Electrochemistry (9701 A Level Physical Chemistry)

Hello future Chemists! This chapter, Electrochemistry, is where we truly bridge the gap between Chemistry and Physics. It deals with how electricity can cause chemical reactions (electrolysis) and how chemical reactions can generate electricity (electrochemical cells).
Don't worry if redox and balancing equations felt challenging before—this section builds on that foundation, but we will break down the complex concepts step-by-step. Mastering electrochemistry is key, as it explains everything from how batteries work to how metals are extracted!

Section 1: The Foundation - Redox Processes (AS Review)

6.1 Redox Processes: Electron Transfer and Oxidation Number

Before diving into cells and circuits, we need to be crystal clear on redox reactions. Remember, these involve simultaneous oxidation and reduction.

Defining Oxidation and Reduction (The Mnemonic)

The easiest way to remember these definitions is the classic mnemonic:
OIL RIG: Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons).

  • Oxidation: Loss of electrons, resulting in an increase in oxidation number.
  • Reduction: Gain of electrons, resulting in a decrease in oxidation number.
Oxidising and Reducing Agents

Agents act opposite to the process they undergo:

  • Oxidising Agent (Oxidant): Causes oxidation by being reduced itself (gains electrons).
  • Reducing Agent (Reductant): Causes reduction by being oxidised itself (loses electrons).

Analogy: Think of a Secret Agent. A secret agent doesn't keep secrets; they reveal them. Similarly, a reducing agent doesn't undergo reduction; it causes reduction by doing the opposite (undergoing oxidation).

Disproportionation Reactions

A disproportionation reaction is a specific redox reaction where the same element in a single substance is simultaneously oxidised and reduced, resulting in two different products with different oxidation states.

Example: In the reaction of chlorine with cold dilute alkali, chlorine (oxidation number \(0\)) is both reduced to \(\text{Cl}^-\) (\(-1\)) and oxidised to \(\text{ClO}^-\) (\(+1\)).

Calculating Oxidation Numbers

The oxidation number (or oxidation state) is a hypothetical charge an atom would have if all bonds were 100% ionic. We use Roman numerals to indicate the magnitude of the oxidation number (e.g., Iron(II) means Fe has an oxidation number of +2).

Here are the rules for calculating oxidation numbers:

  1. Elements in their standard state have an oxidation number of 0 (e.g., \(\text{O}_2\), \(\text{Na}\)).
  2. Ions composed of a single atom have an oxidation number equal to their charge (e.g., \(\text{Cl}^-\) is -1, \(\text{Mg}^{2+}\) is +2).
  3. Group 1 metals are always +1.
  4. Group 2 metals are always +2.
  5. Fluorine is always -1.
  6. Hydrogen is usually +1, except in metal hydrides (e.g., \(\text{NaH}\)) where it is -1.
  7. Oxygen is usually -2, except in peroxides (e.g., \(\text{H}_2\text{O}_2\)) where it is -1.
  8. The sum of all oxidation numbers in a neutral compound must be zero.
  9. The sum of all oxidation numbers in a polyatomic ion must equal the charge of the ion.
Balancing Redox Equations using Oxidation Numbers

To balance complex redox equations, balance the total increase in oxidation number with the total decrease in oxidation number so that the net change in electrons is zero.

Quick Review: Oxidation numbers track electron movement. If the number goes up (becomes more positive), it's oxidation. If it goes down (becomes more negative), it's reduction.

Section 2: Electrolysis (24.1)

Electrolysis is the process of using electrical energy to force a non-spontaneous chemical reaction to occur. This process takes place in an electrolytic cell.

The Electrolytic Cell Setup

  • Electrolyte: The substance (molten or aqueous) containing mobile ions.
  • Electrodes: Conductors dipped into the electrolyte.
  • Power Source: A DC source (battery) provides the electrical push.

