Chemistry - Redox and Electrochemistry

Oxidation and Reduction

OIL RIG

A useful mnemonic for remembering the definitions:

• Oxidation Is Loss (of electrons)
• Reduction Is Gain (of electrons)

Definitions

• Oxidation: Loss of electrons; increase in oxidation number
• Reduction: Gain of electrons; decrease in oxidation number

Oxidising and Reducing Agents

• Oxidising agent: The species that causes oxidation by accepting electrons (it is itself reduced)
• Reducing agent: The species that causes reduction by donating electrons (it is itself oxidised)

Worked Example 1

Identify the species oxidised, reduced, oxidising agent, and reducing agent in:

$\mathrm{Zn} + \mathrm{Cu^{2+}} \to \mathrm{Zn^{2+}} + \mathrm{Cu}$

• $\mathrm{Zn} \to \mathrm{Zn^{2+}} + 2e^-$ (oxidation; $\mathrm{Zn}$ is the reducing agent)
• $\mathrm{Cu^{2+}} + 2e^- \to \mathrm{Cu}$ (reduction; $\mathrm{Cu^{2+}}$ is the oxidising agent)

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Oxidation Numbers

Oxidation numbers are assigned to atoms using the following rules:

1. The oxidation number of an uncombined element is 0.
2. The oxidation number of a simple ion equals its charge.
3. In compounds, the sum of all oxidation numbers equals the overall charge.
4. In most compounds, hydrogen is $+1$ (except in metal hydrides where it is $-1$).
5. In most compounds, oxygen is $-2$ (except in peroxides where it is $-1$, and in $\mathrm{OF_2}$ where it is $+2$).
6. Fluorine is always $-1$ in its compounds.

Worked Example 2

Determine the oxidation numbers of all elements in $\mathrm{KMnO_4}$.

$\mathrm{K}: +1, \quad \mathrm{O}: -2, \quad \mathrm{Mn}: x$

$+1 + x + 4(-2) = 0$

$x + 1 - 8 = 0$

$x = +7$

The oxidation number of manganese in $\mathrm{KMnO_4}$ is $+7$.

Worked Example 3

Determine the oxidation numbers in $\mathrm{Cr_2O_7^{2-}}$.

$2x + 7(-2) = -2$

$2x - 14 = -2$

$2x = 12$

$x = +6$

The oxidation number of chromium in $\mathrm{Cr_2O_7^{2-}}$ is $+6$.

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Balancing Redox Equations

Half-Equation Method

1. Write the two half-equations (oxidation and reduction).
2. Balance atoms other than $\mathrm{O}$ and $\mathrm{H}$.
3. Balance oxygen by adding $\mathrm{H_2O}$.
4. Balance hydrogen by adding $\mathrm{H^+}$.
5. Balance charge by adding electrons.
6. Multiply the half-equations so that the electrons cancel.
7. Add the half-equations and simplify.

Worked Example 4

Balance the reaction between $\mathrm{MnO_4^-}$ and $\mathrm{Fe^{2+}}$ in acidic solution.

Reduction half-equation:

$\mathrm{MnO_4^-} + 8\mathrm{H^+} + 5e^- \to \mathrm{Mn^{2+}} + 4\mathrm{H_2O}$

Oxidation half-equation:

$\mathrm{Fe^{2+}} \to \mathrm{Fe^{3+}} + e^-$

Multiply the oxidation half-equation by 5:

$5\mathrm{Fe^{2+}} \to 5\mathrm{Fe^{3+}} + 5e^-$

Add both half-equations:

$\mathrm{MnO_4^-} + 8\mathrm{H^+} + 5\mathrm{Fe^{2+}} \to \mathrm{Mn^{2+}} + 4\mathrm{H_2O} + 5\mathrm{Fe^{3+}}$

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Displacement Reactions

A displacement reaction occurs when a more reactive metal displaces a less reactive metal from its compound.

