Ch 2 · Standard electrode potentials, standard cell potentials and the Nernst equation · AS & A Level Chemistry 9701
Topic 2 of 2 in Electrochemistry — AS & A Level Chemistry 9701 — 58 lessons.
Introduction
Every redox half-reaction has a tendency to occur as a reduction, measured by its standard electrode potential. Combining two half-cells lets that tendency be used to predict the direction of electron flow, the feasibility of a reaction, and even the free energy released, while the Nernst equation extends these ideas to non-standard concentrations.
Key ideas — Standard electrode potential and the standard hydrogen electrode
The standard electrode (reduction) potential, E^ominus, of a half-cell is the e.m.f. measured when it is connected to a standard hydrogen electrode (SHE), with all solutions at 1\ mathrmmol\ dm^-3, any gas at 1 bar, 298 K, and the SHE arbitrarily assigned E^ominus = 0\ V. The SHE consists of hydrogen gas at 1 bar bubbled over a platinum electrode coated in platinum black, immersed in 1\ mathrmmol\ dm^-3\ H^+(aq). The standard cell potential, E^ominus_cell, is the e.m.f. measured between any two half-cells under these same standard conditions.
🧑🏫 Teacher's explanation
Imagine you are trying to measure the height of a mountain, but you don't have a sea-level reference point. In electrochemistry, we cannot measure the absolute 'pressure' of electrons in a single half-cell. Instead, we measure the difference in electrical potential between two points. To make these measurements consistent across the world, scientists agreed to use a 'zero point'—much like sea level—called the Standard Hydrogen Electrode (SHE). We pretend the SHE has a potential of exactly 0 V.
📖 Definitions
🔍 Worked example — Easy
Calculate the standard cell potential, E^ominus_cell, for a cell consisting of a Zn^2+/Zn half-cell (E^ominus = -0.78 V) and a Cu^2+/Cu half-cell (E^ominus = +0.34 V).
🔍 Worked example — Standard
A redox reaction is proposed: Fe^2+(aq) + Cl_2(g) arrow Fe^3+(aq) + 2Cl^-(aq). Given E^ominus(Fe^3+/Fe^2+) = +0.77 V and E^ominus(Cl_2/Cl^-) = +1.36 V, determine if the reaction is feasible under standard conditions.
🔍 Worked example — Exam-style
Calculate the standard Gibbs free energy change, Δ G^ominus, for the reaction Zn(s) + Cu^2+(aq) arrow Zn^2+(aq) + Cu(s) using E^ominus_cell = 1.10 V and F = 96500 C mol^-1.
⚠️ Common errors
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🎯 Exam question
(a) Calculate the standard cell potential, E^ominus_cell. [2] (b) Write the overall balanced redox equation for the cell reaction. [2] (c) Calculate the standard Gibbs free energy change, Δ G^ominus, for the reaction in part (b). [2] [6 marks]
Key ideas — Combining half-cells: cell potential and electron flow
Two half-cells are connected by a wire (through a voltmeter) and a salt bridge, which completes the circuit by allowing ion flow without the solutions mixing. The standard cell potential is calculated as:
🧑🏫 Teacher's explanation
Think of electrode potential as 'electrical pressure' or 'pulling power'. Every chemical species has a desire to gain electrons. A species with a very high (positive) E^ominus is an electron-hungry predator, while a species with a very low (negative) E^ominus is an electron-donor. When you connect them, the electrons naturally flow from the 'donor' to the 'predator'.
📖 Definitions
🔍 Worked example — Easy
Calculate the standard cell potential (E^ominus_cell) for a cell consisting of a Zinc electrode (E^ominus_Zn^2+/Zn = -0.78 V) and a Copper electrode (E^ominus_Cu^2+/Cu = +0.34 V).
🔍 Worked example — Standard
A cell is made using Ag^+/Ag (E^ominus = +0.80 V) and Fe^2+/Fe (E^ominus = -0.44 V). Determine the direction of electron flow in the external wire.