Key Rule: The power source determines the electrode polarity:

  • Cathode: Connected to the negative terminal (where reduction occurs). Attracts cations (positive ions).
  • Anode: Connected to the positive terminal (where oxidation occurs). Attracts anions (negative ions).

Common Mistake Alert: In ALL cells (electrolytic or electrochemical), oxidation happens at the anode and reduction happens at the cathode. Only the signs change!

Predicting Products of Electrolysis (24.1.1)

1. Electrolysis of Molten Compounds

If the compound is molten (e.g., molten \(\text{NaCl}\)), only two ions are present. The prediction is simple:

  • Cation (\(\text{Na}^+\)) goes to the cathode (reduced to \(\text{Na}\)).
  • Anion (\(\text{Cl}^-\)) goes to the anode (oxidised to \(\text{Cl}_2\)).
2. Electrolysis of Aqueous Solutions (The Competition)

When water is present, there is competition at both electrodes because water itself can be oxidised or reduced. You must use the Redox Series (Standard Electrode Potentials, \(E^\circ\)) to predict the winner.

At the Cathode (Reduction)

Ions competing for reduction are the metal cation and water (\(\text{H}_2\text{O}\)).

The species with the more positive (less negative) \(E^\circ\) value will be reduced most easily.

  • If the metal ion has a high \(E^\circ\) (e.g., \(\text{Cu}^{2+}\)), the metal is produced.
  • If the metal ion has a very negative \(E^\circ\) (e.g., \(\text{Na}^+\)), hydrogen gas is usually produced from the reduction of water.
At the Anode (Oxidation)

Ions competing for oxidation are the anion and water.

The species with the less positive \(E^\circ\) value (the best reducing agent) will be oxidised most easily.

  • If the anion is a halide (\(\text{Cl}^-\), \(\text{Br}^-\), \(\text{I}^-\)), the halogen is usually produced.
  • If the anion is a complex ion like sulfate (\(\text{SO}_4^{2-}\)) or nitrate (\(\text{NO}_3^-\)), oxygen gas is usually produced from the oxidation of water.

Concentration Effect (Halides only): High concentration of a halide ion (like \(\text{Cl}^-\)) can override the water oxidation, making the halide the preferred product, even if its \(E^\circ\) suggests otherwise.

Quantitative Electrolysis (Faraday's Laws) (24.1.3)

The amount of substance liberated during electrolysis is directly proportional to the amount of charge passed.

Key Relationships

1. Charge (Q): The total charge passed through the circuit (measured in Coulombs, C).
\(Q = It\)
Where:

  • \(Q\) = Charge (C)
  • \(I\) = Current (A)
  • \(t\) = Time (s)

2. Faraday Constant (F): The charge carried by one mole of electrons.
\(F = Le\)
Where:

  • \(F\) = Faraday constant (\(96500 \text{ C mol}^{-1}\))
  • \(L\) = Avogadro constant (\(6.02 \times 10^{23} \text{ mol}^{-1}\))
  • \(e\) = Charge on one electron (\(1.60 \times 10^{-19} \text{ C}\))

Determining Avogadro Constant \(L\) and Faraday Constant \(F\) Experimentally

By passing a constant current \(I\) for a measured time \(t\) in an electrolytic cell (such as the electrodeposition of copper using copper electrodes), the mass of metal deposited at the cathode is measured. Knowing the mass change and the charge on a single electron \(e\), the experimental values of \(F\) and \(L\) can be calculated.

Calculations Step-by-Step

To find the mass or volume liberated:

  1. Calculate Q: Use \(Q = It\) (ensure time is in seconds).
  2. Calculate Moles of Electrons (\(n_e\)): Divide the total charge by the Faraday constant:
    \(n_e = \frac{Q}{F}\)
  3. Use the Half-Equation: Determine the mole ratio between the electrons and the substance liberated. For example, for copper:
    \(\text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu}\)
    (Ratio: 2 moles of electrons produce 1 mole of \(\text{Cu}\)).
  4. Calculate Moles of Substance: Use the ratio from step 3.
  5. Calculate Mass or Volume: Convert moles to mass (\(\text{Mass} = \text{Moles} \times M_r\)) or volume (for gas at RTP, \(\text{Volume} = \text{Moles} \times 24 \text{ dm}^3\)).

