Redox Reactions

Oxidation States

Rules for Assigning Oxidation States

1. The oxidation state of an element in its standard state is zero (e.g., Na(s), Cl$_2$(g), S$_8$(s), O$_2$(g)).
2. For a monatomic ion, the oxidation state equals the charge (e.g., Na$^+$ = +1, Ca$^{2+}$ = +2, Cl$^-$ = -1).
3. Oxygen is usually -2 (exceptions: peroxides = -1, OF$_2$ = +2, superoxides = -1/2).
4. Hydrogen is usually +1 (exceptions: metal hydrides like NaH where H = -1).
5. Halogens are usually -1 in compounds (exceptions: when bonded to oxygen or a more electronegative halogen).
6. The sum of all oxidation states in a neutral compound equals zero.
7. The sum of all oxidation states in a polyatomic ion equals the charge of the ion.

Worked Example 1: Assigning Oxidation States

(a) Mn in KMnO$_4$: K = +1, O = -2 (four oxygens = -8). Sum = 0: $+1 + \mathrm{Mn} + 4(-2) = 0$, so Mn = +7.

(b) Cr in K$_2$Cr$_2$O$_7$: K = +1 (two K = +2), O = -2 (seven O = -14). Sum = 0: $+2 + 2\mathrm{Cr} + 7(-2) = 0$, so $2\mathrm{Cr} = +12$, Cr = +6.

(c) S in H$_2$SO$_4$: H = +1 (two H = +2), O = -2 (four O = -8). Sum = 0: $+2 + \mathrm{S} + 4(-2) = 0$, so S = +6.

(d) S in H$_2$SO$_3$: H = +1 (two H = +2), O = -2 (three O = -6). Sum = 0: $+2 + \mathrm{S} + 3(-2) = 0$, so S = +4.

(e) N in HNO$_3$: H = +1, O = -2 (three O = -6). Sum = 0: $+1 + \mathrm{N} + 3(-2) = 0$, so N = +5.

(f) Fe in Fe$_3$O$_4$: O = -2 (four O = -8). Sum = 0: $3\mathrm{Fe} - 8 = 0$, so Fe = +8/3 = +2.67. This is a mixed oxidation state compound (Fe$_2$O$_3$ $\cdot$ FeO, with Fe(III) and Fe(II)).

Oxidation and Reduction

Process	Oxidation State	Electron Transfer
Oxidation	Increases	Loss of electrons (OIL -- Oxidation Is Loss)
Reduction	Decreases	Gain of electrons (RIG -- Reduction Is Gain)

Oxidising agent: The species that causes oxidation (itself gets reduced). Electron acceptor.

Reducing agent: The species that causes reduction (itself gets oxidised). Electron donor.

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Balancing Half-Equations

Steps for Balancing in Acidic Medium

1. Write the unbalanced half-equation with the correct species.
2. Balance all elements except H and O.
3. Balance O by adding $\mathrm{H}_2\mathrm{O}$.
4. Balance H by adding $\mathrm{H}^+$.
5. Balance charge by adding electrons ($e^-$).
6. If necessary, multiply half-equations to make electrons equal, then add them together.

Steps for Balancing in Basic Medium

Follow steps 1--5 for acidic medium, then:

6. Add the same number of OH$^-$ to both sides as the number of H$^+$ present.
7. On the side with both H$^+$ and OH$^-$, combine them to form H$_2$O.
8. Cancel any H$_2$O that appears on both sides.

