A-Level AQA Chemistry Oxidation, Reduction & Redox Equations: The reaction between vanadium(IV

The reaction between vanadium(IV) ions and manganate(VII) ions in acidic solution can be represented by the equation $$5 ext{V}^{4+} + ext{MnO}_4^- + 8 ext{H}^+ \rightarrow 5 ext{V}^{5+} + ext{Mn}^{2+} + 4 ext{H}_2 ext{O}$$ What volume, in dm³, of 0.020 mol dm⁻³ KMnO₄ is needed to oxidise 0.10 mol of vanadium(IV) ions completely? - AQA - A-Level Chemistry - Question 21 - 2022 - Paper 3

Question 21

$The-reaction-between-vanadium(IV)-ions-and-manganate(VII)-ions-in-acidic-solution-can-be-represented-by-the-equation--$$5--ext{V}^{4+}-+--ext{MnO}_4^--+-8--ext{H}^+-\rightarrow-5--ext{V}^{5+}-+--ext{Mn}^{2+}-+-4--ext{H}_2-ext{O}$$--What-volume,-in-dm³,-of-0.020-mol-dm⁻³-KMnO₄-is-needed-to-oxidise-0.10-mol-of-vanadium(IV)-ions-completely?-AQA-A-Level Chemistry-Question 21-2022-Paper 3.png$

The reaction between vanadium(IV) ions and manganate(VII) ions in acidic solution can be represented by the equation $$5 ext{V}^{4+} + ext{MnO}_4^- + 8 ext{H}^+ ... show full transcript

Worked Solution & Example Answer:The reaction between vanadium(IV) ions and manganate(VII) ions in acidic solution can be represented by the equation $$5 ext{V}^{4+} + ext{MnO}_4^- + 8 ext{H}^+ \rightarrow 5 ext{V}^{5+} + ext{Mn}^{2+} + 4 ext{H}_2 ext{O}$$ What volume, in dm³, of 0.020 mol dm⁻³ KMnO₄ is needed to oxidise 0.10 mol of vanadium(IV) ions completely? - AQA - A-Level Chemistry - Question 21 - 2022 - Paper 3

Step 1

Calculate the moles of KMnO₄ needed

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Answer

From the balanced reaction, 5 moles of V(IV) react with 1 mole of KMnO₄. Therefore, the moles of KMnO₄ required for 0.10 moles of V(IV) is:

$ext{Moles of KMnO}_4 = \frac{0.10 ext{ moles of V(IV)}}{5} = 0.020 ext{ moles of KMnO}_4$

Step 2

Calculate the volume of KMnO₄ solution needed

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Answer

Using the concentration of the KMnO₄ solution (0.020 mol dm⁻³), we can find the volume required by the formula:

$ext{Volume (dm}^3 ext{)} = \frac{\text{Moles of solute}}{\text{Concentration}}$

Substituting the values gives:

$ext{Volume} = \frac{0.020 ext{ moles}}{0.020 ext{ mol dm}^{-3}} = 1.0 ext{ dm}^3$

Calculate the moles of KMnO₄ needed

Calculate the volume of KMnO₄ solution needed

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