Part 2: EMF calculations, reaction predictions and their limits

Choose the reduction and oxidation reactions, cancel electrons and calculate the voltage without multiplying potentials.

Subtract reduction potentials once

For spontaneous discharge under standard conditions, choose the more positive reduction potential as the reduction half-reaction. Reverse the less positive half-equation to make oxidation. Subtract the tabulated reduction potential of the oxidised couple from that of the reduced couple.

Multiply half-equations to balance electrons when necessary, but never multiply an electrode potential by a stoichiometric coefficient. Potential is an intensive quantity; doubling the chemical equation doubles charge and energy together, leaving energy per charge unchanged.

Worked copper–silver cell

Use illustrative tabulated values Ag⁺/Ag = +0.80 V and Cu²⁺/Cu = +0.34 V. Silver ions are reduced and copper is oxidised. Double the silver half-equation to cancel the two electrons released by copper.

E°cell = 0.80 − 0.34 = +0.46 V. The equation has two Ag⁺ ions, but the silver potential remains +0.80 V. The corresponding discharge notation is Cu(s) | Cu²⁺(aq) || Ag⁺(aq) | Ag(s).

Test a proposed reaction, not just any pairing

If asked whether a particular reaction is feasible, calculate E°cell for that direction. Positive E°cell supports standard-state thermodynamic feasibility; a negative value favours the reverse standard-state direction. Zero is the equilibrium boundary. It does not tell you how quickly any change occurs.

For Fe³⁺/Fe²⁺ = +0.77 V and I₂/I⁻ = +0.54 V, Fe³⁺ can oxidise I⁻ under the standard-state comparison: 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂, with E°cell = +0.23 V. In the reverse direction the value is −0.23 V. Include states matching the supplied half-equations; aqueous iodine and solid iodine are not automatically the same standard couple.

Why the beaker may not match a simple prediction

Standard potentials apply to standard conditions. Different concentrations, gas pressures or temperatures change actual potentials, and competing reactions, complex formation or precipitation can alter the relevant chemistry. A small standard-potential difference is especially vulnerable to changed conditions.

Even with a positive actual cell potential, a high activation barrier can give an immeasurably slow reaction. An oxide coating may also passivate a surface. A catalyst can improve the current obtainable by speeding electrode reactions but does not change the equilibrium EMF under the same conditions. The Nernst equation is not required by this AQA section.

Explain concentration changes for a specified cell

For Zn + Cu²⁺ → Zn²⁺ + Cu, increasing [Cu²⁺] while holding the other conditions fixed increases the driving force for forward reaction and increases Ecell. Increasing [Zn²⁺] has the opposite effect. Use the actual overall reaction so the explanation remains tied to the cell.

Changing the amount of an existing pure solid electrode does not change its equilibrium potential, provided the required phase remains present. Surface area can affect available current and kinetics; it is not a general way to increase standard EMF.

Temperature also affects electrode potentials, but there is no universal rule that heating every cell lowers its EMF. The temperature dependence of equilibrium EMF relates to the reaction entropy. Use the supplied data or question-specific model rather than inferring voltage from the sign of enthalpy alone.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official AQA mark allocations.

Sources