Mission 1: How can you measure the voltage of half a battery, when a voltmeter needs two connections?

You need from the core lesson

• Oxidation (electron loss) happens at the anode; reduction (electron gain) at the cathode.
• Voltaic cell: a spontaneous reaction produces electrical energy. Anode negative.
• Electrolytic cell: a DC supply drives a non-spontaneous reaction. Anode positive.
• Half-equations balance atoms and charge; electrons lost = electrons gained.

Only differences can be measured

• A voltmeter needs two electrodes, so it measures a potential difference between two half-cells.
• No single half-cell's potential can be measured on its own.
• So one half-cell is chosen as the reference and given the value 0 V.

The standard hydrogen electrode

• Standard hydrogen electrode (SHE): H₂ gas at 100 kPa bubbling over platinum, in 1 mol dm⁻³ H⁺(aq), at 298 K.
• Half-equation: H⁺(aq) + e⁻ ⇌ ½H₂(g). E⦵ = 0.00 V by convention.
• The platinum is inert: it carries electrons but does not react.
• Electron transfer still happens on the platinum. Zero is an agreed value, not "no reaction".

Measuring E⦵ for nickel

• Standard conditions for every half-cell: 298 K, 100 kPa for gases, 1 mol dm⁻³ for ions.
• Connect a Ni²⁺/Ni half-cell to the SHE with a salt bridge and a high-resistance voltmeter.
• The reading is 0.26 V, and nickel is the negative electrode: electrons leave it.
• So E⦵(Ni²⁺/Ni) = −0.26 V (data booklet).
• Negative electrode against the SHE → negative E⦵. Positive electrode → positive E⦵.

Reading a table of E⦵ values

• E⦵ values are standard reduction potentials: each half-equation is written as a reduction.
• More positive E⦵: the species on the left is more easily reduced, a stronger oxidizing agent.
• More negative E⦵: the species on the right is more easily oxidized, a stronger reducing agent.
• Compare signed values, not sizes: −2.93 V is the most negative value here, not the "largest".
• The values you need are in the data booklet.

Mission 2: Why does a lithium coin cell give 3 V when an alkaline AA cell gives 1.5 V?

Cell potential from two E⦵ values

• E⦵cell = E⦵(cathode) − E⦵(anode). The cathode is where reduction happens.
• In a voltaic cell, the half-cell with the more positive E⦵ is the cathode.
• Worked example: Ni(s) + Br₂(l) → Ni²⁺(aq) + 2Br⁻(aq).
• Br₂ is reduced (cathode, +1.09 V); Ni is oxidized (anode, −0.26 V).
• E⦵cell = +1.09 − (−0.26) = +1.35 V.
• Never multiply E⦵ by the coefficients: ½Br₂ or Br₂, the value is +1.09 V.

Spontaneous or not?

• For a given equation: find the species reduced. Its E⦵ is E⦵(cathode).
• E⦵cell positive: spontaneous as written. Negative: the reverse is spontaneous.
• Pb²⁺(aq) + Ni(s) → Pb(s) + Ni²⁺(aq): −0.13 − (−0.26) = +0.13 V. Spontaneous.
• I₂(s) + 2Br⁻(aq) → 2I⁻(aq) + Br₂(l): +0.54 − (+1.09) = −0.55 V. Not spontaneous.
• Its reverse is spontaneous: bromine oxidizes iodide ions.

From E⦵cell to ΔG⦵

• ΔG⦵ = −nFE⦵cell. The equation and F = 96 500 C mol⁻¹ are in the data booklet.
• n = moles of electrons transferred in the balanced overall equation.
• C mol⁻¹ × V = J mol⁻¹. Divide by 1000 for kJ mol⁻¹.
• Worked example: Ni + Br₂, E⦵cell = +1.35 V, n = 2.
• ΔG⦵ = −2 × 96 500 × 1.35 = −260 550 J mol⁻¹ = −261 kJ mol⁻¹.
• E⦵cell positive ↔ ΔG⦵ negative ↔ spontaneous.