🔍 Worked example — Exam-style
Calculate the standard Gibbs free energy change (Δ G^ominus) for the reaction Zn(s) + Cu^2+(aq) arrow Zn^2+(aq) + Cu(s) using E^ominus_cell = 1.10 V. (Use F = 96500 C mol^-1 and n=2).
⚠️ Common errors
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🎯 Exam question
Key ideas — Feasibility, relative reactivity and constructing redox equations
A reaction is feasible when the resulting E^ominus_cell is positive. The half-cell with the more positive E^ominus proceeds as a reduction, so that species acts as the oxidising agent; the half-cell with the more negative E^ominus is reversed to an oxidation, so that species acts as the reducing agent. Comparing E^ominus values across several half-equations ranks species by oxidising or reducing strength: a more positive E^ominus indicates a stronger oxidising agent (more easily reduced), and a more negative E^ominus indicates a stronger reducing agent (more easily oxidised). To construct the overall redox equation, the reduction half-equation and the reversed (oxidation) half-equation are balanced so that electrons lost equal electrons gained, then added together, cancelling the electrons.
🧑🏫 Teacher's explanation
Think of standard electrode potentials (E^ominus) as a 'pressure gauge' for electrons. Just as water flows from high pressure to low pressure, electrons naturally flow from a species with a low (more negative) potential to a species with a high (more positive) potential. The difference in these 'pressures' is what we call the cell potential (E^ominus_cell).\n\nImagine two people playing tug-of-war with an electron. One person is very strong and wants to pull the electron toward them (a strong oxidising agent with a high E^ominus), while the other person is very weak and lets the electron slip away easily (a strong reducing agent). The 'winner' of the tug-of-war is the species that successfully gains the electron, causing a reduction. If the 'pull' from the oxidising side is stronger than the 'push' from the reducing side, the total cell potential is positive, and the reaction happens spontaneously (it is feasible).\n\nTo find the total voltage of a battery, you simply subtract the lower potential from the higher potential: E^ominus_cell = E^ominus_reduction - E^ominus_oxidation.
📖 Definitions
🔍 Worked example — Easy
Calculate the standard cell potential (E^ominus_cell) for a cell consisting of the following half-equations:\nCu^2+(aq) + 2e^- arrow Cu(s), E^ominus = +0.34 V\nZn^2+(aq) + 2e^- arrow Zn(s), E^ominus = -0.76 V
🔍 Worked example — Standard
Construct the overall balanced redox equation for the reaction between Fe^2+(aq) and MnO_4^-(aq) given:\nMnO_4^-(aq) + 8H^+(aq) + 5e^- arrow Mn^2+(aq) + 4H_2O(l)\nFe^2+(aq) arrow Fe^3+(aq) + e^-
🔍 Worked example — Exam-style
Predict whether the reaction 2Ag(s) + Cu^2+(aq) arrow 2Ag^+(aq) + Cu(s) is feasible using the following potentials:\nAg^+(aq) + e^- arrow Ag(s), E^ominus = +0.80 V\nCu^2+(aq) + 2e^- arrow Cu(s), E^ominus = +0.34 V
⚠️ Common errors
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🎯 Exam question
Key ideas — The Nernst equation
Electrode potential is not fixed; it depends on the concentration of the ions involved. Increasing the concentration of the oxidised species in a half-cell makes E more positive (shifts the equilibrium further towards reduction, by Le Chatelier's principle), while increasing the concentration of the reduced species makes E less positive. Quantitatively, the Nernst equation for a metal/metal-ion electrode at 298 K is:
🧑🏫 Teacher's explanation
Imagine a tug-of-war between a metal and its ions. The 'standard electrode potential' (E^ominus) is the strength of the metal's pull when everything is at a standard concentration (1.00 mol dm^-3). However, this strength isn't fixed; it changes based on how many 'opponents' (ions) are in the ring.
