Question

If 10 g of $V_2O_5$ is dissolved in acid and is reduced to $V^{2+}$ by zinc metal, how many mole $I_2$ 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)

0.11 mole of $I_2$

Answer 2

1. Moles of Vanadium:

   • Molar mass of $V_2O_5 = 2 \times 51 + 5 \times 16 = 102 + 80 = 182 \text{ g/mol}$.
   • Moles of $V_2O_5 = \frac{10 \text{ g}}{182 \text{ g/mol}} \approx 0.0549 \text{ mol}$.
   • Since each mole of $V_2O_5$ contains 2 moles of V atoms, moles of V atoms present = $2 \times 0.0549 = 0.1098 \text{ mol}$.

2. First Reduction ($V_2O_5 \rightarrow V^{2+}$):

   • In $V_2O_5$, the oxidation state of V is +5.
   • It is reduced to $V^{2+}$.
   • So, the $0.1098 \text{ mol}$ of V atoms initially at +5 are converted to $0.1098 \text{ mol}$ of $V^{2+}$ ions.

3. Second Oxidation ($V^{2+} \rightarrow VO^{2+}$) and Reduction of $I_2$:

   • The $V^{2+}$ ions are now oxidized to $VO^{2+}$ ions.
   • In $VO^{2+}$, the oxidation state of V is +4 (since O is -2, $x + (-2) = +2 \Rightarrow x = +4$).
   • The change in oxidation state for V is from +2 to +4. This means each V atom loses 2 electrons.
   • Total moles of electrons lost by $V^{2+}$ = Moles of $V^{2+} \times \text{change in O.S.}$ per V atom $= 0.1098 \text{ mol} \times 2 = 0.2196 \text{ mol electrons}$.
   • These electrons are used to reduce $I_2$. The reduction of $I_2$ to $I^-$ is $I_2 + 2e^- \rightarrow 2I^-$.
   • Each mole of $I_2$ accepts 2 moles of electrons.
   • Moles of $I_2$ reduced = $\frac{\text{Total moles of electrons lost by V}}{\text{Electrons accepted per mole of } I_2} = \frac{0.2196 \text{ mol}}{2} = 0.1098 \text{ mol}$.
   • Rounding to two decimal places, this is approximately 0.11 mole.