Chemistry 9701 · AS & A Level · Group 2

Group 2 — practice question

(a(i))[1]

When heated, lithium nitrate, $\text{LiNO}_3$, decomposes in the same pattern as Group 2 nitrates, giving the metal oxide, a brown gas, and oxygen. Write an equation for the decomposition of $\text{LiNO}_3$.

(a(ii))[2]

The other Group 1 nitrates, $\text{MNO}_3$, break down on heating to give the metal nitrite, $\text{MNO}_2$, and oxygen. Their thermal stability rises down the group. Suggest why the thermal stability of $\text{MNO}_3$ increases down the group.

(b)[3]

Acidified manganate(VII) ions, $\text{MnO}_4^{-}$, can be used in the analysis of nitrite-ion solutions, $\text{NO}_2^{-}$, by titration. X is a solution of $\text{NaNO}_2$. $250.0\,\text{cm}^3$ of X is mixed with $50.0\,\text{cm}^3$ of $0.125\,\text{mol dm}^{-3}$ acidified $\text{MnO}_4^{-}\text{(aq)}$. The $\text{MnO}_4^{-}\text{(aq)}$ ions are present in excess; every $\text{NO}_2^{-}$ ion is oxidised. The unreacted $\text{MnO}_4^{-}\text{(aq)}$ needed $22.50\,\text{cm}^3$ of $0.0400\,\text{mol dm}^{-3}$ $\text{Fe}^{2+}\text{(aq)}$ to reach the end-point. The relevant half-equations are: $\text{NO}_2^{-} + \text{H}_2\text{O} \rightleftharpoons \text{NO}_3^{-} + 2\text{H}^+ + 2e^{-}$; $\text{MnO}_4^{-} + 8\text{H}^+ + 5e^{-} \rightleftharpoons \text{Mn}^{2+} + 4\text{H}_2\text{O}$; $\text{Fe}^{2+} \rightleftharpoons \text{Fe}^{3+} + e^{-}$. Calculate the concentration, in $\text{mol dm}^{-3}$, of $\text{NaNO}_2$ in X.

(c(i))[2]

Aqueous manganate(VI) ions, $\text{MnO}_4^{2-}$, are unstable in acidic conditions and undergo disproportionation. The $E^\circ_{\text{cell}}$ for this reaction is $+1.14\,\text{V}$. Construct the overall ionic equation for this disproportionation reaction.

(c(ii))[1]

Suggest and explain how the $E_{\text{cell}}$ value for the disproportionation reaction changes when pH increases.

Worked solution & mark scheme

This 9-mark question has a full step-by-step worked solution and mark scheme. One marking point: Balanced equation is: $2LiNO_3 \rightarrow Li_2O + 2NO_2 + \tfrac{1}{2}O_2$

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