Chemistry 9701 A Level Study Notes: Electrochemistry
Hello future A-grade chemist! Electrochemistry is a fascinating topic that bridges Physics and Chemistry, dealing with how electricity is generated from chemical reactions and how electricity can drive non-spontaneous chemical changes. Don't worry if it seems daunting; we will break down this high-scoring section into clear, manageable steps!
Key Takeaway from the Introduction: Why Study Electrochemistry?
It explains how batteries work (spontaneous reactions making electricity) and how industrial processes like aluminum production occur (using electricity to force reactions). It's all about the movement of electrons and energy transfer.
24.1 Electrolysis: Driving Non-Spontaneous Reactions
What is Electrolysis?
Electrolysis is the process where electrical energy is used to cause a non-spontaneous chemical reaction (a redox reaction) to occur. It involves passing an electric current through an electrolyte (a molten ionic compound or an aqueous ionic solution).
Key Components of an Electrolytic Cell
- Electrolyte: The substance containing free ions that carries the current (liquid).
- Electrodes: Conductors (usually inert like graphite or platinum, or reactive like copper) immersed in the electrolyte.
- Power Source: Provides the electrical energy needed to drive the non-spontaneous reaction.
Memory Aid: PANIC & OIL RIG
You need to remember which reaction happens at which electrode:
1. PANIC: In Electrolytic Cells,
Positive Anode Negative Is Cathode.
2. OIL RIG:
Oxidation Is Loss (of electrons) — occurs at the Anode.
Reduction Is Gain (of electrons) — occurs at the Cathode.
Predicting Products of Electrolysis
Case 1: Electrolysis of Molten Compounds (Simple!)
When a compound is molten (e.g., molten $\text{NaCl}$), only two ions are present. The prediction is straightforward:
- Anode (Positive): The negative ion (anion) is oxidized. (e.g., \(2\text{Cl}^- \to \text{Cl}_2 + 2\text{e}^-\))
- Cathode (Negative): The positive ion (cation) is reduced. (e.g., \(\text{Na}^+ + \text{e}^- \to \text{Na}\))
Case 2: Electrolysis of Aqueous Solutions (The Tricky Part)
In aqueous solutions, you have four possible species that could react: the ions from the salt, plus \(\text{H}^+\) and \(\text{OH}^-\) (or \(\text{H}_2\text{O}\)) from the water. We use the Standard Electrode Potentials ($E^\ominus$) (see Section 24.2) to predict what happens.
Prediction at the Cathode (Reduction)
The species with the more positive $E^\ominus$ (or less negative) will be reduced preferentially.
- Compare the cation (e.g., \(\text{Cu}^{2+}\)) with water/hydrogen ions (\(2\text{H}^+ + 2\text{e}^- \to \text{H}_2\), $E^\ominus = 0.00 \text{ V}$).
- Rule: If the metal cation is below hydrogen in the reactivity series (like $\text{Cu}^{2+}$), the metal is produced. If it is above hydrogen (like $\text{Na}^+$), hydrogen gas is produced.
Prediction at the Anode (Oxidation)
The species that is easiest to oxidize (has the more negative $E^\ominus$ for the reduction half-reaction) will be oxidized preferentially.
- Compare the anion (e.g., \(\text{Cl}^-\) or \(\text{SO}_4^{2-}\)) with the oxidation of water: \(\text{2H}_2\text{O} \to \text{O}_2 + 4\text{H}^+ + 4\text{e}^-\) ($E^\ominus$ for reduction is +1.23 V).
- Rule: For common anions like \(\text{NO}_3^-\) or \(\text{SO}_4^{2-}\), the $E^\ominus$ values are very positive, making them hard to oxidize. Oxygen is usually produced instead from water.
Crucial Exception: Effect of Concentration (24.1.1)
If the electrolyte is concentrated, an ion that would normally be too difficult to oxidize might react instead of water due to its high concentration.
- Example: Electrolysis of concentrated aqueous \(\text{NaCl}\). Although water oxidation ($E^\ominus = +1.23 \text{ V}$) is thermodynamically easier than oxidizing chloride ($E^\ominus = +1.36 \text{ V}$), the high concentration of \(\text{Cl}^-\) ions overcomes this difference, and chlorine gas is produced at the anode.
