The following reactions take place when nitric acid is added to zinc - Scottish Highers Chemistry - Question 4 - 2019

To determine the number of moles of Zn(s) oxidised by one mole of NO$_3^{-}$, we need to analyze the two half-reactions provided.

1. Oxidation Half-Reaction: The oxidation half-reaction shows that one mole of Zn(s) produces two moles of electrons: $\text{Zn}(s) \rightarrow \text{Zn}^{2+} (aq) + 2e^{-}$ Therefore, one mole of Zn(s) will release 2 moles of electrons.

2. Reduction Half-Reaction: The reduction half-reaction indicates that the nitrate ion (NO$_3^{-}$) requires 3 moles of electrons to be reduced to nitric oxide (NO): $\text{NO}_3^{-} (aq) + 4\text{H}^{+} (aq) + 3e^{-} \rightarrow \text{NO}(g) + 2\text{H}_2\text{O}(l)$

3. Mole Ratio Calculation: From the two half-reactions, we can see that:

   • 1 mole of NO$_3^{-}$ requires 3 moles of electrons.
   • 1 mole of Zn produces 2 moles of electrons.

   Now, if one mole of NO$_3^{-}$ requires 3 moles of electrons, then to find out how many moles of Zn are needed: $\text{Moles of Zn} = \frac{\text{Moles of electrons needed}}{\text{Moles of electrons produced by 1 mole of Zn}} = \frac{3}{2} = 1.5$

Thus, one mole of NO$_3^{-}$ oxidises 1.5 moles of Zn(s).