Electrochemistry and Galvanic Cells

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Electrochemistry and Galvanic Cells — Key Facts for WAEC WASSCE

• Electrolyte: A substance that conducts electricity when molten or dissolved in water, via the movement of ions.
• Electrode: A conductor immersed in an electrolyte. The anode is the positive electrode where oxidation occurs (negative ions lose electrons); the cathode is the negative electrode where reduction occurs (positive ions gain electrons).
• Faraday’s Laws of Electrolysis:
  • First Law: The mass $m$ of substance liberated at an electrode is directly proportional to the charge $Q$ passed: $m = \frac{Q}{F} \times \frac{M}{z}$ where $M$ = molar mass, $z$ = number of electrons per ion, $F$ = Faraday constant = 96,500 C mol⁻¹.
  • Second Law: The mass of different substances produced by the same charge is proportional to their equivalent masses.
• Galvanic (Voltaic) Cells: Convert chemical energy to electrical energy. The cell e.m.f. $E°{\text{cell}} = E°{\text{cathode}} - E°_{\text{anode}}$ (reduction potential minus oxidation potential).
• Standard Electrode Potentials: Measured under standard conditions (1 mol dm⁻³, 1 atm, 25°C). $E° > 0$: the species is a stronger oxidising agent (more readily reduced). $E° < 0$: the species is more readily oxidised.

⚡ WAEC Exam Tip: In electrolysis, cations move to the cathode (negative in an electrolytic cell) and anions move to the anode (positive). At the cathode, the cation with the MORE NEGATIVE (or less positive) standard reduction potential is preferentially discharged. At the anode, the anion with the MORE POSITIVE (or less negative) standard oxidation potential is preferentially discharged. These are the ‘competition rules’ WAEC loves to test!

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Electrochemistry and Galvanic Cells — WAEC WASSCE Chemistry Study Guide

Types of Electrochemical Cells

Electrolytic Cells — Electrical energy is used to drive a non-spontaneous chemical reaction. The process is called electrolysis.

Key features:

• Direct current (d.c.) source is required
• The electrolytic cell contains one electrolyte (or two electrodes in the same electrolyte)
• Positive ions migrate to the cathode (negative electrode)
• Negative ions migrate to the anode (positive electrode)
• Oxidation occurs at the anode; reduction occurs at the cathode

Galvanic (Voltaic) Cells — Chemical energy is converted to electrical energy from a spontaneous reaction.

Key features:

• Two half-cells (different metals in their own electrolyte solutions) connected by a salt bridge
• The anode is the electrode where oxidation occurs (negative terminal of the galvanic cell)
• The cathode is the electrode where reduction occurs (positive terminal)
• Electrons flow from anode to cathode through the external wire
• The salt bridge maintains electrical neutrality by allowing ions to migrate

The Daniell Cell

A classic galvanic cell consisting of:

• Anode (oxidation): $\mathrm{Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-}$ (zinc rod in $\mathrm{ZnSO_4}$ solution)
• Cathode (reduction): $\mathrm{Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)}$ (copper rod in $\mathrm{CuSO_4}$ solution)
• Overall: $\mathrm{Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)}$

Cell notation (convention: left = oxidation, right = reduction): $$\mathrm{Zn(s)|Zn^{2+}(aq)||Cu^{2+}(aq)|Cu(s)}$$

The salt bridge (usually $\mathrm{KNO_3}$ or $\mathrm{K_2SO_4}$ in agar-agar gel) completes the circuit by allowing ions to flow, preventing rapid build-up of charge that would stop the cell.

Standard Electrode Potentials and Cell EMF

Standard electrode potentials $E°$ are measured relative to the standard hydrogen electrode (SHE), which is assigned $E° = 0.00$ V.

The standard cell potential: $$E°{\text{cell}} = E°{\text{cathode}} - E°_{\text{anode}}$$

For the Daniell cell: $E°_{\text{cell}} = +0.34\ \text{V} - (-0.76\ \text{V}) = +1.10\ \text{V}$

A positive $E°_{\text{cell}}$ indicates a spontaneous reaction under standard conditions.