Worked Example 2: Balancing in Acidic Medium

Balance: $\mathrm{MnO}_4^- + \mathrm{Fe}^{2+} \to \mathrm{Mn}^{2+} + \mathrm{Fe}^{3+}$ (acidic)

Reduction half-equation (Mn: +7 to +2):

$$\mathrm{MnO}_4^- \to \mathrm{Mn}^{2+}$$

Balance O: $\mathrm{MnO}_4^- \to \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O}$

Balance H: $\mathrm{MnO}_4^- + 8\mathrm{H}^+ \to \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O}$

Balance charge: Left = $-1 + 8 = +7$. Right = $+2$. Add 5 $e^-$ to left:

$$\mathrm{MnO}_4^- + 8\mathrm{H}^+ + 5e^- \to \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O}$$

Oxidation half-equation (Fe: +2 to +3):

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

Multiply by 5 to balance electrons:

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

Combined:

$$\mathrm{MnO}_4^- + 8\mathrm{H}^+ + 5\mathrm{Fe}^{2+} \to \mathrm{Mn}^{2+} + 5\mathrm{Fe}^{3+} + 4\mathrm{H}_2\mathrm{O}$$

Worked Example 3: Balancing in Basic Medium

Balance: $\mathrm{CrO}_4^{2-} + \mathrm{Br}^- \to \mathrm{Cr(OH)}_3 + \mathrm{BrO}_3^-$ (basic)

Reduction half-equation (Cr: +6 to +3):

$$\mathrm{CrO}_4^{2-} \to \mathrm{Cr(OH)}_3$$

Balance O (acidic): $\mathrm{CrO}_4^{2-} + 4\mathrm{H}^+ \to \mathrm{Cr(OH)}_3 + \mathrm{H}_2\mathrm{O}$

Balance H: already balanced (4 H on each side).

Balance charge: Left = $-2 + 4 = +2$. Right = $0$. Add 3 $e^-$ to left:

$$\mathrm{CrO}_4^{2-} + 4\mathrm{H}^+ + 3e^- \to \mathrm{Cr(OH)}_3 + \mathrm{H}_2\mathrm{O}$$

Now convert to basic: add 4 OH$^-$ to both sides:

$$\mathrm{CrO}_4^{2-} + 4\mathrm{H}_2\mathrm{O} + 3e^- \to \mathrm{Cr(OH)}_3 + \mathrm{H}_2\mathrm{O} + 4\mathrm{OH}^-$$

Cancel H$_2$O:

$$\mathrm{CrO}_4^{2-} + 3\mathrm{H}_2\mathrm{O} + 3e^- \to \mathrm{Cr(OH)}_3 + 4\mathrm{OH}^-$$

Oxidation half-equation (Br: -1 to +5):

$$\mathrm{Br}^- + 3\mathrm{H}_2\mathrm{O} \to \mathrm{BrO}_3^- + 6\mathrm{H}^+ + 6e^-$$

Convert to basic: add 6 OH$^-$ to both sides:

$$\mathrm{Br}^- + 3\mathrm{H}_2\mathrm{O} + 6\mathrm{OH}^- \to \mathrm{BrO}_3^- + 6\mathrm{H}_2\mathrm{O} + 6e^-$$

Cancel H$_2$O:

$$\mathrm{Br}^- + 6\mathrm{OH}^- \to \mathrm{BrO}_3^- + 3\mathrm{H}_2\mathrm{O} + 6e^-$$

Multiply reduction by 2, add:

$$2\mathrm{CrO}_4^{2-} + 6\mathrm{H}_2\mathrm{O} + \mathrm{Br}^- + 6\mathrm{OH}^- + 6e^- \to 2\mathrm{Cr(OH)}_3 + 8\mathrm{OH}^- + \mathrm{BrO}_3^- + 3\mathrm{H}_2\mathrm{O} + 6e^-$$

Simplify (cancel 6 $e^-$, subtract common terms):

$$2\mathrm{CrO}_4^{2-} + 3\mathrm{H}_2\mathrm{O} + \mathrm{Br}^- \to 2\mathrm{Cr(OH)}_3 + 2\mathrm{OH}^- + \mathrm{BrO}_3^-$$

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Reactivity Series of Metals

Activity Series

Metals are ranked by their tendency to lose electrons (be oxidised):

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

Key predictions:

• A more reactive metal will displace a less reactive metal from its compound.
• Metals above hydrogen will displace hydrogen from acids.
• The more reactive a metal, the stronger its reducing power (more easily oxidised).
• The more reactive a metal, the more negative its $E^\circ$ value.

