Chemistry - Redox and Electrochemistry

A useful mnemonic for remembering the definitions:

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

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

• 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)

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 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.

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 $+7$.

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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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.

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

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

$$\mathrm{K} \gt \mathrm{Na} \gt \mathrm{Ca} \gt \mathrm{Mg} \gt \mathrm{Al} \gt \mathrm{Zn} \gt \mathrm{Fe} \gt \mathrm{Sn} \gt \mathrm{Pb} \gt (\mathrm{H}) \gt \mathrm{Cu} \gt \mathrm{Ag} \gt \mathrm{Au}$$$$\mathrm{Zn} + \mathrm{CuSO_4} \to \mathrm{ZnSO_4} + \mathrm{Cu}$$ $$\mathrm{Fe} + \mathrm{CuSO_4} \to \mathrm{FeSO_4} + \mathrm{Cu}$$

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

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

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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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).

• 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.

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

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: $2\mathrm{Cl^-} \to \mathrm{Cl_2} + 2e^-$

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}

--- <!-- Breadcrumb Schema for SEO --> <script type="application/ld+json"> { "@context": "https://schema.org", "@type": "BreadcrumbList", "itemListElement": [{"name": "Home", "url": "https://wyattau.com"}, {"name": "dse", "url": "https://dse.wyattau.com"}, {"name": "Chemistry", "url": "https://dse.wyattau.com/chemistry"}, {"name": "6 Redox And Electrochemistry", "url": "https://dse.wyattau.com/chemistry/6-redox-and-electrochemistry"}, {"name": "1_redox And Electrochemistry", "url": "https://dse.wyattau.com/chemistry/6-redox-and-electrochemistry/1_redox-and-electrochemistry"}] } </script> ## 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) --- <!-- Breadcrumb Schema for SEO --> <script type="application/ld+json"> { "@context": "https://schema.org", "@type": "BreadcrumbList", "itemListElement": [{"name": "Home", "url": "https://wyattau.com"}, {"name": "dse", "url": "https://dse.wyattau.com"}, {"name": "Chemistry", "url": "https://dse.wyattau.com/chemistry"}, {"name": "6 Redox And Electrochemistry", "url": "https://dse.wyattau.com/chemistry/6-redox-and-electrochemistry"}, {"name": "1_redox And Electrochemistry", "url": "https://dse.wyattau.com/chemistry/6-redox-and-electrochemistry/1_redox-and-electrochemistry"}] } </script> ## Intuition **Electricity from chemistry:** Electrochemistry is like a chemical battery, redox reactions transfer electrons from one substance to another, and we can harness that electron flow as electricity. **Why it matters:** From car batteries to rechargeable phones, electrochemistry powers modern life. Understanding redox helps you predict which reactions occur and how to control them. **The key insight:** The electrochemical series ranks substances by their tendency to gain or lose electrons, higher in the series means more likely to be oxidized. ## 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. --- <!-- Breadcrumb Schema for SEO --> <script type="application/ld+json"> { "@context": "https://schema.org", "@type": "BreadcrumbList", "itemListElement": [{"name": "Home", "url": "https://wyattau.com"}, {"name": "dse", "url": "https://dse.wyattau.com"}, {"name": "Chemistry", "url": "https://dse.wyattau.com/chemistry"}, {"name": "6 Redox And Electrochemistry", "url": "https://dse.wyattau.com/chemistry/6-redox-and-electrochemistry"}, {"name": "1_redox And Electrochemistry", "url": "https://dse.wyattau.com/chemistry/6-redox-and-electrochemistry/1_redox-and-electrochemistry"}] } </script> ## Summary Table <div class="table-wrap"> | 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 $\mathrm{O_2}$ | </div> --- <!-- Breadcrumb Schema for SEO --> <script type="application/ld+json"> { "@context": "https://schema.org", "@type": "BreadcrumbList", "itemListElement": [{"name": "Home", "url": "https://wyattau.com"}, {"name": "dse", "url": "https://dse.wyattau.com"}, {"name": "Chemistry", "url": "https://dse.wyattau.com/chemistry"}, {"name": "6 Redox And Electrochemistry", "url": "https://dse.wyattau.com/chemistry/6-redox-and-electrochemistry"}, {"name": "1_redox And Electrochemistry", "url": "https://dse.wyattau.com/chemistry/6-redox-and-electrochemistry/1_redox-and-electrochemistry"}] } </script> ## Additional Worked Examples ### Worked Example: Oxidation Numbers in a Polyatomic Ion Determine the oxidation number of nitrogen in the ammonium ion, $\mathrm{NH_4^+}$. <details> <summary>Solution</summary>