Reactivity Series

The reactivity series ranks metals in order of their tendency to lose electrons:

$\mathrm{K} \gt \mathrm{Na} \gt \mathrm{Ca} \gt \mathrm{Mg} \gt \mathrm{Al} \gt \mathrm{Zn} \gt \mathrm{Fe} \gt \mathrm{Sn} \gt \mathrm{Pb} \gt (\mathrm{H}) \gt \mathrm{Cu} \gt \mathrm{Ag} \gt \mathrm{Au}$

Examples

$\mathrm{Zn} + \mathrm{CuSO_4} \to \mathrm{ZnSO_4} + \mathrm{Cu}$

$\mathrm{Fe} + \mathrm{CuSO_4} \to \mathrm{FeSO_4} + \mathrm{Cu}$

Copper cannot displace iron or zinc because copper is less reactive.

Worked Example 5

Will magnesium displace copper from copper(II) sulphate solution? Write the equation if it occurs.

Yes, because magnesium is above copper in the reactivity series.

$\mathrm{Mg} + \mathrm{CuSO_4} \to \mathrm{MgSO_4} + \mathrm{Cu}$

Magnesium is oxidised ($\mathrm{Mg} \to \mathrm{Mg^{2+}} + 2e^-$) and copper(II) ions are reduced ($\mathrm{Cu^{2+}} + 2e^- \to \mathrm{Cu}$).

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Extraction of Metals

Metals Above Carbon in the Reactivity Series

Metals more reactive than carbon ($\mathrm{K}$, $\mathrm{Na}$, $\mathrm{Ca}$, $\mathrm{Mg}$, $\mathrm{Al}$) are extracted by electrolysis of their molten compounds (usually chlorides or oxides).

Example: Extraction of aluminium from $\mathrm{Al_2O_3}$ (Hall-Heroult process)

• $\mathrm{Al_2O_3}$ is dissolved in molten cryolite ($\mathrm{Na_3AlF_6}$) to lower its melting point.
• At the cathode: $\mathrm{Al^{3+}} + 3e^- \to \mathrm{Al}$
• At the anode: $2\mathrm{O^{2-}} \to \mathrm{O_2} + 4e^-$
• The oxygen reacts with the carbon anodes to form $\mathrm{CO_2}$, so the anodes must be replaced periodically.

Metals Below Carbon in the Reactivity Series

Metals less reactive than carbon ($\mathrm{Zn}$, $\mathrm{Fe}$, $\mathrm{Sn}$, $\mathrm{Pb}$, $\mathrm{Cu}$) are extracted by reduction with carbon in a blast furnace.

Example: Extraction of iron from $\mathrm{Fe_2O_3}$

$\mathrm{Fe_2O_3} + 3\mathrm{CO} \to 2\mathrm{Fe} + 3\mathrm{CO_2}$

Carbon monoxide (from the partial combustion of coke) is the reducing agent.

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Rusting and Corrosion Prevention

Rusting of Iron

Rust is hydrated iron(III) oxide ($\mathrm{Fe_2O_3 \cdot xH_2O}$). Rusting requires both water and oxygen. It is an electrochemical process:

• At the anode (iron surface): $\mathrm{Fe} \to \mathrm{Fe^{2+}} + 2e^-$
• At the cathode: $\mathrm{O_2} + 2\mathrm{H_2O} + 4e^- \to 4\mathrm{OH^-}$
• $\mathrm{Fe^{2+}}$ ions are further oxidised to $\mathrm{Fe^{3+}}$ and form rust.

Methods of Prevention

Method	Mechanism
Painting/coating	Barrier: prevents water and oxygen reaching the iron
Oiling/greasing	Barrier: repels water
Galvanising	Coating with zinc; zinc is more reactive and sacrifices itself
Sacrificial anode	Attach a more reactive metal (e.g., Mg or Zn) to iron; it corrodes instead
Alloying	Stainless steel contains Cr, which forms a protective oxide layer

Worked Example 6

Explain why galvanising protects iron even if the zinc coating is scratched.

Zinc is more reactive than iron and is higher in the reactivity series. When the zinc coating is scratched, both zinc and iron are exposed to water and oxygen. Zinc acts as a sacrificial anode and undergoes oxidation in preference to iron. Zinc loses electrons and corrodes, protecting the iron beneath.