Worked Example 4: Displacement Reactions

Predict whether the following reactions will occur:

(a) Zn(s) + CuSO$_4$(aq)

(b) Cu(s) + ZnSO$_4$(aq)

(c) Fe(s) + 2HCl(aq)

(d) Ag(s) + HCl(aq)

Answer:

(a) Yes. Zn is above Cu in the reactivity series, so Zn will displace Cu from CuSO$_4$: $\mathrm{Zn}(s) + \mathrm{Cu}^{2+}(aq) \to \mathrm{Zn}^{2+}(aq) + \mathrm{Cu}(s)$

(b) No. Cu is below Zn, so Cu cannot displace Zn from ZnSO$_4$.

(c) Yes. Fe is above H in the reactivity series: $\mathrm{Fe}(s) + 2\mathrm{H}^+(aq) \to \mathrm{Fe}^{2+}(aq) + \mathrm{H}_2(g)$

(d) No. Ag is below H in the reactivity series, so Ag cannot displace hydrogen from HCl.

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Electrochemical Cells

Galvanic (Voltaic) Cells

A galvanic cell converts chemical energy to electrical energy through a spontaneous redox reaction.

Components:

• Two half-cells: Each consists of an electrode in contact with an electrolyte solution containing its ions.
• Salt bridge: Contains an inert electrolyte (e.g., KNO$_3$, KCl) that allows ion flow to maintain electrical neutrality without mixing the solutions.
• External circuit: Wire connecting the two electrodes through which electrons flow.

Key features:

• Anode: Electrode where oxidation occurs. In a galvanic cell, this is the negative electrode.
• Cathode: Electrode where reduction occurs. In a galvanic cell, this is the positive electrode.
• Electrons flow from anode to cathode through the external circuit.
• Conventional current flows from cathode (positive) to anode (negative) in the external circuit.
• Cations migrate toward the cathode; anions migrate toward the anode through the salt bridge.

Cell Notation (Convention)

$$\mathrm{Anode} \mid \mathrm{Anode electrolyte} \parallel \mathrm{Cathode electrolyte} \mid \mathrm{Cathode}$$

The single vertical line represents a phase boundary. The double vertical line represents the salt bridge. The anode is written on the left (oxidation), and the cathode on the right (reduction).

Example: Zn-Cu Daniel cell:

$$\mathrm{Zn}(s) \mid \mathrm{Zn}^{2+}(aq) \parallel \mathrm{Cu}^{2+}(aq) \mid \mathrm{Cu}(s)$$

• Left (anode): $\mathrm{Zn}(s) \to \mathrm{Zn}^{2+}(aq) + 2e^-$ (oxidation)
• Right (cathode): $\mathrm{Cu}^{2+}(aq) + 2e^- \to \mathrm{Cu}(s)$ (reduction)

Example with inert electrodes:

$$\mathrm{Pt}(s) \mid \mathrm{Fe}^{2+}(aq), \mathrm{Fe}^{3+}(aq) \parallel \mathrm{MnO}_4^-(aq), \mathrm{H}^+(aq), \mathrm{Mn}^{2+}(aq) \mid \mathrm{Pt}(s)$$

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Standard Electrode Potentials ($E^\circ$)

Standard Conditions

• Pressure: $100\mathrm{ kPa}$ for gases
• Concentration: $1.0\mathrm{ mol/L}$ for aqueous species
• Temperature: $298\mathrm{ K}$ ($25\degree\mathrm{C}$)
• Solid electrodes in their standard states

Standard Hydrogen Electrode (SHE)

The SHE is the reference electrode with $E^\circ = 0.00\mathrm{ V}$ by definition:

$$2\mathrm{H}^+(aq, 1\mathrm{ mol/L}) + 2e^- \rightleftharpoons \mathrm{H}_2(g, 100\mathrm{ kPa}) \quad E^\circ = 0.00\mathrm{ V}$$

It consists of a platinum electrode in contact with $1\mathrm{ mol/L}$ H$^+$ and bubbled with H$_2$ gas at $100\mathrm{ kPa}$.