Quantitative Electrolysis: Faraday’s Laws (24.1.3)
These calculations link electrical quantities to the mass or volume of substance produced.
Key Quantities and Relationships (24.1.2)
- Charge (Q): Measured in Coulombs (C).
- Current (I): Measured in Amperes (A).
- Time (t): Measured in seconds (s).
- Faraday Constant (F): The charge carried by 1 mole of electrons. \(F \approx 96500 \text{ C mol}^{-1}\).
- Avogadro Constant (L) and Elementary Charge (e): The Faraday constant is related to the fundamental constants: \(F = L \times e\).
Formulas You Must Know
1. Total Charge Passed: $$Q = I \times t$$
2. Moles of Electrons (\(n_{\text{e}}\)): $$n_{\text{e}} = \frac{Q}{F}$$
3. Calculating Mass or Volume Liberated:
Step 1: Write the balanced half-equation to find the number of electrons (z) required per mole of product.
Example: \(\text{Cu}^{2+} + 2\text{e}^- \to \text{Cu}\). Here, \(z = 2\).
Step 2: Calculate the moles of product (\(n_{\text{prod}}\)) using the ratio: $$n_{\text{prod}} = \frac{n_{\text{e}}}{z}$$
Step 3: Convert moles to mass (using \(M_r\)) or volume (using molar gas volume, \(24.0 \text{ dm}^3\) at $298 \text{ K}$).
Did you know? (24.1.4) You can determine the Avogadro constant ($L$) experimentally by measuring the mass deposited during electrolysis (like copper) and using the known charge of the electron ($e$) and the calculated charge ($Q$) passed.
24.2 Standard Electrode Potentials ($E^\ominus$) and Feasibility
Defining the Potential (24.2.1)
A standard electrode (reduction) potential ($E^\ominus$) is the potential difference measured when an electrode is connected to a Standard Hydrogen Electrode (SHE) under standard conditions ($298 \text{ K}$, $101 \text{ kPa}$, $1.0 \text{ mol}\text{ dm}^{-3}$).
- By convention, all $E^\ominus$ values are written as reduction half-equations (electrons on the left).
Standard Conditions ($\Theta$)
To compare potentials fairly, we use standard conditions:
- Temperature: $298 \text{ K}$ ($25^\circ\text{C}$)
- Pressure (for gases): $101 \text{ kPa}$ (1 atm)
- Concentration (for solutions): $1.0 \text{ mol}\text{ dm}^{-3}$
The Standard Hydrogen Electrode (SHE) (24.2.2)
The SHE is the universal reference point for all half-cells and is arbitrarily assigned an $E^\ominus$ value of $0.00 \text{ V}$.
Reaction: \(\text{2H}^+ (\text{aq}, 1.0 \text{ mol}\text{ dm}^{-3}) + 2\text{e}^- \rightleftharpoons \text{H}_2 (\text{g}, 101 \text{ kPa})\)
It consists of a piece of platinum immersed in $1.0 \text{ mol}\text{ dm}^{-3}$ \(\text{H}^+\) solution, with hydrogen gas bubbled over the platinum at $101 \text{ kPa}$. Platinum is used because it is an inert conductor that allows electrons to transfer.
Electrochemical Cells (Voltaic or Galvanic Cells)
These cells convert chemical energy into electrical energy using spontaneous redox reactions.
Setup and Measurement (24.2.3)
A simple cell requires two half-cells connected by:
- External Circuit: A wire and a voltmeter (to measure the potential difference) for electron flow.
- Salt Bridge: A tube containing an inert electrolyte (like $\text{KNO}_3$) that connects the two solutions. This maintains electrical neutrality by allowing ions to flow, completing the circuit.
To measure the $E^\ominus$ of an unknown half-cell, we simply replace one half-cell (usually the reference electrode) with the SHE.
Calculating Standard Cell Potential ($E_{\text{cell}}^\ominus$) (24.2.4)
The standard cell potential ($E_{\text{cell}}^\ominus$) is the maximum potential difference between the two half-cells under standard conditions.
$$E_{\text{cell}}^\ominus = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}}$$ OR $$E_{\text{cell}}^\ominus = E^\ominus_{\text{RHS}} - E^\ominus_{\text{LHS}}$$
Predicting Feasibility and Reactivity (24.2.5 & 24.2.6)
Feasibility Rule (24.2.5(b))
A reaction is feasible (spontaneous) if the calculated $E_{\text{cell}}^\ominus$ is positive.