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Electrochemical Cells

An electrochemical cell converts chemical energy into electrical energy. It consists of two half-cells connected by a wire (external circuit) and a salt bridge (internal circuit).

Example: Daniell Cell

• Half-cell 1: $\mathrm{Zn}(s) \mid \mathrm{Zn^{2+}}(aq)$ (anode, oxidation)
• Half-cell 2: $\mathrm{Cu}(s) \mid \mathrm{Cu^{2+}}(aq)$ (cathode, reduction)
• Salt bridge: Allows ions to flow, maintaining electrical neutrality

Reactions:

• Anode (oxidation): $\mathrm{Zn} \to \mathrm{Zn^{2+}} + 2e^-$
• Cathode (reduction): $\mathrm{Cu^{2+}} + 2e^- \to \mathrm{Cu}$
• Overall: $\mathrm{Zn} + \mathrm{Cu^{2+}} \to \mathrm{Zn^{2+}} + \mathrm{Cu}$

Electrons flow from the zinc electrode (anode) to the copper electrode (cathode) through the external wire.

Cell Voltage

The cell potential (voltage) is determined by the difference between the standard electrode potentials of the two half-cells:

$E^\circ_{\mathrm{cell}} = E^\circ_{\mathrm{cathode}} - E^\circ_{\mathrm{anode}}$

A positive $E^\circ_{\mathrm{cell}}$ indicates that the reaction is spontaneous.

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Electrolysis

Electrolysis is the use of electrical energy to drive a non-spontaneous chemical reaction.

Key Terms

Term	Definition
Electrolyte	An ionic compound that conducts electricity when molten or dissolved
Anode	Positive electrode; oxidation occurs here
Cathode	Negative electrode; reduction occurs here
Cation	Positive ion; attracted to the cathode
Anion	Negative ion; attracted to the anode

Electrolysis of Molten Compounds

At the cathode: metal cations gain electrons (reduced to metal).

At the anode: non-metal anions lose electrons (oxidised to non-metal).

Example: Electrolysis of molten $\mathrm{NaCl}$

• Cathode: $\mathrm{Na^+} + e^- \to \mathrm{Na}$
• Anode: $2\mathrm{Cl^-} \to \mathrm{Cl_2} + 2e^-$

Electrolysis of Aqueous Solutions

When an aqueous solution is electrolysed, both the dissolved ions and water molecules can be discharged. The product at each electrode depends on the relative reactivity of the ions.

At the cathode (reduction):

• If the metal is more reactive than hydrogen ($\mathrm{K}$, $\mathrm{Na}$, $\mathrm{Ca}$, $\mathrm{Mg}$, $\mathrm{Al}$, $\mathrm{Zn}$): hydrogen gas is produced. $2\mathrm{H_2O} + 2e^- \to \mathrm{H_2} + 2\mathrm{OH^-}$
• If the metal is less reactive than hydrogen ($\mathrm{Cu}$, $\mathrm{Ag}$, $\mathrm{Au}$): the metal is produced. $\mathrm{Cu^{2+}} + 2e^- \to \mathrm{Cu}$

At the anode (oxidation):

• If the anion is a halide ($\mathrm{Cl^-}$, $\mathrm{Br^-}$, $\mathrm{I^-}$): the halogen is produced. $2\mathrm{Cl^-} \to \mathrm{Cl_2} + 2e^-$
• If the anion is anything else ($\mathrm{SO_4^{2-}}$, $\mathrm{NO_3^-}$): oxygen gas is produced. $4\mathrm{OH^-} \to \mathrm{O_2} + 2\mathrm{H_2O} + 4e^-$ (or equivalently: $2\mathrm{H_2O} \to \mathrm{O_2} + 4\mathrm{H^+} + 4e^-$)

Worked Example 7

Predict the products of the electrolysis of aqueous copper(II) sulphate with inert electrodes.

