5.2.3 Redox and Electrode Potentials

An oxidising agent accepts electrons. Because it gains electrons, the oxidising agent is itself reduced.

When you are choosing one from a redox equation, look for the species whose oxidation state falls or whose reduction half-equation runs forwards.

A reducing agent donates electrons. It loses electrons, so the reducing agent is itself oxidised.

If you have half-equations in front of you, the reducing agent is the species on the right-hand side of the reduction half-equation that you reverse for oxidation.

For an unfamiliar half-equation in acidic solution, I use the same order every time: balance the main atom, balance O with H₂O, balance H with H⁺, then balance charge with e⁻.

Finish by checking both atoms and charge. If either side fails, the half-equation is not balanced.

To combine half-equations, make the number of electrons lost equal the number gained, reverse the oxidation half-equation if necessary, then add and cancel.

Multiplying a half-equation changes its coefficients, not the chemistry of the redox couple. Your final equation should have no electrons left and must balance for both atoms and charge.

Track the oxidation numbers or the electrons. A species is oxidised if its oxidation number increases and reduced if it decreases.

For unfamiliar reactions, write the likely oxidation and reduction half-equations first. They usually make the products and electron transfer much easier to see than trying to guess the whole equation at once.

Pipette the Fe²⁺ solution into a conical flask, add excess dilute sulfuric acid, then titrate with potassium manganate(VII) from the burette while swirling.

Manganate(VII) is its own indicator. Near the end point add it dropwise until the first permanent pale pink colour remains.

Titrate iodine with sodium thiosulfate until the iodine colour has become pale, then add starch. The remaining iodine gives a blue-black colour.

Continue adding thiosulfate dropwise until the blue-black colour just disappears. Add the starch near the end point, when only a small amount of iodine remains.

Use a volumetric pipette for the fixed portion and a burette for the titrant. Rinse each with the solution it will contain, make sure the burette jet is full, remove the filling funnel, and swirl the flask as you titrate.

Use a rough titre to find the end point, then repeat accurately and average concordant titres. Record burette readings to 2 decimal places, with the final digit 0 or 5.

Keep the calculation in chemical steps: mean titre → moles of titrant → mole ratio from the balanced redox equation → moles of the substance you want → concentration or mass.

I label the species beside each mole value. That small habit stops a page of numbers becoming hard to follow when the redox ratio is 1:5 or 1:2.

For an unfamiliar titration, do not hunt for one big formula. Write the equations that connect the measured titrant to the substance being analysed, then move through the mole ratios one reaction at a time.

If the sample was diluted or only a portion was titrated, deal with that scale factor explicitly. I’d write something like “25.0 cm³ portion → 250 cm³ original solution: ×10” so the reason for the number is visible.

For the definition, I’d learn this accurately: the standard electrode potential, E°, is the potential of a half-cell measured relative to the standard hydrogen electrode under standard conditions.

At A level those conditions are 298 K, aqueous concentrations of 1.00 mol dm⁻³ and gas pressures of 100 kPa. The standard hydrogen electrode is assigned E° = 0.00 V.

Connect the half-cell to a standard hydrogen electrode through a salt bridge and a high-resistance voltmeter. The voltmeter measures the potential difference between the two half-cells.

The hydrogen reference uses Pt in contact with H₂(g) and H⁺(aq) under standard conditions. Because its E° is defined as 0.00 V, the measured standard cell potential gives the E° of the other half-cell, with the sign set by the polarity.

Draw the two half-cells separately and connect them in two ways: a wire and high-resistance voltmeter for electrons, and a salt bridge for ions. Label the electrode, solution and any gas in each half-cell.

For a metal/metal-ion system the metal can be the electrode. For a gas or a redox pair with no conducting solid, use an inert conductor such as Pt.

If you are measuring a standard cell potential, use standard conditions. For two ions of the same element, OCR also accepts equal concentrations of the two ions.

• two half-cells
• high-resistance voltmeter
• salt bridge
• correct electrode for each redox system
• standard conditions if E° is being measured

If both members of the redox pair are aqueous, there is no conducting solid to carry electrons into or out of the half-cell. Use a platinum electrode as the conducting surface for the electron-transfer reaction.

Platinum is used because it conducts and is inert under the conditions: it takes part in the electrical connection without being consumed in the redox equation.

For a cell producing current spontaneously, the half-cell with the more positive E° runs as reduction and is the positive electrode. The more negative E° half-cell runs in reverse as oxidation and is the negative electrode.

Electrons then flow through the external circuit from the negative electrode to the positive electrode. Keep that electron path separate from the ions moving through the salt bridge.

When both values are written as standard reduction potentials, the quickest route is E°cell = E°(more positive) − E°(more negative).

For Cu²⁺/Cu, +0.34 V, and Zn²⁺/Zn, −0.76 V: E°cell = 0.34 − (−0.76) = +1.10 V.

Choose the reduction with the more positive E° and reverse the other half-equation for oxidation. If the resulting E°cell is positive, the reaction as written is thermodynamically feasible under standard conditions.

When you explain the choice, name the redox couples and the species that are actually reduced and oxidised. “Higher” and “lower” can become unclear when negative values are involved; more positive and more negative are safer.

A positive E°cell tells you about the thermodynamic direction, not how fast the reaction will happen. A reaction can still be extremely slow if it has a large activation energy.

So if the E° values predict reaction but nothing obvious happens, kinetics may be the reason. The electrode-potential calculation has not measured the reaction rate.

E° applies only to standard conditions. Change a concentration or gas pressure and the half-cell equilibrium shifts, so the actual electrode potential, E, changes and the measured cell voltage can change as well.

Use the half-equation to reason out the direction of the shift. The E° value itself stays as the standard reference value.

For OCR, use the storage-cell equations and E° data supplied in the question. During discharge, the more positive reduction occurs at the positive electrode and the other half-reaction runs as oxidation, producing a voltage.

Recharging uses an external power supply to drive the electrode reactions in the opposite directions and restore the chemical state of the cell. OCR supplies the details and equations for named storage cells; your job is to apply the half-equation and E° reasoning to them.

A fuel cell separates the oxidation of the fuel from the reduction of oxygen. Electrons released at the fuel electrode travel through the external circuit before they are accepted at the oxygen electrode, so the redox reaction produces a voltage and current.

Fuel and oxygen must keep being supplied while the cell operates. For a hydrogen–oxygen fuel cell, the overall chemical product is water.

Read the electrolyte first. In acidic conditions, H⁺ can appear in the half-equations; in alkaline conditions, OH⁻ and H₂O are used instead. Then make the electron numbers equal, add the half-equations and cancel anything that appears on both sides.

OCR supplies the fuel-cell equations you need. Use them carefully, then check that the combined equation has the right atoms, charge and complete cancellation of electrons.