Standard Reduction Potentials

All electrode potentials are tabulated as reduction potentials:

Half-Reaction	$E^\circ$ (V)
$\mathrm{Li}^+ + e^- \rightleftharpoons \mathrm{Li}$	-3.04
$\mathrm{K}^+ + e^- \rightleftharpoons \mathrm{K}$	-2.93
$\mathrm{Na}^+ + e^- \rightleftharpoons \mathrm{Na}$	-2.71
$\mathrm{Mg}^{2+} + 2e^- \rightleftharpoons \mathrm{Mg}$	-2.37
$\mathrm{Al}^{3+} + 3e^- \rightleftharpoons \mathrm{Al}$	-1.66
$\mathrm{Zn}^{2+} + 2e^- \rightleftharpoons \mathrm{Zn}$	-0.76
$\mathrm{Fe}^{2+} + 2e^- \rightleftharpoons \mathrm{Fe}$	-0.44
$2\mathrm{H}^+ + 2e^- \rightleftharpoons \mathrm{H}_2$	0.00
$\mathrm{Cu}^{2+} + 2e^- \rightleftharpoons \mathrm{Cu}$	+0.34
$\mathrm{I}_2 + 2e^- \rightleftharpoons 2\mathrm{I}^-$	+0.54
$\mathrm{Ag}^+ + e^- \rightleftharpoons \mathrm{Ag}$	+0.80
$\mathrm{Br}_2 + 2e^- \rightleftharpoons 2\mathrm{Br}^-$	+1.07
$\mathrm{MnO}_4^- + 8\mathrm{H}^+ + 5e^- \rightleftharpoons \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O}$	+1.51
$\mathrm{F}_2 + 2e^- \rightleftharpoons 2\mathrm{F}^-$	+2.87

Calculating Cell EMF

The standard cell potential ($E^\circ_{\mathrm{cell}}$) is the difference between the cathode and anode reduction potentials:

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

Key rules:

• Identify the half-reaction with the more positive $E^\circ$ value as the cathode (reduction).
• The half-reaction with the less positive (more negative) $E^\circ$ is the anode (oxidation).
• The anode half-equation must be reversed (becomes oxidation).
• $E^\circ$ values are intensive properties and are never multiplied, regardless of stoichiometric coefficients.

Spontaneity:

• If $E^\circ_{\mathrm{cell}} \gt 0$: the reaction is spontaneous (galvanic cell).
• If $E^\circ_{\mathrm{cell}} \lt 0$: the reaction is non-spontaneous (requires an external power source).

Worked Example 5: Cell EMF Calculation

Calculate $E^\circ_{\mathrm{cell}}$ for: $\mathrm{Zn}(s) + \mathrm{Cu}^{2+}(aq) \to \mathrm{Zn}^{2+}(aq) + \mathrm{Cu}(s)$

Cathode (reduction): $\mathrm{Cu}^{2+} + 2e^- \to \mathrm{Cu}$, $E^\circ = +0.34\mathrm{ V}$

Anode (oxidation): $\mathrm{Zn} \to \mathrm{Zn}^{2+} + 2e^-$, $E^\circ = -0.76\mathrm{ V}$

$$E^\circ_{\mathrm{cell}} = E^\circ_{\mathrm{cathode}} - E^\circ_{\mathrm{anode}} = 0.34 - (-0.76) = +1.10\mathrm{ V}$$

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

Cell notation: $\mathrm{Zn}(s) \mid \mathrm{Zn}^{2+}(aq) \parallel \mathrm{Cu}^{2+}(aq) \mid \mathrm{Cu}(s)$

Worked Example 6: Predicting Spontaneity

Will the following reaction occur spontaneously under standard conditions?