Finding n, and working backwards

• Scale the half-equations until electrons lost = electrons gained. That number is n.
• 2MnO₄⁻ + 16H⁺ + 10I⁻ → 2Mn²⁺ + 5I₂ + 8H₂O: two MnO₄⁻ each gain 5e⁻, so n = 10.
• Count the electrons once: the same electrons are lost and gained, so do not add the two half-equations' electrons.
• Working backwards: E⦵cell = −ΔG⦵ ÷ nF, with ΔG⦵ in J mol⁻¹.

Mission 3: Electrolyse salt water: why do you get hydrogen at the negative electrode, not sodium?

Water joins the competition

• In aqueous solution, water can be reduced or oxidized too.
• At the cathode, water or the cation is reduced: 2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq) E⦵ = −0.83 V
• At the anode, water or the anion is oxidized: 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻ (reverse of the +1.23 V couple)
• Only one reaction usually wins at each electrode.

Picking the winner from E⦵

• Cathode: the species whose couple has the more positive E⦵ is reduced.
• Anode: the species whose couple has the less positive E⦵ is oxidized.
• Worked example, KI(aq), inert electrodes:
• Cathode: K⁺ (−2.93 V) or water (−0.83 V)? Water → H₂.
• Anode: I⁻ (+0.54 V) or water (+1.23 V)? Iodide → I₂: 2I⁻ → I₂ + 2e⁻.
• Inert electrodes (graphite, platinum) do not react themselves.

Electrolysing water

• Pure water has very few ions, so a little dilute H₂SO₄ is added to carry the current.
• Cathode: 2H⁺(aq) + 2e⁻ → H₂(g). Anode: 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻.
• Sulfate ions are not oxidized. The acid is not used up.
• Overall: 2H₂O(l) → 2H₂(g) + O₂(g), so the gas volumes are H₂ : O₂ = 2 : 1.

NaCl(aq): concentration decides the anode product

• Cathode, at any concentration: water is reduced, not Na⁺ (−2.71 V). H₂ forms.
• Dilute NaCl(aq), inert anode: mainly O₂, as E⦵ predicts (+1.23 V is below +1.36 V for Cl₂/Cl⁻).
• Concentrated NaCl(aq): mainly Cl₂: 2Cl⁻(aq) → Cl₂(g) + 2e⁻.
• E⦵ values apply at 1 mol dm⁻³. In concentrated brine the high [Cl⁻] means Cl⁻ is oxidized in preference to water.
• With Cl⁻ removed and OH⁻ made, concentrated brine leaves NaOH(aq).

Mission 4: When a key is copper-plated in blue copper sulfate solution, why doesn't the blue fade?

CuSO₄(aq) with inert electrodes

• Cathode: Cu²⁺ (+0.34 V) beats water (−0.83 V). Copper is deposited: Cu²⁺(aq) + 2e⁻ → Cu(s).
• Anode: sulfate is not oxidized; water is. 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻.
• Cu²⁺ is used up and H⁺ builds up: the blue fades and the solution turns acidic.

CuSO₄(aq) with copper electrodes

• A copper anode is oxidized before water: Cu(s) → Cu²⁺(aq) + 2e⁻.
• Its couple (+0.34 V) is far less positive than O₂/H₂O (+1.23 V). No oxygen forms.
• The anode gets thinner; the cathode gains the same mass of copper.
• Cu²⁺ is made as fast as it is used, so [Cu²⁺] and the blue colour stay constant.
• The electrode material can change the anode product.

Electroplating

• Electroplating uses electrolysis to coat an object with a thin layer of a metal.
• The object is the cathode (negative): metal ions are reduced onto it.
• The anode is usually a bar of the plating metal; it dissolves.
• The electrolyte contains ions of the plating metal.
• Nickel plating: cathode Ni²⁺(aq) + 2e⁻ → Ni(s); anode Ni(s) → Ni²⁺(aq) + 2e⁻.

The big ideas

• E⦵ is measured against the standard hydrogen electrode, set at 0 V by convention.
• More positive E⦵: stronger oxidizing agent. More negative: stronger reducing agent.
• E⦵cell = E⦵(cathode) − E⦵(anode); positive means spontaneous as written.
• ΔG⦵ = −nFE⦵cell, with n from the balanced overall equation.
• In aqueous electrolysis water competes; E⦵, concentration (NaCl) and electrode material (CuSO₄) decide the products.
• Electroplating: object as cathode, plating-metal anode, plating-metal ions in solution.