Question

If $10g$ of $V_2{O}_5$ is dissolved in acid and is reduced to $V^{2+}$ by zinc metal, how many mole of $I_2$ could be reduced by the resulting solution if it is further oxidised to $V{O}^{2+}$ ions?
[Assume no change in state of $Z{n}^{2+}$ ions] $(V=51,O=16,I=127)$ :

• A. $0.11$ mole of $I_2$
• B. $0.22$ mole of $I_2$
• C. $0.055$ mole of $I_2$
• D. $0.44$ mole of $I_2$

Answer 1

Answer: (A)

$\underset{+5}{V_2{O}_5}+6e^{-}\to \underset{+2}{2V^{2+}}$
Molecular mass of $V_2{O}_5=51\times 2+5\times 16=182$
1 mole of $V_2{O}_5$ produces $2$ moles of $V^{2+}$
$\therefore \frac{10}{182}$ moles of $V_2{O}_5$ produce $2\times \frac{10}{182}$
$=0.11$ moles of $V^{2+}$
$\underset{+2}{V^{2+}}\to \underset{0}{V{O}^{2+}}+2e^{-}$ $\dots (1)$
$\underset{0}{I_2}+2e^{-}\to \underset{-1}{2I^{-}}$ $\dots (2)$
On adding (1) and (2)
$V^{2+}+I_2\to 2I^{-}+V{O}^{2+}$
$1$ mole of $V^{2+}$ reduces $1$ mole of $I_2$
$\therefore 0.11$ moles of $V^{2+}$ reduce $0.11$ moles of $I_2$