Using Standard Potentials to Predict Feasibility

To predict whether a redox reaction will occur spontaneously:

1. Identify the oxidation and reduction half-reactions
2. Use the species with the more positive $E°$ value as the reduction half-reaction
3. Reverse the less positive one to become the oxidation half-reaction
4. Combine and calculate $E°_{\text{cell}}$

Only reactions with $E°_{\text{cell}} > 0$ are spontaneous under standard conditions.

Faraday’s Quantitative Electrolysis

The charge $Q = It$ (current × time in seconds)

Mass of substance deposited: $$m = \frac{Q \times M}{z \times F} = \frac{It \times M}{z \times F}$$

Where $M$ = molar mass of the substance, $z$ = number of electrons per ion (valency), $F$ = 96,500 C mol⁻¹.

Electroplating

Electroplating uses electrolysis to deposit a thin layer of metal on an object:

• The object to be plated is made the cathode
• The metal to be deposited is used as the anode
• The electrolyte is a solution of the plating metal’s salt

Example — Silver plating: Anode = silver; Cathode = object; Electrolyte = $\mathrm{AgNO_3(aq)}$ or $\mathrm{Ag(CN)^-_2}$ (for better quality plating).

For nickel plating, the electrolyte is often $\mathrm{NiSO_4(aq)}$ with some $\mathrm{NiCl_2}$ added to increase anodic dissolution.

Industrial Electrolysis Examples

Process	Electrolyte	Anode	Cathode	Product
Extraction of aluminium	$\mathrm{Al_2O_3}$ in molten cryolite	Carbon (burns to $\mathrm{CO_2}$)	Carbon	Aluminium metal
Chlorine production	$\mathrm{NaCl(aq)}$ (brine)	Titanium	Steel	$\mathrm{Cl_2}$ (anode), $\mathrm{H_2}$ (cathode), $\mathrm{NaOH}$ (solution)
Purification of copper	$\mathrm{CuSO_4(aq)}$	Impure copper	Pure copper	Pure copper at cathode

Competition at the Electrodes During Electrolysis

For a mixture of cations at the cathode: The ion requiring the least energy (most positive reduction potential) is preferentially reduced. For aqueous solutions, $\mathrm{H^+}$ is always in competition with metal ions. If $E°_{\text{metal}} < 0$ (e.g., Zn: $-0.76$ V), hydrogen gas is produced rather than the metal.

For a mixture of anions at the anode: For aqueous solutions, $\mathrm{OH^-}$ (with $E° = +1.36$ V for $\mathrm{O_2 + 4H^+ + 4e^-}$) competes with halide ions. For $\mathrm{Cl^-}$ ($E° = +1.36$ V for $\mathrm{Cl_2 + 2e^-}$), the overpotential means $\mathrm{Cl_2}$ is usually produced preferentially in concentrated solutions.

Comparison Table — Electrolytic vs Galvanic Cells

Feature	Electrolytic Cell	Galvanic (Voltaic) Cell
Energy conversion	Electrical → Chemical	Chemical → Electrical
Spontaneity	Non-spontaneous	Spontaneous
Anode polarity	Positive	Negative
Cathode polarity	Negative	Positive
Power source	Required (d.c.)	Not required
Example	Electroplating	Daniell cell

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Electrochemistry and Galvanic Cells — Comprehensive WAEC Chemistry Notes

Derivation of the Nernst Equation

The standard cell potential $E°{\text{cell}}$ relates to the equilibrium constant $K$ of the overall cell reaction: $$\Delta G° = -nFE°{\text{cell}}$$

But also: $\Delta G° = -RT\ln K$

Therefore: $nFE°{\text{cell}} = RT\ln K$ → $E°{\text{cell}} = \frac{RT}{nF}\ln K$

For non-standard conditions, the Nernst equation gives the actual potential: $$E = E° - \frac{RT}{nF}\ln Q$$

At 25°C (298 K), this simplifies to: $$E = E° - \frac{0.059}{n}\log_{10} Q \quad \text{(in volts)}$$

For a cell reaction at 298 K: $$E_{\text{cell}} = E°{\text{cell}} - \frac{0.059}{n}\log{10} Q$$

Where $n$ = number of moles of electrons transferred, $Q$ = reaction quotient.