• Cathode: Copper is less reactive than hydrogen, so copper is deposited. $\mathrm{Cu^{2+}} + 2e^- \to \mathrm{Cu}$
• Anode: $\mathrm{SO_4^{2-}}$ is not a halide, so oxygen is produced. $4\mathrm{OH^-} \to \mathrm{O_2} + 2\mathrm{H_2O} + 4e^-$

Quantitative Electrolysis

The amount of substance produced or consumed during electrolysis is related to the charge passed:

$Q = It$

$n = \frac{Q}{F} = \frac{It}{F}$

where:

• $Q$ = charge in coulombs (C)
• $I$ = current in amperes (A)
• $t$ = time in seconds (s)
• $F = 96500 \mathrm{ C/mol}$ (Faraday constant)
• $n$ = number of moles of electrons

The mass of substance produced:

$m = \frac{It \times M}{z \times F}$

where $z$ is the number of electrons transferred per ion and $M$ is the molar mass.

Worked Example 8

Calculate the mass of copper deposited when a current of $2.50 \mathrm{ A}$ is passed through copper(II) sulphate solution for $30.0 \mathrm{ minutes}$.

$\mathrm{Cu^{2+}} + 2e^- \to \mathrm{Cu} \quad (z = 2)$

$Q = It = 2.50 \times 30.0 \times 60 = 4500 \mathrm{ C}$

$n(\mathrm{Cu}) = \frac{Q}{zF} = \frac{4500}{2 \times 96500} = 0.02332 \mathrm{ mol}$

$m = n \times M = 0.02332 \times 63.5 = 1.48 \mathrm{ g}$

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Electroplating

Electroplating is the process of depositing a thin layer of one metal onto another using electrolysis.

Setup

• Anode: Made of the plating metal (e.g., silver, nickel, chromium)
• Cathode: The object to be plated
• Electrolyte: A solution containing ions of the plating metal

Example: Silver Plating

• Anode: $\mathrm{Ag}(s)$
• Cathode: Object to be plated
• Electrolyte: $\mathrm{AgNO_3}(aq)$

At the anode: $\mathrm{Ag} \to \mathrm{Ag^+} + e^-$ (silver dissolves)

At the cathode: $\mathrm{Ag^+} + e^- \to \mathrm{Ag}$ (silver deposits on the object)

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Common Pitfalls

• Confusing the anode and cathode in electrolysis vs. electrochemical cells. In an electrolytic cell, the anode is positive; in a galvanic cell, the anode is negative (electrons flow from anode to cathode).
• Forgetting that $2 \times F$ electrons are needed to deposit 1 mole of a divalent metal (e.g., $\mathrm{Cu^{2+}}$, $\mathrm{Zn^{2+}}$). Always check the ion's charge.
• Applying the wrong rule for discharge at the anode in aqueous electrolysis. Halides are discharged in preference to hydroxide; other anions result in oxygen production.
• Assuming the most reactive metal is always deposited at the cathode. In aqueous solution, if the metal is more reactive than hydrogen, hydrogen gas is produced instead.
• Forgetting to convert time to seconds when using $Q = It$.
• Confusing oxidation numbers with ionic charges. Oxidation numbers are assigned by convention; ionic charges are real charges on ions.

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Summary Table

Concept	Key Point
OIL RIG	Oxidation Is Loss, Reduction Is Gain
Oxidising agent	Accepts electrons (is reduced)
Reducing agent	Donates electrons (is oxidised)
Reactivity series	Ranks metals by tendency to lose electrons
Rusting conditions	Requires both water and oxygen
Galvanic cell	Converts chemical energy to electrical energy
Electrolysis	Converts electrical energy to chemical energy
Faraday's law	$m = ItM / (zF)$
Cathode product	Metal (if less reactive than H) or $\mathrm{H_2}$
Anode product	Halogen (if halide present) or $\mathrm{O_2}$

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Additional Worked Examples

Worked Example: Oxidation Numbers in a Polyatomic Ion

Determine the oxidation number of nitrogen in the ammonium ion, $\mathrm{NH_4^+}$.

Solution

$x + 4(+1) = +1$

$x + 4 = +1$

$x = -3$

The oxidation number of nitrogen in $\mathrm{NH_4^+}$ is $-3$.

Worked Example: Cell Potential and Spontaneity

A cell is constructed from a $\mathrm{Mg}^{2+}/\mathrm{Mg}$ half-cell ($E^\circ = -2.37 \mathrm{ V}$) and a $\mathrm{Fe}^{2+}/\mathrm{Fe}$ half-cell ($E^\circ = -0.44 \mathrm{ V}$). Calculate the cell potential, write the overall equation, and state whether the reaction is spontaneous.

