Summary — What This Chapter Covers

• Vanadium chemistry: four oxidation states (+2 to +5), each with its own colour, reduced step-by-step by zinc.
• Chromium chemistry: oxidation states +2, +3, +6; the famous yellow ⇌ orange chromate/dichromate equilibrium.
• Ions in aqueous solution: how metal-aqua ions react with hydroxide and ammonia — precipitates, ligand substitution, and amphoteric behaviour.
• Catalysts: heterogeneous (different phase, surface adsorption theory) vs homogeneous (same phase, intermediate species).
• The Contact Process: V₂O₅ as a heterogeneous catalyst that changes oxidation state mid-reaction.
• Catalytic converters: Pt/Rh catalysts converting car exhaust pollutants.
• Homogeneous catalysis: Fe²⁺/Fe³⁺ catalysing the iodide–peroxodisulfate reaction.
• Autocatalysis: Mn²⁺ speeding up its own formation in the manganate(VII)–oxalate reaction.

1. Vanadium Chemistry

Vanadium is unusual because it comfortably exists in four different oxidation states, and — brilliantly for students trying to remember it — every single one has a different, vivid colour. Think of it like a traffic light system, except it has four "lights" instead of three, and they cycle from yellow all the way down to purple as vanadium keeps gaining electrons.

Why does zinc do this?

Zinc metal is a strong reducing agent — it's very willing to lose its own electrons (Zn → Zn²⁺ + 2e⁻), and those electrons have to go somewhere. In acidic conditions, vanadium(V) grabs them one pair at a time, stepping down through +4, +3, and finally +2. Each step is its own separate redox reaction, and each one has a colour change you could watch happen in a test tube — genuinely one of the most visually striking reactions in A Level chemistry.

Using standard electrode potentials (E°) to predict reactions

This is the technique the exam is really testing — not memorising vanadium chemistry, but knowing how to use E° values to work out which reaction happens. Here's the golden rule, stated as simply as possible:

Worked example — reducing V⁵⁺ to V⁴⁺ using zinc:

1. Zn²⁺(aq) + 2e⁻ ⇌ Zn(s)   E° = −0.76 V
2. VO₂⁺(aq) + 2H⁺(aq) + e⁻ ⇌ VO²⁺(aq) + H₂O(l)   E° = +1.00 V

The vanadium equation has the more positive E°, so it runs forward (reduction). The zinc equation is reversed (oxidation): Zn(s) ⇌ Zn²⁺(aq) + 2e⁻. To balance electrons, double the vanadium equation (×2 gives 2 electrons to match zinc's 2 electrons):

2. Chromium Chemistry

Chromium follows the exact same E° logic as vanadium, but it's tested in two different contexts: redox reactions (chromium changing oxidation state, using zinc or H₂O₂) and a completely separate acid-base equilibrium between chromate and dichromate ions that students very commonly mix up with a redox reaction.

Reduction of Cr(VI) to Cr(III) by zinc — acidic conditions

Dichromate(VI) has the most positive E°, so it's reduced. Zinc is reversed and oxidised. Balancing electrons (dichromate needs 6e⁻, zinc gives 2e⁻ each, so triple the zinc equation):

Oxidation of Cr(III) to Cr(VI) by H₂O₂ — alkaline conditions

This time chromium is going the other way — being oxidised from +3 up to +6. This only happens in alkaline conditions (notice OH⁻ appears in the equation, not H⁺). H₂O₂ has the more positive E° here, so it's reduced; the chromium equation is reversed:

The Chromate ⇌ Dichromate Equilibrium (NOT a redox reaction!)

This is the part everyone confuses. Chromate (CrO₄²⁻, yellow) and dichromate (Cr₂O₇²⁻, orange) can interconvert:

Look closely: chromium's oxidation number is +6 on both sides. No electrons are transferred, no oxidation state changes — this is a plain acid-base equilibrium (Le Chatelier's principle applies, not electrode potentials), even though the colour change makes it look exactly like a redox reaction.

3. Ions in Aqueous Solution

Every transition metal ion in water isn't actually "naked" — it's surrounded by six water molecules acting as ligands, forming an octahedral complex like [Fe(H₂O)₆]²⁺. When you add a base like NaOH or NH₃, two very different things can happen depending on how much you add, and this is the single most-tested idea in this section.

