Questions on the electrochemical series and predicting feasibility

Question 1

Explain the observations in the three test-tube experiments described in the first table, using the E° values given in the second table, and write an equation for each reaction that takes place.

Accepted Answer

Indicative content

in experiment 1 chlorine is reduced and bromide is oxidised, and E°cell = 1.36 − 1.09 = +0.27 V, so the reaction is feasible
Cl₂(aq) + 2Br⁻(aq) → 2Cl⁻(aq) + Br₂(aq); the orange-yellow colour is the bromine formed in the solution
in experiment 2 iron(III) is reduced and iodide is oxidised, and E°cell = 0.77 − 0.54 = +0.23 V, so the reaction is feasible
2Fe³⁺(aq) + 2I⁻(aq) → 2Fe²⁺(aq) + I₂(aq), with the iron half-equation doubled and its E° not doubled; the brown colour is iodine
in experiment 3, for Fe³⁺ to oxidise Br⁻, E°cell = 0.77 − 1.09 = −0.32 V, which is negative, so no reaction is feasible and nothing is seen
the oxidising agents rank Cl₂ > Br₂ > Fe³⁺ > I₂, and each oxidises only the reduced forms of the couples with less positive E° values

Levels

Level 3 (5–6 marks) all three experiments are explained with the correct E°cell for each, the two feasible reactions have balanced equations with the colour observed attributed to the product, the third is shown to be not feasible from its negative E°cell, and the order of oxidising power is drawn from the data
Level 2 (3–4 marks) two experiments are explained with E°cell and an equation, or all three are taken on with one E°cell or one equation wrong or absent
Level 1 (1–2 marks) one reaction is said to happen, with a colour noted; E° values are quoted and not combined, or the more negative value is taken as the stronger oxidising agent

Question 2

Calculate E°cell for the reaction of acidified dichromate(VI) ions with iron(II) ions, and write the overall ionic equation. E°(Cr₂O₇²⁻/Cr³⁺) = +1.33 V; E°(Fe³⁺/Fe²⁺) = +0.77 V.

Accepted Answer

M1 dichromate(VI) is reduced, so its half-cell is on the right: E°cell = 1.33 − 0.77 B1 the dichromate half-equation takes six electrons, so six Fe²⁺ ions are oxidised for each Cr₂O₇²⁻ B1 Cr₂O₇²⁻(aq) + 14H⁺(aq) + 6Fe²⁺(aq) → 2Cr³⁺(aq) + 7H₂O(l) + 6Fe³⁺(aq); ignore state symbols A1 E°cell = +0.56 V

Question 3

Explain why aluminium shows little reaction with dilute hydrochloric acid at room temperature, although E°(Al³⁺/Al) = −1.66 V predicts that it should react, and give one other reason why a reaction predicted from E° values may not be observed.

Accepted Answer

B1 E°(Al³⁺/Al) is more negative than E°(H⁺/H₂) = 0.00 V, so the reduction of H⁺ by aluminium is feasible, E°cell = +1.66 V B1 the aluminium is covered by a thin, unreactive layer of aluminium oxide that keeps the acid off the metal B1 so the rate is negligible; E° predicts whether a reaction can happen, not how fast it goes B1 the conditions may not be standard, so the actual electrode potentials differ from the E° values used; allow a high activation energy for a different reaction

Question 4

Explain the effect on the EMF of the cell Cu(s) | Cu²⁺(aq) || Ag⁺(aq) | Ag(s), set up with standard solutions, of (a) adding solid potassium chloride to the silver half-cell, and (b) adding water to the copper half-cell.

Accepted Answer

B1 (a) chloride precipitates silver chloride, so [Ag⁺] falls and Ag⁺(aq) + e⁻ ⇌ Ag(s) shifts to the left B1 (a) the silver electrode, on the right, becomes less positive, so the EMF falls B1 (b) [Cu²⁺] falls, so Cu²⁺(aq) + 2e⁻ ⇌ Cu(s) shifts to the left and the copper electrode becomes less positive B1 (b) copper is the left-hand electrode, so E right − E left increases and the EMF rises

Question 5

Deduce which of magnesium, iron, lead and silver reduce tin(II) ions to tin under standard conditions, using the excerpt of the electrochemical series, and comment on the reliability of the prediction for lead.

Accepted Answer

B1 Sn²⁺ is reduced, so the tin half-cell is on the right and E°cell = −0.14 − E°(metal) B1 magnesium, +2.23 V, and iron, +0.30 V, give positive values, so both reduce Sn²⁺ B1 silver gives −0.94 V and does not reduce Sn²⁺ B1 lead gives −0.01 V, not feasible, but so close to zero that a small change in concentration could reverse the prediction

Question 6

Evaluate the conclusion of a student who calculates from E° values that acidified dichromate(VI) does not oxidise chloride ions, and concludes that dichromate(VI) can never be used to make chlorine. E°(Cr₂O₇²⁻/Cr³⁺) = +1.33 V; E°(Cl₂/Cl⁻) = +1.36 V.