$\mathrm{Fe}^{3+}(aq) + \mathrm{I}^-(aq) \to \mathrm{Fe}^{2+}(aq) + \frac{1}{2}\mathrm{I}_2(s)$

Cathode (reduction -- more positive $E^\circ$): $\mathrm{Fe}^{3+} + e^- \to \mathrm{Fe}^{2+}$, $E^\circ = +0.77\mathrm{ V}$

Anode (oxidation -- less positive $E^\circ$): $2\mathrm{I}^- \to \mathrm{I}_2 + 2e^-$, so $E^\circ = +0.54\mathrm{ V}$ (reduction potential of I$_2$)

$$E^\circ_{\mathrm{cell}} = 0.77 - 0.54 = +0.23\mathrm{ V}$$

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

Electrochemical Series vs Reactivity Series

Electrochemical Series	Reactivity Series
Based on $E^\circ$ values (quantitative)	Based on observed reactions (qualitative)
Measures tendency to be reduced	Measures tendency to lose electrons (be oxidised)
More positive $E^\circ$ = stronger oxidising agent	More reactive metal = stronger reducing agent
$E^\circ$ for Li$^+$/Li = -3.04 V (least easily reduced)	K is the most reactive metal

The two series are consistent: a more reactive metal has a more negative $E^\circ$ value.

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Electrolysis

Principles

Electrolysis uses electrical energy to drive a non-spontaneous redox reaction.

Key components:

• Electrolyte: Ionic compound in molten or aqueous solution that conducts electricity.
• Anode (positive): Oxidation occurs (anions lose electrons).
• Cathode (negative): Reduction occurs (cations gain electrons).

Electrolysis of Molten Salts

Example: Molten NaCl

• Cathode: $\mathrm{Na}^+ + e^- \to \mathrm{Na}(l)$ (reduction)
• Anode: $2\mathrm{Cl}^- \to \mathrm{Cl}_2(g) + 2e^-$ (oxidation)
• Overall: $2\mathrm{NaCl}(l) \to 2\mathrm{Na}(l) + \mathrm{Cl}_2(g)$

Electrolysis of Aqueous Solutions

When water is present, both the cation and water compete for reduction at the cathode, and both the anion and water compete for oxidation at the anode.

Cathode (reduction):

Cation	Product
K$^+$, Na$^+$, Ca$^{2+}$, Mg$^{2+}$, Al$^{3+}$	H$_2$ (from water)
Zn$^{2+}$, Fe$^{2+}$, Ni$^{2+}$, Sn$^{2+}$, Pb$^{2+}$	Metal (variable, depends on conditions)
Cu$^{2+}$, Ag$^+$, Au$^{3+}$	Metal

Anode (oxidation):

Anion	Product
Halides (Cl$^-$, Br$^-$, I$^-$)	Halogen (Cl$_2$, Br$_2$, I$_2$)
SO$_4^{2-}$, NO$_3^-$, CO$_3^{2-}$	O$_2$ (from water)

At the anode (from water): $2\mathrm{H}_2\mathrm{O} \to \mathrm{O}_2 + 4\mathrm{H}^+ + 4e^-$

At the cathode (from water): $2\mathrm{H}_2\mathrm{O} + 2e^- \to \mathrm{H}_2 + 2\mathrm{OH}^-$

:::warning[Exam Tip] The rules above are simplified. In reality, the product at an electrode depends on the relative electrode potentials and concentrations (overpotential effects). For IB exams, use the rules as stated. :::

Faraday's Laws

Faraday's First Law: The amount of substance produced at an electrode is directly proportional to the quantity of charge passed.

$$m = \frac{Q \cdot M}{n \cdot F}$$

Faraday's Second Law: When the same quantity of charge is passed through different electrolytes, the masses of substances produced are proportional to their equivalent masses (molar mass divided by number of electrons transferred).