Key Takeaway for Electrolysis: Electrolysis uses electricity to force chemical change. Products are predicted by comparing electrode potentials (Redox Series) or by simple charge movement in molten salts. Calculations rely on the relationship between charge, current, time, and the mole ratio of electrons (Faraday's laws).

Section 3: Standard Electrode Potentials and Cells (24.2)

Electrochemical cells (galvanic or voltaic cells) generate electrical energy from spontaneous chemical reactions. This is the basis of a battery!

Definitions and Standard Conditions (24.2.1)

Standard Electrode Potential, \(E^\circ\)

The potential difference measured when an electrode is connected to the Standard Hydrogen Electrode (SHE) under standard conditions.

  • It is always written as a reduction potential (electrons on the left).
  • A more positive \(E^\circ\) means the species is more easily reduced (it is a better oxidising agent).
  • A more negative \(E^\circ\) means the species is more easily oxidised (it is a better reducing agent).

Standard Conditions: (Remember these are crucial for the symbol \(E^\circ\))

  • Temperature: \(298 \text{ K}\) (\(25^\circ \text{C}\))
  • Pressure: \(101 \text{ kPa}\) (\(1 \text{ atm}\))
  • Concentration: \(1.0 \text{ mol dm}^{-3}\) for all aqueous ions.

The Standard Hydrogen Electrode (SHE) (24.2.2)

The reference point for all \(E^\circ\) measurements.

  • It is assigned a potential of exactly zero volts: \(E^\circ = 0.00 \text{ V}\).
  • The half-reaction is: \(2\text{H}^+(\text{aq}) + 2\text{e}^- \rightleftharpoons \text{H}_2(\text{g})\).
  • Setup involves hydrogen gas at \(101 \text{ kPa}\) bubbling over an inert platinum electrode immersed in a \(1.0 \text{ mol dm}^{-3}\) solution of \(\text{H}^+\) ions.

Measuring Electrode Potentials (24.2.3)

To measure the \(E^\circ\) of any half-cell, you connect it via an external circuit (with a high-resistance voltmeter) and a salt bridge to the SHE.

  • Salt Bridge: Allows ions to flow to complete the circuit and maintain electrical neutrality (usually filter paper soaked in saturated \(\text{KNO}_3\)).
  • Voltmeter: Measures the potential difference, which is the standard cell potential (\(E^\circ_{\text{cell}}\)).

This method is used for:

  • Metals/non-metals in contact with their ions (e.g., \(\text{Cu}/\text{Cu}^{2+}\)).
  • Ions of the same element in different oxidation states (e.g., \(\text{Fe}^{3+}/\text{Fe}^{2+}\) in solution, using a platinum electrode).

Calculating Standard Cell Potential (\(E^\circ_{\text{cell}}\)) (24.2.4)

The standard cell potential is calculated using the standard reduction potentials of the two half-cells:
\(E^\circ_{\text{cell}} = E^\circ_{\text{reduction}} - E^\circ_{\text{oxidation}}\)

Trick: Since oxidation occurs at the less positive electrode and reduction at the more positive electrode:
\(E^\circ_{\text{cell}} = E^\circ_{\text{RHS}} - E^\circ_{\text{LHS}}\)
(Where RHS is the positive terminal/reduction electrode, and LHS is the negative terminal/oxidation electrode).

Polarity: The half-cell with the more positive \(E^\circ\) value is the cathode (positive terminal). The half-cell with the less positive \(E^\circ\) value is the anode (negative terminal).

Predicting Feasibility (Spontaneity) (24.2.5, 24.2.6)

The sign of the cell potential tells you if the reaction is spontaneous (feasible).

