Chemistry (9701) | Electrolysis

⚡️ The Power of Push: Comprehensive Study Notes on Electrolysis (9701 A Level Chemistry)

Hello future Chemists! This chapter, Electrolysis, links together many core ideas you’ve already studied—redox, stoichiometry, and energy—and shows how we can use electricity to force chemical reactions to happen. Why is this important? Because electrolysis is vital for industry, producing essential materials like aluminium, chlorine, and pure copper!

Don't worry if it seems tricky at first. We will break down the process into easy steps, starting with the basics of what makes an electrolytic cell work, and then moving onto predicting the products and performing quantitative calculations.

1. What is Electrolysis? The Chemical 'Push'

Definition: Electrolysis is the chemical decomposition (breaking down) of an electrolyte using an electric current.

1.1 The Electrolytic Cell Setup

Electrolysis happens inside an electrolytic cell. You need three things:

• Electrolyte: A substance containing free-moving ions, usually a molten ionic compound or an aqueous solution. This conducts electricity by the movement of ions.
• Electrodes: Two electrical conductors (usually carbon/graphite or metals) dipped into the electrolyte.
• External Power Source: This provides the energy needed to drive the non-spontaneous reaction. It pushes electrons.

1.2 Key Terminology and Electrode Reactions

Remember the difference between the two types of electrodes:

• Anode: Connected to the positive terminal of the power supply.
  Reaction: Oxidation (loss of electrons) always occurs here.
  Ions Attracted: Anions (negatively charged ions).
• Cathode: Connected to the negative terminal of the power supply.
  Reaction: Reduction (gain of electrons) always occurs here.
  Ions Attracted: Cations (positively charged ions).

🔥 Memory Aid: AN OX and RED CAT

AN OX: ANode = OXidation
RED CAT: REDuction = CAThode

Key Takeaway 1:

Electrolysis uses electrical energy to force a non-spontaneous redox reaction. Oxidation happens at the anode, and reduction happens at the cathode.

2. Predicting Products: Who Gets to React?

This is the most crucial part of qualitative electrolysis. The product formed at each electrode depends on which ion (or water molecule) is the easiest to oxidize or reduce.

2.1 Electrolysis of Molten Compounds

When an ionic substance is melted (molten), only two ions are present. This is the simplest case!

Example: Electrolysis of molten Lead(II) Bromide (\(PbBr_2\))

• Cations available: \(Pb^{2+}\)
• Anions available: \(Br^{-}\)

At the Cathode (Reduction): \(Pb^{2+}(l) + 2e^{-} \rightarrow Pb(l)\) (Lead metal is produced)
At the Anode (Oxidation): \(2Br^{-}(l) \rightarrow Br_2(g) + 2e^{-}\) (Bromine gas is produced)

2.2 Electrolysis of Aqueous Solutions (The Competition)

When the electrolyte is aqueous, water (\(H_2O\)) is also present, meaning it can compete with the dissolved ions to be oxidized or reduced. We use the position in the redox series (linked to $E^\theta$ values) to decide the winner.

2.2.1 Cathode Prediction (Reduction)

Two species are competing for reduction: the metal cation (\(M^{n+}\)) and the water molecule.

Rule: The species that is a stronger oxidizing agent (or requires less energy/is easier to reduce) will react.

• If the metal ion (\(M^{n+}\)) is from a highly reactive metal (like Group 1, Group 2, or Aluminium), it is difficult to reduce.
  Winner: Water is reduced, forming hydrogen gas (\(H_2\)) and hydroxide ions (\(OH^{-}\)).
  Half-Equation (Water Reduction): \(2H_2O(l) + 2e^{-} \rightarrow H_2(g) + 2OH^{-}(aq)\)
• If the metal ion (\(M^{n+}\)) is from a less reactive metal (like Copper, Silver, Gold), it is easier to reduce than water.
  Winner: The metal ion is reduced, forming the solid metal.
  Half-Equation (Metal Reduction): \(M^{n+}(aq) + ne^{-} \rightarrow M(s)\)

Don't worry if this seems tricky at first: Just remember that reactive metals (those high up in the reactivity series) don't want to gain electrons back, so water takes its place.

2.2.2 Anode Prediction (Oxidation)

Three species might compete for oxidation: the anion (\(A^{n-}\)), water, and the electrode itself (if it is reactive). We generally assume the electrodes are inert (unreactive, e.g., platinum or carbon) unless specified otherwise.

Competition 1: Anion vs. Water (Using Inert Electrodes)

• Case A: Non-Halide Anions (\(SO_4^{2-}\), \(NO_3^{-}\))
  These polyatomic ions are very stable and difficult to oxidize.
  Winner: Water is oxidized, producing oxygen gas (\(O_2\)) and hydrogen ions (\(H^{+}\)).
  Half-Equation (Water Oxidation): \(2H_2O(l) \rightarrow O_2(g) + 4H^{+}(aq) + 4e^{-}\)
• Case B: Halide Anions (\(Cl^{-}\), \(Br^{-}\), \(I^{-}\)) — Concentration Matters!
  Halide ions are relatively easy to oxidize, but concentration plays a role due to overvoltage effects.
  Concentrated Halide: Halogen gas is produced. (e.g., \(2Cl^{-}(aq) \rightarrow Cl_2(g) + 2e^{-}\))
  Dilute Halide: Water is oxidized (producing \(O_2\)).

Did you know? This phenomenon where concentration can overcome the natural tendency (like chloride being oxidized even though water *should* be easier) is known as the effect of concentration on electrode potential, which allows us to get chlorine instead of oxygen commercially.