Solution

Iron has the more positive $E^\circ$, so it undergoes reduction (cathode).

$E^\circ_{\mathrm{cell}} = -0.44 - (-2.37) = 1.93 \mathrm{ V}$

Cathode (reduction): $\mathrm{Fe}^{2+} + 2e^- \to \mathrm{Fe}$

Anode (oxidation): $\mathrm{Mg} \to \mathrm{Mg}^{2+} + 2e^-$

Overall: $\mathrm{Mg} + \mathrm{Fe}^{2+} \to \mathrm{Mg}^{2+} + \mathrm{Fe}$

Since $E^\circ_{\mathrm{cell}} = +1.93 \mathrm{ V} \gt 0$, the reaction is spontaneous.

Worked Example: Electrolysis of Aqueous $\mathrm{CuSO_4}$ with Active Electrodes

Predict the products when aqueous $\mathrm{CuSO_4}$ is electrolysed using copper electrodes (instead of inert electrodes).

Solution

At the cathode: Copper is less reactive than hydrogen, so copper is deposited.

$\mathrm{Cu}^{2+} + 2e^- \to \mathrm{Cu}$

At the anode: Since the anode is made of copper (not inert), the copper anode itself dissolves rather than $\mathrm{SO_4^{2-}}$ or $\mathrm{OH^-}$ being discharged.

$\mathrm{Cu} \to \mathrm{Cu}^{2+} + 2e^-$

The concentration of $\mathrm{CuSO_4}$ remains constant because copper dissolves from the anode at the same rate it deposits at the cathode. This is the principle of electrolytic refining of copper.

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Exam-Style Practice Questions

Question 1: Determine the oxidation number of sulfur in $\mathrm{H_2SO_4}$.

$2(+1) + x + 4(-2) = 0$

$2 + x - 8 = 0$

$x = +6$

Question 2: Balance the following redox equation:

$\mathrm{MnO_4^-} + \mathrm{SO_3^{2-}} \to \mathrm{Mn^{2+}} + \mathrm{SO_4^{2-}}$ (in acidic solution)

Reduction: $\mathrm{MnO_4^-} + 8\mathrm{H^+} + 5e^- \to \mathrm{Mn^{2+}} + 4\mathrm{H_2O}$

Oxidation: $\mathrm{SO_3^{2-}} + \mathrm{H_2O} \to \mathrm{SO_4^{2-}} + 2\mathrm{H^+} + 2e^-$

Multiply reduction by 2 and oxidation by 5:

$2\mathrm{MnO_4^-} + 16\mathrm{H^+} + 5\mathrm{SO_3^{2-}} + 5\mathrm{H_2O} \to 2\mathrm{Mn^{2+}} + 8\mathrm{H_2O} + 5\mathrm{SO_4^{2-}} + 10\mathrm{H^+}$

Simplify: $2\mathrm{MnO_4^-} + 6\mathrm{H^+} + 5\mathrm{SO_3^{2-}} \to 2\mathrm{Mn^{2+}} + 3\mathrm{H_2O} + 5\mathrm{SO_4^{2-}}$

Question 3: Explain why aluminium is extracted by electrolysis rather than by reduction with carbon.

Aluminium is more reactive than carbon in the reactivity series. Carbon cannot reduce aluminium oxide because aluminium has a greater affinity for oxygen than carbon does. Therefore, electrolysis of molten $\mathrm{Al_2O_3}$ is required.

Question 4: Predict the products of electrolysis of concentrated aqueous $\mathrm{NaCl}$ with inert electrodes.

Cathode: $\mathrm{Na}$ is more reactive than $\mathrm{H}$, so $\mathrm{H_2}$ is produced: $2\mathrm{H_2O} + 2e^- \to \mathrm{H_2} + 2\mathrm{OH^-}$

Anode: $\mathrm{Cl^-}$ is a halide and is present in high concentration (concentrated), so $\mathrm{Cl_2}$ is produced: $2\mathrm{Cl^-} \to \mathrm{Cl_2} + 2e^-$

Question 5: A current of $0.500 \mathrm{ A}$ is passed through molten lead(II) bromide for $965 \mathrm{ s}$. Calculate the mass of lead produced.

