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Write down the half equations and overall balanced equation for the reaction between oxalate ions and permanganate ions in acidic solution, and calculate the concentration of KMnO4 solution.

Chemistry

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Part b(i)

Write down the half equation for

A: Oxidation:

C2O42−(aq)→2CO2(g)+2e−\text{C}_2\text{O}_4^{2-}(\text{aq}) \rightarrow 2\text{CO}_2(\text{g}) + 2\text{e}^-

B: Reduction:

MnO4−(aq)+8H+(aq)+5e−→Mn2+(aq)+4H2O(l)\text{MnO}_4^{-}(\text{aq}) + 8\text{H}^{+}(\text{aq}) + 5\text{e}^- \rightarrow \text{Mn}^{2+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l})

Part b(ii)· 2 steps

Write down the overall balanced equation

  1. Multiply the oxidation half-equation by 5 and the reduction half-equation by 2 to balance the electrons.

    5C2O42−(aq)→10CO2(g)+10e−5\text{C}_2\text{O}_4^{2-}(\text{aq}) \rightarrow 10\text{CO}_2(\text{g}) + 10\text{e}^- 2MnO4−(aq)+16H+(aq)+10e−→2Mn2+(aq)+8H2O(l)2\text{MnO}_4^{-}(\text{aq}) + 16\text{H}^{+}(\text{aq}) + 10\text{e}^- \rightarrow 2\text{Mn}^{2+}(\text{aq}) + 8\text{H}_2\text{O}(\text{l})

  2. Add the two half-equations and cancel out the electrons.

    5C2O42−(aq)+2MnO4−(aq)+16H+(aq)→10CO2(g)+2Mn2+(aq)+8H2O(l)\boxed{5\text{C}_2\text{O}_4^{2-}(\text{aq}) + 2\text{MnO}_4^{-}(\text{aq}) + 16\text{H}^{+}(\text{aq}) \rightarrow 10\text{CO}_2(\text{g}) + 2\text{Mn}^{2+}(\text{aq}) + 8\text{H}_2\text{O}(\text{l})}

Part b(iii)· 3 steps

Calculate the concentration of the KMnO4_4 solution.

  1. Calculate the moles of oxalic acid (H2C2O4\text{H}_2\text{C}_2\text{O}_4).

    Volume of oxalic acid = 50 cm3=0.050 dm350 \text{ cm}^3 = 0.050 \text{ dm}^3.
    Concentration of oxalic acid = 0.25 M0.25 \text{ M}.

    Moles of C2O42−=Concentration×Volume=0.25 mol/dm3×0.050 dm3=0.0125 mol\text{Moles of C}_2\text{O}_4^{2-} = \text{Concentration} \times \text{Volume} = 0.25 \text{ mol/dm}^3 \times 0.050 \text{ dm}^3 = 0.0125 \text{ mol}

  2. Use the mole ratio from the balanced equation to find the moles of MnO4−\text{MnO}_4^{-}.

    From the balanced equation, 2 moles of MnO4−2 \text{ moles of MnO}_4^{-} react with 5 moles of C2O42−5 \text{ moles of C}_2\text{O}_4^{2-}.

    Moles of MnO4−=0.0125 mol C2O42−×2 mol MnO4−5 mol C2O42−=0.005 mol\text{Moles of MnO}_4^{-} = 0.0125 \text{ mol C}_2\text{O}_4^{2-} \times \frac{2 \text{ mol MnO}_4^{-}}{5 \text{ mol C}_2\text{O}_4^{2-}} = 0.005 \text{ mol}

  3. Calculate the concentration of the KMnO4\text{KMnO}_4 solution.

    Volume of KMnO4\text{KMnO}_4 solution = 31.0 cm3=0.0310 dm331.0 \text{ cm}^3 = 0.0310 \text{ dm}^3.

    Concentration of KMnO4=Moles of MnO4−Volume of KMnO4=0.005 mol0.0310 dm3≈0.16129 M\text{Concentration of KMnO}_4 = \frac{\text{Moles of MnO}_4^{-}}{\text{Volume of KMnO}_4} = \frac{0.005 \text{ mol}}{0.0310 \text{ dm}^3} \approx 0.16129 \text{ M}

    Rounding to three significant figures:

    0.161 M\boxed{0.161 \text{ M}}

Part c(i)

What do you understand by the term bond energy?

Bond energy can provide information about the energy changes that accompany a chemical reaction.

Bond energy is the amount of energy required to break one mole of a specific type of covalent bond in the gaseous state. It is also the energy released when one mole of that same bond is formed.

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