Question

A current of 2 ampere passing for 5 hours deposits 22.2 g of tin, the oxidation state of tin is:

[At. wt. = 119]

• A. 0
• B. 3
• C. 2
• D. 4

Answer 1

Answer: (C)

2

$I=2amp$

$t=5hr\times 3600Sec/hr$
$W=22.2g$ of tin

$\text{Moles of electrons}=\frac{2\times 5\times 3600}{96500}=0.37$

$\text{Moles of tin produced}=\frac{22.2}{119}=0.186$

$\therefore \text{Moles of tin}\times n=\text{moles of electrons}$

$0.186\times n=0.37$

$\therefore n=2$

$\therefore \text{Oxidation state of tin}=2$

Hence, option C is correct.