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9476 / 2027
The Periodic Table

Lesson 5 of 6

Group 2 and Group 17 chemistry

Use electrode potentials and explain thermal stability.

A-Level 9476 (2026-2027)

  • I can use standard potentials to compare reactivity.Syllabus

    Syllabus 9476, 5(f). Use standard potentials to compare reactivity.

  • I can explain thermal-stability trends.Syllabus

    Syllabus 9476, 5(g). Explain thermal-stability trends.

Read reduction potentials in their written direction

Group 2 metals reduce other species; halogens oxidise them.

For M2+(aq) + 2e- ⇌ M(s), a more negative standard reduction potential means the metal is more readily oxidised in a suitable paired reaction. Group 2 metals are reducing agents because they lose electrons. The standard potentials become generally more negative from Mg towards Ba, supporting greater reducing power down the group.

For X2 + 2e- ⇌ 2X-(aq), a more positive standard reduction potential means the halogen is a stronger oxidising agent. The usual values are approximately +1.36 V for chlorine, +1.07 V for bromine and +0.54 V for iodine, with the standard physical states specified in the data. Oxidising strength therefore decreases Cl2 > Br2 > I2.

Worked examplePredict a displacement using E degrees

Use +1.36 V for Cl2/Cl- and +1.07 V for Br2/Br- to assess Cl2 + 2Br- → 2Cl- + Br2.

Try it first. Then show the worked stepsHide the worked steps
  1. Chlorine is reduced; bromide is oxidised.
  2. Ecell = E(reduction at cathode) - E(reduction for the reverse oxidation pair) = 1.36 - 1.07.
  3. Ecell = +0.29 V, so the written reaction is thermodynamically favourable under standard conditions.
Answer

Chlorine can oxidise bromide to bromine. Reversing the reaction gives -0.29 V and is not favourable under those standard conditions.

Check your understanding

E for Cl2/Cl- is +1.36 V, for Br2/Br- +1.07 V and for I2/I- +0.54 V. Which halogen is the strongest oxidising agent?

  1. Bromine, as its E is in the middle
  2. Iodine, as it has the least positive E
  3. Chlorine, as it has the most positive E
Show the answer

Chlorine, as it has the most positive E

A more positive reduction potential means the species gains electrons more readily, so it is a stronger oxidising agent.

Two thermal trends need two different explanations

Carbonates depend on cation polarisation; hydrogen halides depend on their covalent bond strength.

Make a guess. You are not marked.

Magnesium carbonate and barium carbonate are heated. Which needs the higher temperature to decompose?

  1. Barium carbonate, as Ba2+ polarises it less
  2. Magnesium carbonate, as Mg2+ holds it tighter
  3. Both at the same temperature, as both are Group 2
Show the answer

Barium carbonate, as Ba2+ polarises it less

Ba2+ is larger than Mg2+, so it has a lower charge density and polarises the carbonate ion less. The carbonate stays intact to a higher temperature.

Group 2 carbonates decompose on heating: MCO3(s) → MO(s) + CO2(g). Down Mg to Ba, the cation radius increases while its charge remains +2. Charge density and polarising power decrease. The larger cation distorts the large carbonate ion less, so the carbonate is harder to decompose and its thermal stability increases.

A complete carbonate explanation
  1. Same charge, larger cation down the group

    Mg2+ is smaller than Ba2+, so Mg2+ has greater charge density.

  2. Polarise the large anion

    The smaller cation distorts the carbonate electron cloud more strongly, destabilising the carbonate relative to decomposition products.

  3. Predict the heating result

    Magnesium carbonate decomposes more readily; barium carbonate requires stronger heating.

For hydrogen halides, the relevant reaction is 2HX(g) ⇌ H2(g) + X2(g). From HCl to HBr to HI, the halogen atom becomes larger, the H-X bond becomes longer and the bond energy decreases. Thermal stability therefore decreases in that order: HI decomposes most readily.

Choose the mechanism that matches the substance
Trend down the groupKey reasoningDirection of stability
MgCO3 to BaCO3Lower cation charge density and less polarisation of carbonate.Increases
HCl to HILonger, weaker H-X covalent bond.Decreases

Check your understanding

Iodine is less volatile than chlorine, but HI is less thermally stable than HCl. Is that a contradiction?

  1. No. Both depend on the I-I bond, which is strong in I2 but broken in HI.
  2. No. Volatility is about forces between X2 molecules; stability is about the H-X bond.
  3. Yes. Stronger attractions in iodine should make HI more stable as well, not less.
Show the answer

No. Volatility is about forces between X2 molecules; stability is about the H-X bond.

These are different changes that involve different interactions. Stronger forces between I2 molecules lower volatility; a weaker H-I bond lowers thermal stability.