📖 Definitions
🔍 Worked example — Easy
Calculate the electrode potential for the Ag^+ / Ag(s) half-cell at 298 K if [Ag^+] = 0.100 mol dm^-3. Given E^ominus = +0.80 V and the half-equation is Ag^+(aq) + e^- leftharpoons Ag(s).
🔍 Worked example — Standard
For the half-reaction Fe^2+(aq) + 2e^- leftharpoons Fe(s) with E^ominus = -0.44 V, calculate E when [Fe^2+] = 0.001 mol dm^-3.
🔍 Worked example — Exam-style
A cell is composed of a Zn/Zn^2+ electrode (E^ominus = -0.76 V) and a Cu/Cu^2+ electrode (E^ominus = +0.34 V). Calculate the standard cell potential E^ominus_cell and determine the direction of electron flow.
⚠️ Common errors
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🎯 Exam question
Key ideas — Linking ΔG⦵, n, F and E⦵cell
The standard free energy change of a cell reaction is linked to the standard cell potential by:
🧑🏫 Teacher's explanation
Think of a chemical reaction like a ball rolling down a hill. The 'hill' is the difference in electrical potential between two electrodes. The steeper the hill (the higher the E^ominus_cell), the more 'energy' is released when the ball rolls down. In electrochemistry, we use the cell potential to measure this 'steepness'.
📖 Definitions
🔍 Worked example — Easy
Calculate the standard free energy change (Δ G^ominus) for a cell with E^ominus_cell = 1.10 V, where n = 2 moles of electrons are transferred. (Use F = 96500 C mol^-1)
🔍 Worked example — Standard
A redox reaction has a standard cell potential of E^ominus_cell = 0.45 V. If the reaction involves the transfer of 1 mole of electrons, calculate Δ G^\ominus in kJ mol^-1.
🔍 Worked example — Exam-style
For the reaction Zn(s) + Cu^2+(aq) arrow Zn^2+(aq) + Cu(s), the E^ominus_cell is 1.10 V. Calculate Δ G^\ominus per mole of Zn reacted. (Given E^ominus(Zn^2+/Zn) = -0.78 V and E^ominus(Cu^2+/Cu) = 0.34 V)
⚠️ Common errors
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🎯 Exam question
Worked example
Question. E^ominus[mathrmCu^2+/Cu] = +0.34\ V; E^ominus[mathrmZn^2+/Zn] = -0.76\ V. (a) Calculate E^ominus_cell for a cell combining these two half-cells and state the direction of electron flow. [3] (b) Write the overall cell equation. [2] (c) Calculate Δ G^ominus for this reaction (F = 9.65 × 10^4\ mathrmC\ mol^-1). [3]
🔍 Worked example — Easy
Calculate the standard cell potential (E^ominus_cell) for a cell consisting of a silver half-cell (E^ominus[Ag^+/Ag] = +0.80 V) and a magnesium half-cell (E^ominus[Mg^2+/Mg] = -2.37 V), assuming the silver electrode acts as the cathode.
🔍 Worked example — Standard
Given E^ominus[Fe^3+/Fe^2+] = +0.77 V and E^ominus[Fe^2+/Fe] = -0.44 V, calculate the standard cell potential for a cell where Fe(s) is the anode and Fe^3+(aq) is the cathode.
🔍 Worked example — Exam-style
A cell is composed of Zn(s)|Zn^2+(aq) || Cu^2+(aq)|Cu(s). Given E^ominus[Cu^2+/Cu] = +0.34 V and E^ominus[Zn^2+/Zn] = -0.76 V, calculate the standard Gibbs free energy change (Δ G^ominus) for the reaction Zn(s) + Cu^2+(aq) arrow Zn^2+(aq) + Cu(s) in kJ mol^-1. Use F = 9.65 × 10^4 C mol^-1.
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