Question

$2.68\times {10}^{-3}$ moles of a solution containing an ion $A^{n+}$ require $1.61\times {10}^{-3}$ moles $Mn{O}_4^{-}$ for the oxidation of $A^{n+}$ to $A{O}_3^{-}$ in acid medium. What is the value of n?

• A. 2
• B. 4
• C. 3
• D. 1

Answer 1

The correct option is A 2

$Step1:\text{Write reduction and oxidation half.}$

$\text{Reduction:}Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O...\left(1\right)$
$\text{Oxidation:}{A}^{n+}+3H_{2}O\to A{O}_3^{-}+6H^{+}+(5-n)e^{-}...\left(2\right)$

$Step2:\text{To find out value of n.}$

$\because$ in a redox reaction
$\text{number of electrons lost = number of electrons gained}$
$\therefore \text{multiply oxidant of equation (1) by (5 - n) and reductant of equation (2) by 5}$
$(5-n)Mn{O}_4^{-}=5A^{n+}i.e,\text{(5 - n) moles of}Mn{O}_4^{-}\text{will oxidise 5 moles of}{A}^{n+}$
$\therefore 1.61\times {10}^{-3}\text{moles of}Mn{O}_4^{-}\text{will oxidise}{A}^{n+}=\frac{5}{(5-n)}\times 1.61\times {10}^{-3}\text{moles}...\left(3\right)$
$\text{But the moles of}{A}^{n+}\text{actually oxidised}=2.68\times {10}^{-3}\text{moles}...\left(4\right)$
On equating equation (3) and (4)
$\frac{5}{(5-n)}\times 1.61\times {10}^{-3}=2.68\times {10}^{-3}$
$n=\frac{5.335}{2.68}=2$