  • If \(E^\circ_{\text{cell}}\) is positive (\(+\)), the reaction is feasible (spontaneous).
  • If \(E^\circ_{\text{cell}}\) is negative (\(-\)), the reaction is non-feasible (it requires energy input, like in electrolysis).

We can deduce relative strengths:

The species on the left side of the half-equation with the most positive \(E^\circ\) is the strongest oxidising agent.

The species on the right side of the half-equation with the most negative \(E^\circ\) is the strongest reducing agent.

Commercial Cells and Fuel Cells (24.2.7)

Standard electrode potential principles apply directly to modern commercial energy storage devices:

  • Rechargeable Cells: Secondary cells where the spontaneous discharge redox reactions can be reversed by applying an external electrical potential.
  • Hydrogen-Oxygen Fuel Cells: Devices that convert chemical energy directly into electrical energy by reacting continuous supplies of hydrogen and oxygen fuels without combustion, producing only water as a chemical byproduct.

The Gibbs Free Energy Connection (24.2.10)

There is a direct link between the electrical energy produced by a cell and the thermodynamic feasibility (\(\Delta G^\circ\)).
\(\Delta G^\circ = -nE^\circ_{\text{cell}}F\)

Where:

  • \(\Delta G^\circ\) = Standard Gibbs Free Energy Change (\(\text{J mol}^{-1}\))
  • \(n\) = Number of moles of electrons transferred in the balanced overall equation.
  • \(E^\circ_{\text{cell}}\) = Standard cell potential (V).
  • \(F\) = Faraday constant (\(96500 \text{ C mol}^{-1}\)).

Since \(F\) and \(n\) are positive, the negative sign confirms the link:

  • For a spontaneous reaction, \(E^\circ_{\text{cell}}\) must be positive, making \(\Delta G^\circ\) negative (feasible).

Concentration Effects (The Nernst Equation) (24.2.8, 24.2.9)

Standard potentials \(E^\circ\) are only valid under standard conditions. If concentrations change, the electrode potential \(E\) changes.

Qualitative Prediction (What happens?)

We use Le Chatelier's principle on the half-equation. Consider the equilibrium:
\(\text{Oxidised species} + z\text{e}^- \rightleftharpoons \text{Reduced species}\)

  • Increasing [Oxidised species]: The equilibrium shifts right (reduction is favoured). The potential \(E\) becomes more positive.
  • Increasing [Reduced species]: The equilibrium shifts left (oxidation is favoured). The potential \(E\) becomes more negative.

Example: For \(\text{Cu}^{2+} + 2\text{e}^- \rightleftharpoons \text{Cu}\). If you increase the concentration of \(\text{Cu}^{2+}\), reduction becomes easier, so the potential increases (becomes more positive).

Quantitative Calculation (Using Nernst)

The Nernst equation allows you to calculate the new potential \(E\) at non-standard concentrations:
\(E = E^\circ + \left( \frac{0.059}{z} \right) \log \left( \frac{[\text{oxidised species}]}{[\text{reduced species}]} \right)\)

  • \(E\) is the electrode potential under non-standard conditions.
  • \(z\) is the number of electrons transferred (e.g., 2 for \(\text{Cu}^{2+}/\text{Cu}\)).

Note: You must be able to use this formula for specific examples like \(\text{Cu}^{2+}(\text{aq}) + 2\text{e}^- \rightleftharpoons \text{Cu}(\text{s})\) or \(\text{Fe}^{3+}(\text{aq}) + \text{e}^- \rightleftharpoons \text{Fe}^{2+}(\text{aq})\).

Key Takeaway for Cells: Electrochemical cells run spontaneously, generating a positive \(E^\circ_{\text{cell}}\). You use the standard reduction potentials to determine which half-reaction undergoes reduction (the more positive \(E^\circ\)) and calculate feasibility. Concentration changes affect the potential according to Le Chatelier's principle, quantified by the Nernst equation.