• B. 0.214A
• C. 0.428A
• D. 0.027A

Answer 1

The correct option is A 0.054A

$[PtC{l}_6{]}^{2-}$ or $P{t}^{4+}+4e^{-}\to Pt$
4 $\times$ 96500C of electricity will deposit Pt = 195g.
$\therefore$ 195g of Pt will be deposited by 96500 $\times$ 4C
$\therefore$ 0.195g of Pt will be deposited b y
$\frac{96500\times 4\times 0.195}{195}$
Currrent (I) $=\frac{4\times 96500\times 0.195}{195\times 2\times 60\times 60}=0.054A$