Competition 2: Reactive Electrodes

If the anode is made of a reactive metal (like copper, $Cu$), the metal electrode itself is usually oxidized rather than the anion or water. This is commonly seen in refining metals.

Half-Equation (Reactive Anode): \(Cu(s) \rightarrow Cu^{2+}(aq) + 2e^{-}\)

Key Takeaway 2:

In aqueous solutions, water competes. At the cathode, only ions of unreactive metals are reduced. At the anode, water is oxidized unless the anion is concentrated halide or the electrode is reactive.

3. Quantitative Electrolysis: How Much Product?

Quantitative electrolysis is about answering "how much" product is formed based on the amount of electricity passed.

3.1 Charge, Current, and Time (24.1.3a)

The total electric charge ($Q$) passed through the cell is calculated using the current ($I$) and the time ($t$):

$$Q = I \times t$$

• $Q$ = Charge, measured in Coulombs (C).
• $I$ = Current, measured in Amperes (A).
• $t$ = Time, measured in seconds (s).

Crucial Tip: Always convert time to seconds before calculating $Q$. If $I$ is in Amperes and $t$ is in seconds, $Q$ will be in Coulombs.

3.2 The Faraday Constant ($F$) (24.1.2)

The Faraday Constant (\(F\)) is the quantity of charge carried by one mole of electrons.

$$F = 96500 \text{ C mol}^{-1}$$

This constant links the electrical quantity ($Q$) to the chemical quantity (moles of electrons, \(n_e\)).

The syllabus also requires you to understand the relationship between the Faraday constant ($F$), the Avogadro constant ($L$), and the charge on an electron ($e$):

$$F = L \times e$$

Think of it this way: the total charge of 1 mole of electrons ($F$) is the charge of a single electron ($e$) multiplied by the number of electrons in a mole ($L$).

3.3 Calculations: Mass/Volume Liberated (24.1.3b)

This calculation requires combining your knowledge of $Q = It$ with mole ratios from the half-equations.

Step-by-Step Guide to Quantitative Calculations:

1. Calculate Charge (Q): Use $Q = I \times t$ (remember $t$ in seconds).
2. Calculate Moles of Electrons (\(n_e\)): Use the Faraday constant: \(n_e = \frac{Q}{F}\) (where $F = 96500 \text{ C mol}^{-1}$).
3. Use Stoichiometry: Write the balanced half-equation for the substance being liberated (e.g., \(Cu^{2+} + 2e^{-} \rightarrow Cu\)). Use the mole ratio (moles of electrons : moles of product) to find the moles of substance produced.
4. Calculate Final Quantity:
   • If mass is needed: Mass = moles $\times$ \(M_r\).
   • If volume (of gas) is needed: Volume = moles $\times$ Molar Gas Volume (usually $24.0 \text{ dm}^{3}$ at room temp, or use $pV=nRT$ if given $T$ and $P$).

Key Takeaway 3:

Quantitative electrolysis is a three-step stoichiometric process: Charge ($Q$) $\rightarrow$ Moles of $e^{-}$ $\rightarrow$ Moles of product. Remember $Q = I \times t$ and $F = 96500 \text{ C mol}^{-1}$.

4. Experimental Determination of the Avogadro Constant ($L$)

The syllabus requires you to describe how the Avogadro constant ($L$) can be determined using an electrolytic method (24.1.4).

4.1 The Method

This experiment involves electrolysing a solution like aqueous silver nitrate (\(AgNO_3\)) or copper sulfate (\(CuSO_4\)) and carefully measuring the resulting mass change.

We rely on the equation: $$L = \frac{F}{e}$$

If we can determine $F$ experimentally (the charge per mole of electrons) and we already know the fundamental charge of an electron ($e$), we can calculate $L$.

4.2 Step-by-Step Procedure (e.g., using silver)

1. Measure Current (I) and Time (t): Pass a constant current (I) through the \(AgNO_3\) solution for a known time ($t$).
2. Measure Mass Change: Weigh the cathode (e.g., a platinum electrode) accurately before and after the experiment. The difference in mass ($\Delta m$) is the mass of silver deposited.
3. Calculate Moles of Silver (\(n_{Ag}\)): Divide the mass deposited ($\Delta m$) by the molar mass of silver (\(A_r\)). $$n_{Ag} = \frac{\Delta m}{A_r}$$
4. Determine Moles of Electrons (\(n_e\)): Use the half-equation: \(Ag^{+}(aq) + e^{-} \rightarrow Ag(s)\).
   In this case, the mole ratio is 1:1, so \(n_e = n_{Ag}\).
5. Calculate Total Charge Passed (Q): $$Q = I \times t$$
6. Determine the Faraday Constant (F): Since $F$ is the charge required to deposit one mole of product (using the electron ratio), we calculate: $$F = \frac{\text{Total Charge }(Q)}{\text{Moles of electrons }(n_e)}$$
7. Calculate Avogadro Constant (L): Use the known value of the electron charge ($e$, typically provided) and the calculated value of $F$: $$L = \frac{F}{e}$$

This experiment demonstrates the link between macro-scale measurements (current, time, mass) and fundamental constants (F, L, e).

Key Takeaway 4:

The Avogadro constant ($L$) can be found experimentally by measuring the mass of metal deposited ($\Delta m$) by a known charge ($Q = It$), using the relationships $n_e = \Delta m/A_r$ and $F = Q/n_e$.