Accepted Answer

B1 E°cell = 1.33 − 1.36 = −0.03 V, so the reaction is not feasible under standard conditions and the calculation is correct B1 the value is very close to zero, so a small change in conditions can reverse the prediction B1 with concentrated hydrochloric acid and heat, higher [H⁺] and [Cl⁻] make the dichromate electrode more positive and the chlorine electrode less positive B1 chlorine is then released, so 'never' is not justified; the prediction applies only under standard conditions

Question 7

Identify the strongest oxidising agent and the strongest reducing agent among the species in the half-equations in the table, and state what is meant by the term oxidising agent.

Accepted Answer

B1 Br₂ is the strongest oxidising agent, as the oxidised form with the most positive E°; reject Br⁻ B1 Zn is the strongest reducing agent, as the reduced form with the most negative E°; reject Zn²⁺ B1 an oxidising agent is a species that accepts electrons and is itself reduced

Question 8

Calculate E°cell for the reaction Sn²⁺(aq) + 2Fe³⁺(aq) → Sn⁴⁺(aq) + 2Fe²⁺(aq), and state whether it is feasible under standard conditions. E°(Sn⁴⁺/Sn²⁺) = +0.15 V; E°(Fe³⁺/Fe²⁺) = +0.77 V.

Accepted Answer

M1 iron(III) is reduced, so the iron half-cell is on the right: E°cell = 0.77 − 0.15 A1 E°cell = +0.62 V B1 E°cell is positive, so the reaction is feasible under standard conditions

Question 9

Calculate E°cell for 2Ag⁺(aq) + Cu(s) → 2Ag(s) + Cu²⁺(aq). A student gives 2 × 0.80 − 0.34 = +1.26 V. E°(Ag⁺/Ag) = +0.80 V; E°(Cu²⁺/Cu) = +0.34 V.

Accepted Answer

M1 silver ions are reduced, so the silver half-cell is on the right: E°cell = 0.80 − 0.34 M1 the E° of silver is not doubled when its half-equation is doubled, so the student's value is wrong A1 E°cell = +0.46 V

Question 10

Explain, using E° values, why manganate(VII) titrations are acidified with sulfuric acid and not hydrochloric acid. E°(MnO₄⁻/Mn²⁺) = +1.51 V; E°(Cl₂/Cl⁻) = +1.36 V.

Accepted Answer

B1 the manganate(VII) E° is more positive, so MnO₄⁻ oxidises Cl⁻ to Cl₂, with E°cell = +0.15 V B1 some manganate(VII) would then react with chloride ions as well as with the substance being analysed B1 so the titre would be too large; sulfate ions are not oxidised by manganate(VII)

Question 11

Explain the effect on the electrode potential of a Cu²⁺/Cu half-cell of increasing the concentration of Cu²⁺ ions above 1.00 mol dm⁻³.

Accepted Answer

B1 the half-cell equilibrium is Cu²⁺(aq) + 2e⁻ ⇌ Cu(s) B1 increasing [Cu²⁺] shifts the position of equilibrium to the right, so the electrode takes electrons more readily B1 the electrode potential becomes more positive than +0.34 V

Question 12

Explain why gold is found in nature as the metal, whereas sodium is found only in compounds. E°(Au³⁺/Au) = +1.50 V; E°(Na⁺/Na) = −2.71 V; E°(O₂/H₂O) = +1.23 V.

Accepted Answer

B1 the very negative E° of sodium means sodium is a strong reducing agent, readily oxidised to Na⁺ by oxygen and water B1 the very positive E° of gold means Au³⁺ is readily reduced and gold atoms are very hard to oxidise B1 oxygen's E° of +1.23 V is less positive than gold's +1.50 V, so oxidising gold with oxygen is not feasible and gold stays as the metal

Question 13

Suggest, using E° values, what is seen when white copper(I) sulfate is added to water. Cu⁺(aq) + e⁻ ⇌ Cu(s), E° = +0.52 V; Cu²⁺(aq) + e⁻ ⇌ Cu⁺(aq), E° = +0.15 V.

Accepted Answer

B1 Cu⁺ is reduced in the first half-equation and oxidised in the reverse of the second, E°cell = 0.52 − 0.15 = +0.37 V, so the reaction is feasible B1 2Cu⁺(aq) → Cu(s) + Cu²⁺(aq), a disproportionation B1 a brown or pink solid of copper forms in a blue solution of copper(II) ions

Question 14

Suggest why pale green iron(II) sulfate solution left open to the air slowly turns yellow-brown, and write the equation. O₂(g) + 4H⁺(aq) + 4e⁻ ⇌ 2H₂O(l), E° = +1.23 V; Fe³⁺(aq) + e⁻ ⇌ Fe²⁺(aq), E° = +0.77 V.

Accepted Answer

B1 the oxygen half-equation is more positive, so dissolved oxygen oxidises Fe²⁺ to Fe³⁺, E°cell = +0.46 V B1 4Fe²⁺(aq) + O₂(g) + 4H⁺(aq) → 4Fe³⁺(aq) + 2H₂O(l) B1 Fe³⁺ ions are yellow-brown, and the colour builds up slowly because the reaction is slow at room temperature

Question 15

Suggest why the EMF of a zinc–copper cell falls as it delivers a current, until no current flows.

