When a sample of hydrated lithium ethanedioate, Li₂C₂O₄•H₂O, is gently heated, two gaseous products are formed and a white solid residue remains. The residue is added to HNO₃(aq). A gas is produced that turns limewater milky. Complete the equation for the decomposition of Li₂C₂O₄•H₂O. Li₂C₂O₄•H₂O → +
- (a)When a sample of hydrated lithium ethanedioate, Li₂C₂O₄•H₂O, is gently heated, two gaseous products are formed and a white solid residue remains. The residue is added to HNO₃(aq). A gas is produced that turns limewater milky. Complete the equation for the decomposition of Li₂C₂O₄•H₂O. Li₂C₂O₄•H₂O → +[1]
- (b)The trend in the decomposition temperatures of the Group 2 ethanedioates is similar to that of the Group 2 nitrates. Suggest which of CaC₂O₄ and BaC₂O₄ will decompose at the lower temperature. Explain your answer.[2]
- (c)Potassium iron(III) ethanedioate, K₃[Fe(C₂O₄)₃], dissolves in water to form a green solution. Explain why transition elements can form coloured complexes.[3]
- (d)The anhydrous iron(III) compound K₃[Fe(C₂O₄)₃] decomposes on heating to form a mixture of K₂[Fe(C₂O₄)₂], K₂C₂O₄ and CO₂. Complete the equation for the decomposition of K₃[Fe(C₂O₄)₃]. K₃[Fe(C₂O₄)₃] → K₂[Fe(C₂O₄)₂] + K₂C₂O₄ + CO₂[1]
- (e)The [Fe(C₂O₄)₃]³⁻ complex ion shows stereoisomerism. Complete the three-dimensional diagrams in Fig. 3.1 to show the two stereoisomers of [Fe(C₂O₄)₃]³⁻. The C₂O₄²⁻ ligand can be represented using O O. [Figure 3.1][2]
- (f)Buffer solutions are used to regulate pH. Write two equations to describe how a solution containing HC₂O₄⁻ ions acts as a buffer solution when small amounts of acid or alkali are added.[2]
- (g)A fuel cell is an electrochemical cell that can be used to generate electrical energy by using oxygen to oxidise a fuel. Ethanedioic acid, (COOH)₂, dissolved in an alkaline electrolyte is being investigated as a fuel. The relevant standard electrode potentials, Eº, for the cell are shown. O₂(g) + 2H₂O(l) + 4e⁻ ⇌ 4OH⁻(aq) Eº = +0.40V 2CO₂(g) + 2e⁻ ⇌ C₂O₄²⁻(aq) Eº = −0.59 V Use these equations to deduce the overall cell reaction. Calculate the value of E°cell. overall cell reaction E°cell = V[2]
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Mark scheme
Marking points from the official mark scheme (May/June 2024). The mark or mark type is shown in brackets.
- [1] Li₂C₂O₄•H₂O → Li₂CO₃ + H₂O + CO
- [1] M1 CaC₂O₄ AND Ca²⁺ has a smaller ionic radius / Ca²⁺ has a higher charge density
- [1] M2 anion / C₂O₄²⁻ becomes more polarised / distorted
- [1] M1 d orbital(s) of different energy / d-d splitting occurs
- [1] M2 electron(s) promoted / excited
- [1] M3 light is absorbed AND colour seen is complementary
- [1] 2K₃Fe(C₂O₄)₃ → 2K₂Fe(C₂O₄)₂ + K₂C₂O₄ + 2CO₂
- [2] [Diagrams showing two enantiomers of tris(oxalato)ferrate(III) ion]
- [1] M1 HC₂O₄⁻ + H⁺ → H₂C₂O₄
- [1] M2 HC₂O₄⁻ + OH⁻ → C₂O₄²⁻ + H₂O
- [1] M1 2C₂O₄²⁻ + O₂(g) + 2H₂O(l) → 4CO₂(g) + 4OH⁻(aq)
- [1] M2 Ecell = 0.40 – (−0.59) = 0.99 V
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