CAIE A-Level Chemistry 28.2.10 Calculations for Other Redox Systems

Practise transferring half-equation and mole-ratio methods to unfamiliar redox titrations supplied with suitable data.

Exam points

derive the electron and reactant mole ratio from balanced half-equations or the equation supplied
convert titre volume and concentration to moles, then scale for aliquots or dilution where required
use the resulting metal amount to calculate solution concentration, sample purity or percentage by mass

[Maximum number: 3]

An impure sample of a vanadium(V) compound of mass 0.250 g is dissolved in aqueous acid. This solution contains $\mathrm{VO}_{3}{ }^{-}$ ions.

An excess of zinc is added to this solution. All the $\mathrm{VO}_{3}{ }^{-}$ ions are reduced to $\mathrm{V}^{2+}$ ions and Zn atoms are oxidised to $\mathrm{Zn}^{2+}$ ions.

The unreacted zinc is removed and the resulting solution is titrated with acidified $\mathrm{MnO}_{4}^{-}$ .
The end-point is reached when $22.5 \mathrm{~cm}^{3}$ of $0.0750 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{MnO}_{4}^{-}$ is added.
A redox reaction takes place and all the $\mathrm{V}^{2+}$ reacts forming $\mathrm{VO}_{3}{ }^{-}$ .

3 \mathrm{MnO}_{4}^{-}+5 \mathrm{~V}^{2+}+3 \mathrm{H}_{2} \mathrm{O} \rightarrow 3 \mathrm{Mn}^{2+}+5 \mathrm{VO}_{3}^{-}+6 \mathrm{H}^{+}

Calculate the percentage by mass of vanadium in the 0.250 g of impure sample.

Assume the impurities do not contain any vanadium ions.
Show your working.
percentage of vanadium =

M1: moles $\mathrm{MnO}_{4}{ }^{-}=0.075 \times 0.0225=1.6875 \times 10^{-3}$

M2: moles $\mathrm{V}^{2+}=5 / 3 \times 1.6875 \times 10^{-3}=2.8125 \times 10^{-3}$

M3: mass $\mathrm{V}=50.9 \times 2.8125 \times 10^{-3}=0.143 \mathrm{~g}$

% of $\mathrm{V}=0.143 / 0.250 \times 100=57.3 \mathrm{~min} 2 \mathrm{sf}$

All question bank results loaded

28.2.10—Calculations: other redox systems question 1