Where:

• $m$ = mass of substance produced (g)
• $Q$ = total charge (C) = $I \times t$
• $I$ = current (A)
• $t$ = time (s)
• $M$ = molar mass (g/mol)
• $n$ = number of moles of electrons transferred per mole of product
• $F$ = Faraday constant = $96\,500\mathrm{ C/mol}$

Worked Example 7: Electrolysis Calculation

A current of $2.50\mathrm{ A}$ is passed through molten Al$_2$O$_3$ for $30.0$ minutes. Calculate the mass of aluminium produced. ($M_r(\mathrm{Al}) = 27.0\mathrm{ g/mol}$)

Cathode half-reaction: $\mathrm{Al}^{3+} + 3e^- \to \mathrm{Al}$, so $n = 3$.

$$Q = I \times t = 2.50 \times 30.0 \times 60 = 4500\mathrm{ C}$$$$m = \frac{Q \times M}{n \times F} = \frac{4500 \times 27.0}{3 \times 96500} = \frac{121500}{289500} = 0.420\mathrm{ g}$$

Worked Example 8: Electrolysis of Aqueous CuSO$_4$

A current of $0.500\mathrm{ A}$ is passed through $200\mathrm{ mL}$ of $0.500\mathrm{ mol/L}$ CuSO$_4$(aq) using inert graphite electrodes for 2.00 hours.

(a) Write the half-equations for the reactions at each electrode.

(b) Calculate the mass of copper deposited at the cathode.

(c) Calculate the volume of gas produced at the anode at STP.

Answer:

(a)

• Cathode: $\mathrm{Cu}^{2+} + 2e^- \to \mathrm{Cu}(s)$ (Cu$^{2+}$ is below H in the reactivity series)
• Anode: $2\mathrm{H}_2\mathrm{O} \to \mathrm{O}_2(g) + 4\mathrm{H}^+(aq) + 4e^-$ (SO$_4^{2-}$ is not a halide, so water is oxidised)

(b)

$$Q = 0.500 \times 2.00 \times 3600 = 3600\mathrm{ C}$$

For Cu: $n = 2$, $M = 63.5\mathrm{ g/mol}$.

$$m = \frac{3600 \times 63.5}{2 \times 96500} = \frac{228600}{193000} = 1.18\mathrm{ g}$$

(c) For O$_2$: $n = 4$ electrons per mole of O$_2$.

$$n(\mathrm{O}_2) = \frac{Q}{n \times F} = \frac{3600}{4 \times 96500} = 0.00933\mathrm{ mol}$$

At STP ($22.4\mathrm{ L/mol}$):

$$V = 0.00933 \times 22.4 = 0.209\mathrm{ L} = 209\mathrm{ mL}$$

Worked Example 9: Faraday's Second Law

The same charge is passed through two separate electrolytic cells: one containing molten NaCl and the other containing molten Al$_2$O$_3$. If $2.30\mathrm{ g}$ of Na is produced in the first cell, what mass of Al is produced in the second?

Answer:

For Na: $\mathrm{Na}^+ + e^- \to \mathrm{Na}$, $n_1 = 1$, $M_1 = 23.0\mathrm{ g/mol}$. Equivalent mass of Na = $23.0/1 = 23.0\mathrm{ g/equiv}$.

For Al: $\mathrm{Al}^{3+} + 3e^- \to \mathrm{Al}$, $n_2 = 3$, $M_2 = 27.0\mathrm{ g/mol}$. Equivalent mass of Al = $27.0/3 = 9.0\mathrm{ g/equiv}$.

By Faraday's Second Law:

$$\frac{m(\mathrm{Na})}{m(\mathrm{Al})} = \frac{\mathrm{Equiv. mass of Na}}{\mathrm{Equiv. mass of Al}}$$$$\frac{2.30}{m(\mathrm{Al})} = \frac{23.0}{9.0}$$$$m(\mathrm{Al}) = \frac{2.30 \times 9.0}{23.0} = 0.900\mathrm{ g}$$

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Rusting and Corrosion Prevention

The Rusting Process

Iron rusting is an electrochemical process requiring both water and oxygen:

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

Cathode (reduction): $\mathrm{O}_2(g) + 2\mathrm{H}_2\mathrm{O}(l) + 4e^- \to 4\mathrm{OH}^-(aq)$

The Fe$^{2+}$ ions further react:

$$4\mathrm{Fe}^{2+}(aq) + \mathrm{O}_2(g) + 6\mathrm{H}_2\mathrm{O}(l) \to 4\mathrm{Fe(OH)}_3(s)$$

$\mathrm{Fe(OH)}_3$ dehydrates to form rust: $\mathrm{Fe}_2\mathrm{O}_3 \cdot n\mathrm{H}_2\mathrm{O}$.

