May/June 2026 Paper 42

2026 · 9 questions · 56 parts · 100 marks

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1(a)(i)1(a)(ii)1(b)(i)1(b)(ii)1(c)1(d)(i)1(d)(ii)1(e)1(f)(i)1(f)(ii)1(g)1(h)

–/20 mks

2(a)2(b)2(c)2(d)(i)2(d)(ii)2(d)(iii)

–/10 mks

3(a)(i)3(a)(ii)3(a)(iii)3(b)3(c)

–/10 mks

4(a)4(b)4(c)

–/6 mks

5(a)5(b)(i)5(b)(ii)5(b)(iii)5(c)

–/11 mks

6(a)(i)6(a)(ii)6(b)6(c)(i)6(c)(ii)6(d)6(e)(i)6(e)(ii)

–/14 mks

7(a)(i)7(a)(ii)7(b)(i)7(b)(ii)7(c)7(d)(i)7(d)(ii)

–/11 mks

8(a)8(b)(i)8(b)(ii)8(b)(iii)8(c)

–/8 mks

9(a)9(b)(i)9(b)(ii)9(c)(i)9(c)(ii)

–/10 mks

5(a)Electrode Potentials Cell Free EnergyEasy2 marks

Define standard electrode potential,

E^{⊖}

, including a description of standard conditions.

Definition

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Standard conditions

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5(b)(i)Electrode Potentials Cell Free EnergyMedium3 marks

Table 5.1 shows the standard electrode potentials for three redox systems.

redox system	electrode reaction	E / V ⦵⦵
1	Fe $^{3 +}$ + e $^{-}$ Fe $^{2 +}$	+0.77
2	NO $_{3}$ – + 10H $^{+}$ + 8e $^{-}$ NH $_{4}$ + + 3H $_{2}$ O	+0.87
3	Zn $^{2 +}$ + 2e $^{-}$ Zn	–0.76

Use the information in Table 5.1 to write equations for two reactions that are feasible. Explain why these reactions are feasible.

Two equations

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Explanation

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5(b)(ii)Electrode Potentials Cell Free EnergyMedium1 mark

Suggest why the feasible reactions in 5(b)(i) may not take place under standard conditions.

Answer

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5(b)(iii)Electrode Potentials Cell Free EnergyHard3 marks

An electrochemical cell is set up to measure

E^{⊖}

for redox system 2 in Table 5.1 (NO

_{3}^{-}

+ 10H

^{+}

+ 8e

^{-} ⇌

_{4}^{+}

+ 3H

_{2}

E^{⊖} = + 0.87 V

Complete Figure 5.1 to show a labelled diagram of this electrochemical cell. Include all necessary substances and relevant pieces of apparatus needed to measure

E^{⊖}

. It is not necessary to state the conditions used.

Draw with the mouse or a finger.

5(c)Electrode Potentials Cell Free EnergyMedium2 marks

A different electrochemical cell is set up. The cell reaction for this electrochemical cell is shown.

C r + 3 M n^{3 +} \to 3 M n^{2 +} + C r^{3 +} Δ G^{⊖} = - 645.6 kJ mo l^{- 1}

Calculate the standard cell potential,

E_{ce l l}^{⊖}

, for this cell reaction.

E_{ce l l}^{⊖}