3.1.11 Electrode Potentials

Standard electrode potentials are tabulated using reduction half-equations, with electrons on the left. Keep that convention when you copy data, even if one half-equation will run in reverse in the operating cell.

This lets you compare values consistently. Decide the actual oxidation/reduction direction only after you have written both tabulated reduction couples.

For conventional cell notation for a spontaneous cell, put the oxidation half-cell on the left and the reduction half-cell on the right. Use | for a phase boundary and || for the salt bridge.

Species in the same phase are separated by commas. Add an inert Pt electrode when a half-cell contains no conducting solid that can provide an electrode surface.

Think of the standard hydrogen electrode as the zero point of the electrode-potential scale: E° is defined as 0.00 V. Every other standard electrode potential is measured relative to that reference.

Under standard conditions the hydrogen electrode uses H₂ at 100 kPa, H⁺ at 1.00 mol dm⁻³ and a platinum electrode at 298 K.

Electrode potential depends on the position of the half-cell equilibrium, so temperature, gas pressure and ion concentration can change it. E° refers specifically to standard conditions.

AQA does not require the Nernst equation here. You do need to understand why a potential measured away from 298 K, 100 kPa or 1.00 mol dm⁻³ need not equal the tabulated E° value.

Those conditions define a standard electrode potential: 298 K, 100 kPa for gases and 1.00 mol dm⁻³ for aqueous ions. If the conditions differ, label the value E. The superscript ° is reserved for standard conditions.

Write both reduction half-equations before deciding the cell direction. That makes the signs, electron flow and E°cell calculation much less error-prone.

When you read a table of standard electrode potentials, a more positive E° means a greater tendency for the reduction half-equation to proceed as written. That makes the species on the left of a very positive couple a relatively strong oxidising agent under standard conditions.

Conversely, a species on the right of a very negative reduction couple is readily oxidised and can act as a strong reducing agent.

The half-equation with the more positive E° has the greater tendency to proceed as reduction. Reverse the other half-equation for oxidation, then combine them and check E°cell is positive for the proposed direction.

With tabulated reduction potentials, a convenient check is E°cell = E°(more positive reduction) − E°(less positive reduction). A positive result supports thermodynamic feasibility for the direction you have written under standard conditions.

Use tabulated reduction potentials consistently: E°cell = E°(reduction) − E°(oxidation couple, still quoted as its tabulated reduction potential).

I find cell-potential calculations safest if you leave the data-booklet half-equations as reductions, identify which one actually runs backwards, and then calculate the potential difference. Never multiply E° when you scale a half-equation.

Cell notation compresses the chemical cell into one line. Put the oxidation half-cell on the left, reduction half-cell on the right, | at phase boundaries and || at the salt bridge.

When you interpret notation, expand each side back into its half-equation and identify where electrons are produced and consumed. That is safer than trying to infer the chemistry from punctuation alone.

Trace what moves in an electrochemical cell: electrons go through the external circuit while ions move through the salt bridge to maintain electrical neutrality. Do not send the electrons through the salt bridge.

A spontaneous redox reaction separated into two half-cells forces electrons through the external circuit, allowing chemical free energy to be delivered as electrical work.

For the simplified lithium cell, Li is oxidised: Li → Li⁺ + e⁻. At the other electrode, Li⁺ + CoO₂ + e⁻ → Li⁺[CoO₂]⁻ consumes the electron.

Add the supplied half-equations to obtain the overall discharge chemistry. Keep the simplified course equations as given. A more detailed commercial-cell mechanism is outside the chemistry being tested here.

Separate the three cell types by what happens after discharge: a primary cell is essentially used once, a rechargeable cell can be driven electrically back towards its charged state, and a fuel cell keeps working while fresh reactants are supplied from outside.

Those are different operating ideas: rechargeability depends on reversibility of the cell chemistry, while a fuel cell keeps running because fresh fuel and oxidant continue to arrive.

A fuel cell keeps supplying current because fuel and oxidant are fed in continuously. You do not recharge it by electrically forcing the discharge reaction backwards; fresh reactants arrive and products are removed.

For a hydrogen–oxygen fuel cell, the overall chemical product is water. The practical evaluation still has to include how the hydrogen is produced, stored and transported.

For the alkaline H₂/O₂ fuel-cell half-equations, keep OH⁻ and H₂O in the species as required by the alkaline conditions. Balance the electrons before you add the two half-equations.

When the half-equations are combined, electrons cancel and the overall reaction simplifies to hydrogen reacting with oxygen to form water.

When you evaluate a battery or fuel cell, judge the actual electrochemical system, not the label. Look at things such as energy density, lifetime, rechargeability, raw materials, toxicity, fire risk, infrastructure and end-of-life handling.

A strong answer links each claimed advantage or disadvantage to the chemistry or engineering feature that causes it.

Use the supplied half-equations or electrode data to identify the discharge direction first. In a rechargeable cell, an external power supply drives the electrode reactions in the reverse direction during charging.

Keep discharge and recharge equations separate on the page. Reversing one half-cell but not the other produces an impossible mixed description of the cell.

If one potential or Ecell is missing, use Ecell = E(reduction) − E(oxidation couple) algebraically with the values quoted as reduction potentials.

Do not change the sign of a tabulated E° just because you have reversed its half-equation in your working. The subtraction in the cell-emf equation already handles the operating direction.

Trace the electron flow: oxidation releases electrons at one electrode and reduction consumes them at the other. Those electrons travel through the external circuit while ions move through the electrolyte or salt bridge to maintain charge balance.

Do not send electrons through the salt bridge. Its job is ionic conduction and charge balance between the half-cells.