Methods of Corrosion Prevention

Method	Description	How It Works
Coating (painting, oiling)	Physical barrier between Fe and O$_2$/H$_2$O	Prevents contact with the corrosive environment
Galvanising	Coating iron with a layer of zinc	Zn acts as a sacrificial anode (more reactive than Fe); even if scratched, Zn oxidises preferentially
Tin plating	Coating iron with tin	Provides a physical barrier; however, if scratched, Fe becomes the anode and rusts faster (since Sn is less reactive)
Sacrificial anode (cathodic protection)	Attaching a more reactive metal (Mg, Zn) to Fe	The more reactive metal oxidises in preference to Fe, protecting it
Alloying	Adding Cr and/or Ni to Fe to make stainless steel	Chromium forms a passive oxide layer that protects the surface
Electrochemical protection	Applying a negative voltage to the iron structure	Makes the iron the cathode, preventing its oxidation

:::warning[Exam Tip] The key distinction: galvanising (Zn coating) provides both barrier and sacrificial protection. Tin plating provides only barrier protection and actually accelerates rusting if the coating is damaged, because Sn is less reactive than Fe and Fe becomes the anode. :::

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Fuel Cells

A fuel cell converts the chemical energy of a fuel (usually H$_2$) directly into electrical energy. Unlike batteries, fuel cells consume fuel continuously and can operate as long as fuel is supplied.

Hydrogen-Oxygen Fuel Cell (Alkaline)

Anode (oxidation): $2\mathrm{H}_2(g) + 4\mathrm{OH}^-(aq) \to 4\mathrm{H}_2\mathrm{O}(l) + 4e^-$

Cathode (reduction): $\mathrm{O}_2(g) + 2\mathrm{H}_2\mathrm{O}(l) + 4e^- \to 4\mathrm{OH}^-(aq)$

Overall: $2\mathrm{H}_2(g) + \mathrm{O}_2(g) \to 2\mathrm{H}_2\mathrm{O}(l)$

Hydrogen-Oxygen Fuel Cell (Acidic)

Anode (oxidation): $2\mathrm{H}_2(g) \to 4\mathrm{H}^+(aq) + 4e^-$

Cathode (reduction): $\mathrm{O}_2(g) + 4\mathrm{H}^+(aq) + 4e^- \to 2\mathrm{H}_2\mathrm{O}(l)$

Overall: $2\mathrm{H}_2(g) + \mathrm{O}_2(g) \to 2\mathrm{H}_2\mathrm{O}(l)$

Advantages and Disadvantages

Advantages	Disadvantages
High efficiency (~60--80%)	H$_2$ is expensive to produce and store
No greenhouse gas emissions (only H$_2$O)	H$_2$ infrastructure is limited
Quiet operation	Fuel cells use expensive catalysts (Pt)
Continuous operation (refuel instead of recharge)	H$_2$ production often relies on fossil fuels

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Common Pitfalls

1. Anode/Cathode sign: In a galvanic cell, the anode is negative (oxidation) and the cathode is positive (reduction). In an electrolytic cell, the anode is positive and the cathode is negative. Remember: galvanic = spontaneous, electrolytic = forced.

2. $E^\circ$ values are not multiplied: $E^\circ$ is an intensive property. When balancing electrons, multiply the half-equation but NOT the $E^\circ$ value.

3. Cell notation direction: The anode is always written on the left, cathode on the right. Do not reverse this convention.

4. Electrolysis of aqueous solutions: Know the discharge rules. Na$^+$ and K$^+$ are never discharged from aqueous solution; H$_2$ is produced instead. Halides are discharged in preference to SO$_4^{2-}$.

5. Faraday constant: $F = 96\,500\mathrm{ C/mol}$ (not $96\,500\mathrm{ C}$). It has units of coulombs per mole of electrons.

