Question

If $10 \,g$ 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 $VO^{2+}$ ions?
[Assume no change in state of $Zn^{2+}$ 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 $V^{2+}$
$\underset{+2}{V^{2+}} \to \underset{0}{VO^{2+}}+2e^{-}$ $\dots(1)$
$\underset{0}{I_{2}}+2e^{-} \to \underset{-1}{2I^{-}}$ $\dots(2)$
On adding (1) and (2)
$V^{2+}+I_{2} \to 2I^{-}+VO^{2+}$
$1$ mole of $V^{2+}$ reduces $1$ mole of $I_{2}$
$\therefore 0.11$ moles of $V^{2+}$ reduce $0.11$ moles of $I_{2}$