x + 4(+1) = +1

$$$$

x + 4 = +1

$$$$

x = -3

The oxidation number of nitrogen in $\mathrm{NH_4^+}$ is $-3$. </details> ### 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. <details> <summary>Solution</summary> 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. </details> ### 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). <details> <summary>Solution</summary> 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. </details> --- <!-- Breadcrumb Schema for SEO --> <script type="application/ld+json"> { "@context": "https://schema.org", "@type": "BreadcrumbList", "itemListElement": [{"name": "Home", "url": "https://wyattau.com"}, {"name": "dse", "url": "https://dse.wyattau.com"}, {"name": "Chemistry", "url": "https://dse.wyattau.com/chemistry"}, {"name": "6 Redox And Electrochemistry", "url": "https://dse.wyattau.com/chemistry/6-redox-and-electrochemistry"}, {"name": "1_redox And Electrochemistry", "url": "https://dse.wyattau.com/chemistry/6-redox-and-electrochemistry/1_redox-and-electrochemistry"}] } </script> <details> <summary>Exam-Style Practice Questions</summary> **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. </details> --- <!-- Breadcrumb Schema for SEO --> <script type="application/ld+json"> { "@context": "https://schema.org", "@type": "BreadcrumbList", "itemListElement": [{"name": "Home", "url": "https://wyattau.com"}, {"name": "dse", "url": "https://dse.wyattau.com"}, {"name": "Chemistry", "url": "https://dse.wyattau.com/chemistry"}, {"name": "6 Redox And Electrochemistry", "url": "https://dse.wyattau.com/chemistry/6-redox-and-electrochemistry"}, {"name": "1_redox And Electrochemistry", "url": "https://dse.wyattau.com/chemistry/6-redox-and-electrochemistry/1_redox-and-electrochemistry"}] } </script> ## 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_ <details> <summary>Solution</summary> In $\mathrm{H_2SO_4}$: $2(+1) + x + 4(-2) = 0 \Rightarrow x = +6$ In $\mathrm{SO_3^{2-}}$: $x + 3(-2) = -2 \Rightarrow x - 6 = -2 \Rightarrow x = +4$ </details> **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_ <details> <summary>Solution</summary> $\mathrm{C}$ 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. </details> **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_ <details> <summary>Solution</summary> 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: $\mathrm{Cr_2O_7^{2-}} + 14\mathrm{H^+} + 6\mathrm{I^-} \to 2\mathrm{Cr^{3+}} + 7\mathrm{H_2O} + 3\mathrm{I_2}$ </details> **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_ <details> <summary>Solution</summary> 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^-

</details> **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_ <details> <summary>Solution</summary>

\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}

</details> **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_ <details> <summary>Solution</summary> 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: $\mathrm{Zn} + 2\mathrm{Ag^+} \to \mathrm{Zn^{2+}} + 2\mathrm{Ag}$ </details> **Problem 7:** Explain why aluminium is extracted by electrolysis rather than reduction with carbon. _If you get this wrong, revise: Extraction of Metals_ <details> <summary>Solution</summary> 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. </details> **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_ <details> <summary>Solution</summary> 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. </details> **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_ <details> <summary>Solution</summary>

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}

</details> **Problem 10:** Explain why iron pipes corrode faster when connected to copper plumbing. _If you get this wrong, revise: Rusting and Corrosion Prevention_ <details> <summary>Solution</summary> 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. </details> ## Worked Examples ### Example 1: Balancing redox **Problem.** Balance: MnO$_4^-$ + Fe$^{2+}$ + H$^+$ $\to$ Mn$^{2+}$ + Fe$^{3+}$ + H$_2$O. **Solution.** Oxidation: Fe$^{2+}$ $\to$ Fe$^{3+}$ + $e^-$. Reduction: MnO$_4^-$ + 8H$^+$ + 5$e^-$ $\to$ Mn$^{2+}$ + 4H$_2$O. Multiply oxidation by 5: MnO$_4^-$ + 5Fe$^{2+}$ + 8H$^+$ $\to$ Mn$^{2+}$ + 5Fe$^{3+}$ + 4H$_2$O. $\blacksquare$ ### Example 2: Faraday's law **Problem.** What mass of Cu is deposited when $2.5 \mathrm{ A}$ flows through CuSO$_4$ for 30 minutes? **Solution.** $Q = 2.5 \times 1800 = 4500 \mathrm{ C}$. $n(e^-) = 4500/96500 = 0.0466 \mathrm{ mol}$. $n(\text{Cu}) = 0.0466/2 = 0.0233 \mathrm{ mol}$. Mass $= 0.0233 \times 63.5 = 1.48 \mathrm{ g}$. $\blacksquare$ ## Summary - Oxidation numbers: assign using standard rules; redox involves changes in oxidation number. - Balancing: split into half-equations, balance atoms, charge with electrons, then combine. - Electrolysis: $Q = It = nF$; Faraday's laws relate charge to amount deposited. ## Cross-References <div class="table-wrap"> | Topic | Site | Link | | ------------------ | ------- | ------------------------------------------------------------------------------------------------------------ | | [Electrochemistry] | A-Level | [View](https://alevel-sciences.wyattau.com/docs/alevel/chemistry/electrochemistry) | | [Electrochemistry] | IB | [View](https://ib.wyattau.com/chemistry/9-redox/3_electrochemistry/) | | [Electrochemistry] | DSE | [View](https://dse.wyattau.com/chemistry/6-redox-and-electrochemistry/1_redox-and-electrochemistry/) | | [Redox Reactions] | IB | [View](https://ib.wyattau.com/chemistry/9-redox/1_redox-reactions/) | </div> | [Redox Reactions] | DSE | [View](https://dse.wyattau.com/chemistry/6-redox-and-electrochemistry/1_redox-and-electrochemistry/) | ## See Also - [Redox And Electrochemistry](./) - [DSE Chemistry](/) - [Chemistry - Acids, Bases, and Electrochemistry](/chemistry/5-acids-bases/1_acids-bases-and-electrochemistry/)