Step 1: Limited OH⁻ or limited NH₃ → deprotonation, not substitution

It looks like hydroxide ions are swapping in for water ligands, but what's actually happening is more subtle: the OH⁻ (or NH₃, acting as a base) is pulling a H⁺ ion off two of the water ligands, turning them into hydroxide ligands still attached to the metal. This forms an insoluble, neutral precipitate.

Step 2: Excess OH⁻ (only for Cr³⁺, and to a lesser extent Zn²⁺) → amphoteric behaviour

Chromium(III) hydroxide is special: it can act as both an acid and a base — this is called amphoteric behaviour.

Step 3: Excess NH₃ → true ligand substitution

This time it's genuinely different. With Cu²⁺, Co²⁺, and Cr³⁺, excess ammonia actually swaps out water ligands for ammonia ligands (NH₃ molecules are similar in size to H₂O and uncharged, so they fit the same "slots").

Summary table — what you actually need memorised

4. Catalysts

Heterogeneous Catalysis

A heterogeneous catalyst is in a different physical state from the reactants — almost always a solid catalyst working on gaseous or dissolved reactants. Picture it like a crowded dance floor (the reactant gas) with a few designated "meeting spots" on the wall (the catalyst's active sites) where partners can pair up more easily than by randomly bumping into each other.

Strength of adsorption matters: too weak (e.g. silver) and reactants barely stick, so concentration at the surface stays low. Too strong (e.g. tungsten) and the products get "stuck" and can't desorb, blocking the active site. Metals like nickel and platinum hit the sweet spot — the "Goldilocks zone" of adsorption strength — which is why they're the go-to industrial catalysts.

The Contact Process (manufacture of sulfuric acid)

What makes this a beautiful exam example is that the catalyst's oxidation state actually changes mid-cycle and then changes back — proof that variable oxidation states aren't just a vanadium party trick, they're genuinely useful chemistry:

Catalytic Converters

Finely divided platinum and rhodium, spread over a ceramic honeycomb structure to maximise surface area (these metals are expensive, so you want maximum active-site coverage per gram used). The same three-step adsorption theory applies:

Homogeneous Catalysis

Here the catalyst is in the same phase as the reactants — usually everything dissolved in aqueous solution. The defining feature is that the catalyst forms a genuine intermediate species with its own formula, rather than just providing a surface.

Classic example — Fe²⁺ catalysing the reaction between iodide and peroxodisulfate ions. Without a catalyst, this reaction is slow because both I⁻ and S₂O₈²⁻ are negatively charged and repel each other, so successful collisions are rare:

Adding Fe²⁺ opens up a faster two-step pathway, because now a negative ion is reacting with a positive ion (much more favourable collisions) in each step:

Notice the elegant symmetry: Fe²⁺ gets oxidised in step 1 and then reduced right back in step 2, ending the cycle exactly where it started — that's what makes it a catalyst rather than just a reactant.

5. Autocatalysis

Autocatalysis is a special case of homogeneous catalysis where a product of the reaction catalyses that same reaction as it forms. This creates a very distinctive rate graph shape: the reaction actually speeds up partway through, before eventually slowing as reactants run out — the opposite of what a normal rate graph does at the start.

The Mn²⁺ produced doesn't just sit there — it becomes the catalyst for the rest of the reaction, cycling between +2 and +3 oxidation states:

At the very start of the reaction there's no Mn²⁺ around yet, so the reaction crawls along slowly and uncatalysed. But as soon as a little bit of Mn²⁺ is produced, it kicks off this catalytic cycle and the reaction accelerates dramatically — hence the "slow start, fast middle" shape on a concentration-time graph.

What to Memorise

Concepts Checklist

Exam Tips — Common Mistakes & What Examiners Look For

What examiners actually reward in mark schemes

• Correctly balanced electrons before combining half-equations — this is checked carefully.
• State symbols and correct use of ⇌ vs → (equilibrium vs one-way reaction).
• Explicitly stating oxidation number changes when asked to justify whether something is redox.
• For catalyst questions: mentioning "alternative reaction pathway" and "lower activation energy" — vague answers like "it speeds up the reaction" without this reasoning lose marks.
• For heterogeneous catalysis: using the specific vocabulary — adsorption, active site, desorption — rather than generic descriptions.