Using the Nernst Equation

For the Daniell cell: $\mathrm{Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)}$, $n = 2$

$$E = 1.10 - \frac{0.059}{2}\log_{10}\frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]}$$

If $[\text{Zn}^{2+}] = [\text{Cu}^{2+}]$ (standard conditions), $E = 1.10$ V.

As the cell discharges, $[\text{Cu}^{2+}]$ decreases and $[\text{Zn}^{2+}]$ increases, so $\log(Q)$ changes and $E$ decreases — this is why a battery’s voltage drops as it discharges.

Faraday’s First Law — Detailed Derivation

One Faraday ($F$) of charge (96,500 C) deposits one mole of a univalent substance (e.g., 1 mole of $\mathrm{Ag^+}$, or $\frac{1}{2}$ mole of $\mathrm{Cu^{2+}}$).

For a divalent ion $\mathrm{Cu^{2+} + 2e^- \rightarrow Cu}$: $z = 2$ To deposit 1 mole of Cu, the charge required is $2F = 193,000$ C.

General formula: $m = \frac{ItM}{zF} = \frac{QM}{zF}$

Electrochemical Series (Standard Reduction Potentials)

The electrochemical series arranges metals (and other redox couples) in order of decreasing tendency to be reduced:

$$\mathrm{Li^+ + e^- \rightarrow Li} \quad E° = -3.04\ \text{V}$$ $$\mathrm{Zn^{2+} + 2e^- \rightarrow Zn} \quad E° = -0.76\ \text{V}$$ $$\mathrm{2H^+ + 2e^- \rightarrow H_2} \quad E° = 0.00\ \text{V}$$ $$\mathrm{Cu^{2+} + 2e^- \rightarrow Cu} \quad E° = +0.34\ \text{V}$$ $$\mathrm{Ag^+ + e^- \rightarrow Ag} \quad E° = +0.80\ \text{V}$$ $$\mathrm{F_2 + 2e^- \rightarrow 2F^-} \quad E° = +2.87\ \text{V}$$

Applications of the Electrochemical Series

1. Predicting spontaneous reactions: Any metal higher in the series will displace a metal lower in the series from its salt solution. This is why zinc displaces copper from $\mathrm{CuSO_4}$ solution (zinc is more reactive).
2. Predicting products of electrolysis: The ion with the more favourable (more positive for reduction, less negative) potential is discharged first.
3. Corrosion: A metal higher in the series corrodes preferentially. Iron corrodes because its $E°$ is lower than many metals — it is oxidised in the presence of oxygen and moisture. ‘Sacrificial anodes’ (e.g., magnesium blocks attached to iron pipes) protect iron by being oxidised instead (zinc $E° = -0.76$ V, more negative than iron’s $-0.44$ V for $\mathrm{Fe^{2+} + 2e^- \rightarrow Fe}$).
4. Predicting whether a metal reacts with acid: Only metals with $E° < 0$ (more negative than hydrogen) react with dilute $\mathrm{HCl}$ or $\mathrm{H_2SO_4}$ to produce hydrogen gas. Copper ($E° = +0.34$ V) does not react with dilute acids.

Concentration Cells

A concentration cell is a galvanic cell where both electrodes are made of the same material but are in solutions of different concentrations. For example: $$\mathrm{Zn(s)|Zn^{2+}(1\ mol\ dm^{-3})||Zn^{2+}(0.01\ mol\ dm^{-3})|Zn(s)}$$

The cell potential is zero when concentrations are equal ($Q = 1$, so $E = E° - 0 = 0$). As the concentrations equalise (the cell ‘discharges’), the voltage drops to zero.

