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A-Level Redox, electrode potentials and electrolysis
What the A-Level syllabus expects for Redox, electrode potentials and electrolysis, and how to practise it.
What the syllabus expects
- Describe and account for a redox change through electron transfer or through a shift in oxidation number.
- Define these terms:
- standard electrode (redox) potential
- standard cell potential
- Describe how the standard hydrogen electrode is set up.
- Describe how standard electrode potentials are measured for:
- a metal or a non-metal in contact with its ions in aqueous solution
- ions of one element in two different oxidation states
- Find a standard cell potential by putting two standard electrode potentials together.
- Use standard cell potentials to:
- explain or deduce which way electrons flow in a simple cell
- predict whether a reaction is spontaneous
- Appreciate where standard cell potentials fall short as predictors of spontaneity.
- Assemble redox equations from the appropriate half-equations.
Scope: see also Section 13 - Quote and apply the relationship delta G-standard = minus n F E-standard to an electrochemical cell, including finding E-standard for combined half-reactions.
- Predict qualitatively how an electrode potential shifts as the aqueous ion's concentration changes.
- Set out the appeal of developing newer cells, among them improved batteries for electric vehicles and the H2/O2 fuel cell, on grounds of lower mass, smaller size and a higher voltage.
- Quote the relationship F = Le, which ties the electron's charge to the Avogadro constant and to the Faraday constant.
- Predict what substance electrolysis will set free by weighing the electrolyte's state, molten or aqueous, where the species sit in the redox series, and their concentration.
- Calculate:
- the quantity of charge that passes during electrolysis
- the mass or the volume of substance set free during electrolysis
- Using each electrode reaction, account for these industrial processes:
- anodising aluminium
- purifying copper electrolytically
Scope: the technical details are not required
How it's examined
Questions on this topic most often ask you to calculate, determine, explain, suggest. About 8% of the past-paper style questions in Rae's bank for this subject sit in this topic.
Worked examples
Example 1 (3 marks)
A breathalyser built as an electrochemical cell oxidises the ethanol in a person's breath, producing a current of 0.20 A that flows for 45 s. Work out the mass of alcohol contained in the exhaled breath.
Show the worked answer
Charge passed Q = It = 0.20 x 45 = 9.0 C. Moles of electrons = Q/F = 9.0 / 96500 = 9.33e-5 mol. In the fuel cell ethanol is oxidised to ethanoic acid, transferring 4 electrons per molecule (CH3CH2OH + H2O -> CH3COOH + 4H+ + 4e-), so moles of ethanol = 9.33e-5 / 4 = 2.33e-5 mol. Mass = 2.33e-5 x 46.0 = 1.07e-3 g (about 1.1 mg).
Example 2 (4 marks)
Aluminium, the crust's most abundant metal, can be anodised once its oxide layer is removed. (i) Complete Table 2.1 with the half-equations for anodising an aluminium object; the anode half-equation takes the form ...Al + ...H2O -> ...Al2O3 + ...H+ + ...e, and give the cathode half-equation too. [2] (ii) If 3.50 g of the protective aluminium oxide layer forms over 2 hours, calculate the current used. [2]
Show the worked answer
(i) Anode (oxidation, forming the oxide layer), balancing Al, O, H and charge: 2Al + 3H2O -> Al2O3 + 6H+ + 6e- (check: Al 2=2, O 3=3, H 6=6, charge 0 = +6-6 = 0) Cathode (reduction of H+ to hydrogen): 2H+ + 2e- -> H2 (ii) M(Al2O3) = 2(27.0) + 3(16.0) = 102 g mol⁻¹. Moles Al2O3 = 3.50 / 102 = 0.03431 mol. Each Al2O3 requires 6 electrons (from the anode equation): moles e- = 6 x 0.03431 = 0.2059 mol. Charge Q = n x F = 0.2059 x 96500 = 1.987 x 10⁴ C. Time t = 2 h = 7200 s. Current I = Q / t = 19870 / 7200 = 2.76 A.
Example 3 (2 marks)
Work out the mass of ethane, C2H6, formed when a 2.0 A current passes through sodium ethanoate solution for 55 minutes.
Show the worked answer
Kolbe electrolysis of ethanoate: 2CH3COO- -> C2H6 + 2CO2 + 2e-, so 2 mol electrons give 1 mol C2H6. Charge Q = I x t = 2.0 A x (55 x 60) s = 2.0 x 3300 = 6600 C. Moles of electrons = Q/F = 6600/96500 = 0.0684 mol. Moles of C2H6 = 0.0684/2 = 0.0342 mol. Mr(C2H6) = 30.0. Mass = 0.0342 x 30.0 = 1.03 g.
More worked questions on this topic
- Predict, with explanation, how the electrode potential of Cd2+(aq)/Cd(s) changes when aqueous s (2 marks)
- Acidified KMnO4 and acidified K2Cr2O7 both serve as oxidants in organic chemistry. By referring (2 marks)
- In a small-scale trial the researchers passed a current of 0.150 A for 1200 seconds. Step 1 gav (2 marks)
- A dilute solution of potassium iodide is subjected to electrolysis using inert carbon (graphite (2 marks)
- In Kolbe electrolysis, carboxylate ions couple at the anode with loss of CO2: 2R-CO2- -> R-R + (2 marks)
- After being absorbed into the blood, ethanol is carried to the body's organs. Because the amoun (2 marks)
More A-Level H2 Chemistry topics
The make-up of the atom and how its electrons are arranged · How atoms bond and how that governs a substance's behaviour · Ideal gases and working with gas mixtures · Competing definitions of acids and bases · Trends in the elements across a period and down a group · The mole and reacting-quantity calculations · all of A-Level H2 Chemistry