Deducing Electron Flow and Polarity (24.2.5(a))
In a galvanic cell, the reaction always flows from the less positive $E^\ominus$ to the more positive $E^\ominus$.
- The half-cell with the more negative $E^\ominus$ undergoes oxidation (where electrons are lost). This electrode is the Negative Pole.
- The half-cell with the more positive $E^\ominus$ undergoes reduction (where electrons are gained). This electrode is the Positive Pole.
- Electron flow is always from the Negative electrode (oxidation) to the Positive electrode (reduction).
Redox Strength (24.2.6)
We can use $E^\ominus$ values to compare the relative strength of oxidising and reducing agents:
- Strong Oxidising Agents: Have large positive $E^\ominus$ values (easy to reduce/gain electrons). Example: \(\text{F}_2\).
- Strong Reducing Agents: Have large negative $E^\ominus$ values (easy to oxidize/lose electrons). Example: \(\text{Li}\).
Analogy: Think of $E^\ominus$ values as "electron greed". A highly positive $E^\ominus$ means the species is very "greedy" and will happily steal electrons from almost anything else.
The Nernst Equation (24.2.8 & 24.2.9)
Standard electrode potentials ($E^\ominus$) are only valid at standard concentration ($1.0 \text{ mol}\text{ dm}^{-3}$). The Nernst equation tells us how the electrode potential ($E$) changes when concentrations are non-standard.
$$E = E^{\ominus} + \frac{0.059}{z} \log_{10} \frac{\text{[oxidised species]}}{\text{[reduced species]}}$$
- \(E\): Electrode potential under non-standard conditions.
- \(E^\ominus\): Standard electrode potential.
- \(z\): Number of moles of electrons transferred in the half-reaction.
- [oxidised species] / [reduced species]: The ratio of concentrations. (Note: Pure solids/liquids are usually omitted from this ratio).
Qualitative Predictions (24.2.8)
If the concentration of the reactant (the oxidised species) is increased, the reduction becomes more favourable, and the $E$ value becomes more positive.
Example: For the half-cell $\text{Cu}^{2+} + 2\text{e}^- \rightleftharpoons \text{Cu}$, increasing \([\text{Cu}^{2+}]\) pushes the equilibrium to the right (Le Chatelier's Principle), making the reduction easier, hence $E$ increases (becomes more positive).
Linking Electrochemistry to Thermodynamics (24.2.10)
Electrochemistry is directly related to the Gibbs Free Energy change ($\Delta G^\ominus$), which determines thermodynamic feasibility.
$$\Delta G^\ominus = -nFE_{\text{cell}}^\ominus$$
- \(n\): Total moles of electrons transferred in the balanced overall cell equation.
- \(F\): Faraday constant (\(96500 \text{ C mol}^{-1}\)).
- \(E_{\text{cell}}^\ominus\): Standard cell potential.
We already learned that a reaction is feasible if $\Delta G^\ominus$ is negative.
Since $n$ and $F$ are always positive, the negative sign in the equation confirms the link:
- If \(E_{\text{cell}}^\ominus\) is positive, then $\Delta G^\ominus$ is negative. The reaction is feasible (spontaneous).
- If \(E_{\text{cell}}^\ominus\) is negative, then $\Delta G^\ominus$ is positive. The reaction is non-feasible (non-spontaneous).
Quick Review: $E^\ominus$, $\Delta G^\ominus$, and Feasibility
| $E_{\text{cell}}^\ominus$ sign | $\Delta G^\ominus$ sign | Feasibility |
|---|---|---|
| Positive | Negative | Feasible / Spontaneous |
| Negative | Positive | Non-feasible / Non-spontaneous |
Common Mistake to Avoid: When calculating $E_{\text{cell}}^\ominus$, ensure you use the formula $E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}}$, where both $E^\ominus$ values are the standard reduction potentials provided in data sheets. Do not change the sign of the oxidation potential before subtracting.