Using the Nernst equation: $$E = 0 - \frac{0.059}{n}\log\frac{[\text{reduced}]}{[\text{oxidised}]}$$

For the concentration cell: $E = -\frac{0.059}{2}\log\frac{0.01}{1} = +\frac{0.059}{2} \times 2 = +0.059$ V

Primary, Secondary, and Fuel Cells

Type	Rechargeable	Example	Reaction
Primary	No	Leclanché (dry cell)	$\mathrm{Zn \rightarrow Zn^{2+} + 2e^-}$ (anode); $\mathrm{MnO_2 + NH_4^+ + e^- \rightarrow MnO(OH) + NH_3}$ (cathode)
Secondary	Yes	Lead-acid battery	Charging reverses reactions: $\mathrm{Pb + PbO_2 + 2H_2SO_4 \rightarrow 2PbSO_4 + 2H_2O}$
Fuel cell	Continuous (as long as fuel supplied)	Hydrogen-oxygen fuel cell	$\mathrm{2H_2 + O_2 \rightarrow 2H_2O}$ (electrical energy + water as product)

Lead-Acid Battery Details

The lead-acid battery is a secondary (rechargeable) cell. During discharge:

• Anode (oxidation): $\mathrm{Pb + H_2SO_4 \rightarrow PbSO_4 + 2H^+ + 2e^-}$
• Cathode (reduction): $\mathrm{PbO_2 + H_2SO_4 + 2H^+ + 2e^- \rightarrow PbSO_4 + 2H_2O}$

Each 2 V cell consists of one $\mathrm{Pb/PbSO_4}$ electrode and one $\mathrm{PbO_2/PbSO_4}$ electrode in $\mathrm{H_2SO_4}$. Six cells in series give a 12 V car battery.

Hydrogen-Oxygen Fuel Cell

The fuel cell converts chemical energy directly to electrical energy with high efficiency (no thermal energy losses like in combustion engines). The overall reaction: $\mathrm{2H_2 + O_2 \rightarrow 2H_2O}$ with $E° = +1.23$ V.

WAEC Past Question Patterns

Typical WAEC questions include:

1. Writing cell notation and half-equations for galvanic cells
2. Calculating $E°_{\text{cell}}$ from standard reduction potentials
3. Applying Faraday’s laws to calculate mass deposited or charge required
4. Predicting products of electrolysis of various electrolytes (including molten salts vs aqueous solutions)
5. Using the Nernst equation for non-standard cell potentials
6. Explaining industrial electrolytic processes (aluminium extraction, chlorine production)
7. Explaining the role of the salt bridge in a galvanic cell
8. Distinguishing between primary and secondary cells

⚡ WAEC Exam Tip: In electrolysis of aqueous sodium chloride solution (brine), the products are hydrogen gas at the cathode (NOT sodium metal!) and chlorine gas at the anode. This is because sodium has a much more negative reduction potential ($-2.71$ V) than water ($-0.83$ V for $\mathrm{2H_2O + 2e^- \rightarrow H_2 + 2OH^-}$), so water is reduced instead. For the anode, $\mathrm{Cl^-}$ ($+1.36$ V for $\mathrm{Cl_2 + 2e^-}$) is oxidised in preference to water ($+1.36$ V is close; in practice overpotential effects favour $\mathrm{Cl_2}$ in concentrated solutions). Always consider BOTH the standard potentials AND the actual conditions!

Numerical Worked Example (WAEC-style)

Calculate the mass of copper deposited when a current of 5.0 A is passed through a solution of copper(II) sulfate for 30 minutes. (Cu = 63.5, F = 96,500 C mol⁻¹)

Solution: $Q = It = 5.0 \times 30 \times 60 = 9,000$ C

For $\mathrm{Cu^{2+} + 2e^- \rightarrow Cu}$: $z = 2$, $M = 63.5$ g mol⁻¹

$m = \frac{QM}{zF} = \frac{9000 \times 63.5}{2 \times 96500} = \frac{571500}{193000} = 2.96$ g

Answer: Mass of copper deposited ≈ 3.0 g

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