$\mathrm{Pb^{2+}} + 2e^- \to \mathrm{Pb} \quad (z = 2)$

$Q = 0.500 \times 965 = 482.5 \mathrm{ C}$

$n(\mathrm{Pb}) = \frac{482.5}{2 \times 96500} = 2.50 \times 10^{-3} \mathrm{ mol}$

$m = 2.50 \times 10^{-3} \times 207 = 0.518 \mathrm{ g}$

Question 6: Explain why a block of magnesium attached to an underground iron pipeline prevents the pipeline from rusting.

Magnesium is more reactive than iron. When attached to the iron pipeline, magnesium acts as a sacrificial anode. It undergoes oxidation preferentially ($\mathrm{Mg} \to \mathrm{Mg^{2+}} + 2e^-$), supplying electrons to the iron and preventing iron from losing electrons. As long as magnesium remains, the iron is protected from corrosion.

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Problem Set

Problem 1: Determine the oxidation number of sulfur in $\mathrm{H_2SO_4}$ and in $\mathrm{SO_3^{2-}}$.

If you get this wrong, revise: Oxidation Numbers

Solution

In $\mathrm{H_2SO_4}$: $2(+1) + x + 4(-2) = 0 \Rightarrow x = +6$

In $\mathrm{SO_3^{2-}}$: $x + 3(-2) = -2 \Rightarrow x - 6 = -2 \Rightarrow x = +4$

Problem 2: In the reaction $\mathrm{Fe_2O_3} + 3\mathrm{CO} \to 2\mathrm{Fe} + 3\mathrm{CO_2}$, identify the species oxidised, reduced, the oxidising agent, and the reducing agent.

If you get this wrong, revise: Oxidising and Reducing Agents

Solution

$\mathrm{C}$ in $\mathrm{CO}$: oxidation number $+2 \to +4$ in $\mathrm{CO_2}$ (oxidation). $\mathrm{CO}$ is the reducing agent.

$\mathrm{Fe}$ in $\mathrm{Fe_2O_3}$: oxidation number $+3 \to 0$ in $\mathrm{Fe}$ (reduction). $\mathrm{Fe_2O_3}$ is the oxidising agent.

Problem 3: Balance the following redox equation in acidic solution:

$\mathrm{Cr_2O_7^{2-}} + \mathrm{I^-} \to \mathrm{Cr^{3+}} + \mathrm{I_2}$

If you get this wrong, revise: Balancing Redox Equations

Solution

Reduction: $\mathrm{Cr_2O_7^{2-}} + 14\mathrm{H^+} + 6e^- \to 2\mathrm{Cr^{3+}} + 7\mathrm{H_2O}$

Oxidation: $2\mathrm{I^-} \to \mathrm{I_2} + 2e^-$

Multiply oxidation by 3: $6\mathrm{I^-} \to 3\mathrm{I_2} + 6e^-$

Combine: $\mathrm{Cr_2O_7^{2-}} + 14\mathrm{H^+} + 6\mathrm{I^-} \to 2\mathrm{Cr^{3+}} + 7\mathrm{H_2O} + 3\mathrm{I_2}$

Problem 4: Predict the products of electrolysis of aqueous $\mathrm{CuSO_4}$ using inert (carbon) electrodes. Write half-equations.

If you get this wrong, revise: Electrolysis of Aqueous Solutions

Solution

Cathode: $\mathrm{Cu}$ is below $\mathrm{H}$ in the reactivity series, so copper is deposited.

$\mathrm{Cu^{2+}} + 2e^- \to \mathrm{Cu}$

Anode: $\mathrm{SO_4^{2-}}$ is not a halide, so oxygen is produced.

$4\mathrm{OH^-} \to \mathrm{O_2} + 2\mathrm{H_2O} + 4e^-$

Problem 5: A current of $0.500 \mathrm{ A}$ is passed through molten $\mathrm{PbBr_2}$ for $965 \mathrm{ s}$. Calculate the mass of lead deposited.