Accepted Answer

B1 [Cu²⁺] falls, so Cu²⁺(aq) + 2e⁻ ⇌ Cu(s) shifts left and the copper electrode becomes less positive B1 [Zn²⁺] rises, so Zn²⁺(aq) + 2e⁻ ⇌ Zn(s) shifts right and the zinc electrode becomes less negative B1 the difference between the two potentials falls, and is zero when the cell reaction reaches equilibrium

Question 16

Describe how the anticlockwise rule predicts the reaction between the Zn²⁺/Zn half-cell, E° = −0.76 V, and the Cu²⁺/Cu half-cell, E° = +0.34 V.

Accepted Answer

B1 the half-equations are written as reductions with the more negative, Zn²⁺/Zn, above the more positive, Cu²⁺/Cu B1 an anticlockwise loop runs forwards along the lower equation, Cu²⁺ to Cu, and backwards along the upper, Zn to Zn²⁺ B1 so Cu²⁺ oxidises Zn: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

Question 17

Compare zinc, iron and copper as reducing agents, using E°(Zn²⁺/Zn) = −0.76 V, E°(Fe²⁺/Fe) = −0.44 V and E°(Cu²⁺/Cu) = +0.34 V, and say which of them react with H⁺ ions in a dilute acid.

Accepted Answer

B1 the more negative the E°, the stronger the reducing agent, so zinc is strongest, then iron, then copper B1 zinc and iron have E° values below 0.00 V, so each reduces H⁺ to hydrogen, E°cell +0.76 V and +0.44 V B1 copper's E° is above 0.00 V, so its reaction with H⁺ would have E°cell = −0.34 V and is not feasible

Question 18

Calculate the electrode potential at 298 K of a silver half-cell in which [Ag⁺] = 0.010 mol dm⁻³, and the EMF of the cell it makes with a standard copper half-cell. Use the Nernst equation, E = E° + (0.059/z) log₁₀[Ag⁺]. E°(Ag⁺/Ag) = +0.80 V; E°(Cu²⁺/Cu) = +0.34 V.

Accepted Answer

M1 z = 1, so E = 0.80 + 0.059 × log₁₀(0.010) = 0.80 + 0.059 × (−2) A1 E(Ag⁺/Ag) = +0.68 V A1 the silver electrode is still the more positive, so EMF = 0.68 − 0.34 = +0.34 V; allow error carried forward

Question 19

Calculate ΔG°, in kJ mol⁻¹, for the reaction Cl₂(g) + 2Br⁻(aq) → 2Cl⁻(aq) + Br₂(aq), for which E°cell = +0.27 V. F = 96 500 C mol⁻¹.

Accepted Answer

M1 two electrons are transferred in the equation as written, so n = 2 M1 ΔG° = −nFE°cell = −2 × 96 500 × 0.27 = −52 110 J mol⁻¹ A1 ΔG° = −52 kJ mol⁻¹, to the two significant figures of E°cell; allow −52.1; a positive value is rejected

Question 20

Explain why doubling the equation for a cell reaction doubles its ΔG° but leaves its E°cell unchanged.

Accepted Answer

B1 ΔG° = −nFE°cell, and ΔG° refers to the amounts in the equation as written B1 doubling the equation doubles n, the number of moles of electrons transferred, so ΔG° doubles B1 E°cell is energy per unit of charge, a property of the two couples that does not depend on how many electrons are written, so it is unchanged

Question 21

Explain why a reaction with E°cell = +0.80 V goes essentially to completion, whereas one with E°cell = +0.02 V gives an equilibrium mixture.

Accepted Answer

B1 ln K, and the total entropy change, are proportional to E°cell B1 a large positive E°cell gives a very large K, so at equilibrium almost no reactant remains B1 an E°cell close to zero gives K close to 1, so reactants and products are both present in significant amounts at equilibrium

Experiment	Observation
1: chlorine water added to potassium bromide solution	colourless solution turns orange-yellow
2: iron(III) chloride solution added to potassium iodide solution	solution turns brown
3: iron(III) chloride solution added to potassium bromide solution	no change

Half-equation	E° / V
Cl₂(aq) + 2e⁻ ⇌ 2Cl⁻(aq)	+1.36
Br₂(aq) + 2e⁻ ⇌ 2Br⁻(aq)	+1.09
Fe³⁺(aq) + e⁻ ⇌ Fe²⁺(aq)	+0.77
I₂(aq) + 2e⁻ ⇌ 2I⁻(aq)	+0.54

Half-equation	E° / V
Zn²⁺(aq) + 2e⁻ ⇌ Zn(s)	−0.76
Fe²⁺(aq) + 2e⁻ ⇌ Fe(s)	−0.44
Ag⁺(aq) + e⁻ ⇌ Ag(s)	+0.80
Br₂(aq) + 2e⁻ ⇌ 2Br⁻(aq)	+1.09