A-LevelChemistryElectrochemistry9701/42 · Feb/Mar 2023 · Q217 Marks

Hypophosphorous acid is an inorganic acid. The conjugate base of hypophosphorous acid is H2PO2⁻.

(a)Give the formula of hypophosphorous acid.[1]
(b)(i)In an experiment, an alkaline HPO3²⁻/H2PO2⁻ half-cell is constructed with [H2PO2⁻] = 0.050 mol dm⁻³. All other ions are at their standard concentration. Predict how the value of E of this half-cell differs from its E° value. Explain your answer.[2]
(b)(ii)The Cr³⁺/Cr half-cell has a standard electrode potential of −0.74 V. An electrochemical cell consists of an alkaline HPO3²⁻/H2PO2⁻ half-cell and a Cr³⁺/Cr half-cell. Calculate the standard cell potential, E°cell. E°cell = .............................................. V[1]
(b)(iii)Complete the diagram in Fig. 2.1 to show how the standard electrode potential of the Cr³⁺/Cr half-cell can be measured relative to that of the standard hydrogen electrode. Identify the chemicals, conditions and relevant pieces of apparatus. [Figure 2.1][3]
(b)(iv)Label Fig. 2.1 to show: • which is the positive electrode • the direction of electron flow in the external circuit.[1]
(b)(v)H2PO2⁻ reduces Ni²⁺ to Ni in alkaline conditions. Use equation 1 to construct the ionic equation for this reaction. equation 1 HPO3²⁻ + 2H2O + 2e⁻ ⇌ H2PO2⁻ + 3OH⁻[1]
(c)(i)The volume of H2 was measured under room conditions. Use the molar volume of gas, Vm, and the data from experiment 1 to calculate the rate of reaction in mol dm⁻³ s⁻¹. rate of reaction = .............................................. mol dm⁻³ s⁻¹[1]
(c)(ii)The rate equation was found to be: rate = k [H2PO2⁻(aq)] [OH⁻(aq)]² Show that the data in Table 2.1 is consistent with the rate equation.[2]
(c)(iii)State the units of the rate constant, k, for the reaction.[1]
(c)(iv)The experiment is repeated using a large excess of OH⁻(aq). Under these conditions, the rate equation is: rate = k₁ [H2PO2⁻(aq)] k₁ = 8.25 × 10⁻⁵ s⁻¹ Calculate the value of the half-life, t½, of the reaction. t½ = .............................................. s[1]
(c)(v)Describe how an increase in temperature affects the value of the rate constant, k₁.[1]
(d)A student suggests that the reaction between H2PO2⁻(aq) and OH⁻(aq) might happen more quickly in the presence of a heterogeneous catalyst. Describe the mode of action of a heterogeneous catalyst.[2]

This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.

Mark scheme

Marking points from the official mark scheme (Feb/Mar 2023). The mark or mark type is shown in brackets.

[1] H3PO2
[1] electrode potential E would become more positive / less negative (than E°)
[1] lower [H2PO2⁻] AND shifts equilibrium to the right-hand side
[1] +1.57 -0.74 = (+)0.83 (V)
[3] Pt(s)/C/graphite • Cr(s) • H+(aq)/any strong acid • Cr3+(aq) • H2(g) • voltmeter/V • salt bridge labelled conditions of 1 atm AND 1 mol dm-3/1M/1mol/dm3 • other liquid level (salt bridge) and wire to electrode any three [1] any six [2] all nine [3]
[1] Pt electrode positive AND flow of electrons anticlockwise (to the SHE)
[1] H2PO2⁻ + 3OH⁻ + Ni2⁺ → HPO3²⁻ + 2H2O + Ni
[1] (6.4 / 24000) ÷ 60 = 4.44 × 10⁻⁶ (mol dm⁻³ s⁻¹) min 2sf
[1] [H2PO2] doubles /×2 from experiments 1 to 2 volume of H2 produced doubles / ×2 (: first order wrt [H2PO2]) ORA
[1] [H2PO2] × 3 and [OH-] × ½ from experiments 1 to 3 volume of H2 produced falls to ¼ original (if first order wrt [H2PO2¯] then must be second order wrt [OH-]) ORA
[2] ALLOW input data into rate equation and show k is the same k =2.8 × 10⁻⁶ / k = 6.7 × 10⁻² (1/15) / k = 4 for all experiments [2]
[1] mol⁻² dm⁶ s⁻¹
[1] t½ = 0.693/8.25 × 10⁻⁵ = 8400 (s) OR t½ = In2 / 8.25 × 10⁻⁵ = 8401.8 (s)
[1] (k₁) increases (with temperature)
[1] reactants adsorb (to surface of catalyst)
[1] bonds (in reactant) weaken
[1] (reaction occurs and the) products are desorbed OR reaction occurs and substances are desorbed any two [1] all three [2]