If you get this wrong, revise: Quantitative Electrolysis

Solution

$\mathrm{Pb^{2+}} + 2e^- \to \mathrm{Pb} \quad (z = 2)$

$Q = 0.500 \times 965 = 482.5 \mathrm{ C}$

$n(\mathrm{Pb}) = \frac{Q}{zF} = \frac{482.5}{2 \times 96500} = 2.50 \times 10^{-3} \mathrm{ mol}$

$m = 2.50 \times 10^{-3} \times 207 = 0.518 \mathrm{ g}$

Problem 6: A cell is made from $\mathrm{Zn}^{2+}/\mathrm{Zn}$ ($E^\circ = -0.76 \mathrm{ V}$) and $\mathrm{Ag^+}/\mathrm{Ag}$ ($E^\circ = +0.80 \mathrm{ V}$). Calculate the cell potential and write the overall equation.

If you get this wrong, revise: Cell Voltage

Solution

Silver has the more positive $E^\circ$ (cathode, reduction):

$E^\circ_{\mathrm{cell}} = 0.80 - (-0.76) = 1.56 \mathrm{ V}$

Cathode: $2\mathrm{Ag^+} + 2e^- \to 2\mathrm{Ag}$

Anode: $\mathrm{Zn} \to \mathrm{Zn^{2+}} + 2e^-$

Overall: $\mathrm{Zn} + 2\mathrm{Ag^+} \to \mathrm{Zn^{2+}} + 2\mathrm{Ag}$

Problem 7: Explain why aluminium is extracted by electrolysis rather than reduction with carbon.

If you get this wrong, revise: Extraction of Metals

Solution

Aluminium is more reactive than carbon in the reactivity series. Carbon cannot reduce $\mathrm{Al_2O_3}$ because aluminium has a stronger affinity for oxygen than carbon. Electrolysis of molten $\mathrm{Al_2O_3}$ (dissolved in cryolite) is required to force the reduction.

Problem 8: Why does the mass of the anode decrease when copper is purified by electrolysis using copper electrodes and $\mathrm{CuSO_4}$ solution?

If you get this wrong, revise: Electrolysis of Aqueous Solutions and Electroplating

Solution

The impure copper anode dissolves: $\mathrm{Cu} \to \mathrm{Cu^{2+}} + 2e^-$. Copper ions go into solution, so the anode loses mass. At the cathode, pure copper deposits: $\mathrm{Cu^{2+}} + 2e^- \to \mathrm{Cu}$. Impurities fall to the bottom as "anode sludge." This is the principle of electrolytic refining.

Problem 9: Calculate the volume of hydrogen gas (at r.t.p.) produced when a current of $3.00 \mathrm{ A}$ is passed through dilute $\mathrm{H_2SO_4}$ for $10.0 \mathrm{ minutes}$.

If you get this wrong, revise: Quantitative Electrolysis

Solution

$Q = 3.00 \times 10.0 \times 60 = 1800 \mathrm{ C}$

Cathode: $2\mathrm{H^+} + 2e^- \to \mathrm{H_2} \quad (z = 2)$

$n(\mathrm{H_2}) = \frac{Q}{zF} = \frac{1800}{2 \times 96500} = 9.33 \times 10^{-3} \mathrm{ mol}$

$V = 9.33 \times 10^{-3} \times 24.0 = 0.224 \mathrm{ dm^3} = 224 \mathrm{ cm^3}$

Problem 10: Explain why iron pipes corrode faster when connected to copper plumbing.

If you get this wrong, revise: Rusting and Corrosion Prevention

Solution

Iron is more reactive than copper. When in electrical contact in the presence of water and oxygen, iron acts as the anode and copper as the cathode. Electrons flow from iron to copper. At the anode: $\mathrm{Fe} \to \mathrm{Fe^{2+}} + 2e^-$ (iron corrodes faster). At the cathode: $\mathrm{O_2} + 2\mathrm{H_2O} + 4e^- \to 4\mathrm{OH^-}$. This sets up an electrochemical cell that accelerates corrosion of the iron.