6. Oxidation state of O in peroxides: In H$_2$O$_2$ and Na$_2$O$_2$, oxygen has an oxidation state of -1, not -2. In OF$_2$, oxygen is +2.

7. Oxidising/reducing agents: The oxidising agent is the species that gets reduced. The reducing agent is the species that gets oxidised. Do not confuse the agent with the process.

8. Electron flow vs current flow: Electrons flow from anode to cathode (negative to positive in the external circuit of a galvanic cell). Conventional current flows in the opposite direction.

9. n in Faraday calculations: n is the moles of electrons per mole of product, not the moles of product. For Al$^{3+}$ to Al, n = 3. For Cu$^{2+}$ to Cu, n = 2. Always check the half-equation.

10. Standard conditions: $E^\circ$ values apply only at standard conditions (1 mol/L, 100 kPa, 298 K). Changing concentration shifts the actual cell potential (Nernst equation), though IB typically does not require Nernst equation calculations.

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Practice Problems

Question 1: Oxidation States

Determine the oxidation state of the underlined element in each compound:

(a) K\underline{\mathrm{2}}$$\mathrm{Cr}_2$$\underline{\mathrm{O}}_7 (b) \underline{\mathrm{H}}_2$$\underline{\mathrm{O}}_2 (c) \underline{\mathrm{N}}$$\mathrm{H}_4^+ (d) \underline{\mathrm{Mn}}$$\mathrm{O}_4^{2-} (e) \underline{\mathrm{S}}$$\mathrm{O}_3^{2-}

Answer:

(a) K = +1 (two K = +2). O = -2 (seven O = -14). $+2 + 2\mathrm{Cr} - 14 = 0$, $2\mathrm{Cr} = +12$, Cr = +6.

(b) H = +1 (two H = +2). $+2 + 2\mathrm{O} = 0$, $2\mathrm{O} = -2$, O = -1. (Hydrogen peroxide.)

(c) H = +1 (four H = +4). $+4 + \mathrm{N} = +1$ (charge of ion), N = -3.

(d) O = -2 (four O = -8). $\mathrm{Mn} - 8 = -2$, Mn = +6.

(e) O = -2 (three O = -6). $\mathrm{S} - 6 = -2$, S = +4.

Question 2: Balancing Redox Equations

Balance the following redox equation in acidic medium:

$\mathrm{Cr}_2\mathrm{O}_7^{2-} + \mathrm{SO}_3^{2-} \to \mathrm{Cr}^{3+} + \mathrm{SO}_4^{2-}$

Answer:

Reduction (Cr: +6 to +3):

$\mathrm{Cr}_2\mathrm{O}_7^{2-} \to 2\mathrm{Cr}^{3+}$

Balance O: $\mathrm{Cr}_2\mathrm{O}_7^{2-} \to 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O}$

Balance H: $\mathrm{Cr}_2\mathrm{O}_7^{2-} + 14\mathrm{H}^+ \to 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O}$

Balance charge: Left = $-2 + 14 = +12$. Right = $+6$. Add 6 $e^-$ to left:

$\mathrm{Cr}_2\mathrm{O}_7^{2-} + 14\mathrm{H}^+ + 6e^- \to 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O}$

Oxidation (S: +4 to +6):

$\mathrm{SO}_3^{2-} + \mathrm{H}_2\mathrm{O} \to \mathrm{SO}_4^{2-} + 2\mathrm{H}^+ + 2e^-$

Multiply oxidation by 3:

$3\mathrm{SO}_3^{2-} + 3\mathrm{H}_2\mathrm{O} \to 3\mathrm{SO}_4^{2-} + 6\mathrm{H}^+ + 6e^-$

Add and simplify:

$\mathrm{Cr}_2\mathrm{O}_7^{2-} + 14\mathrm{H}^+ + 3\mathrm{SO}_3^{2-} + 3\mathrm{H}_2\mathrm{O} \to 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O} + 3\mathrm{SO}_4^{2-} + 6\mathrm{H}^+$

Simplify (cancel 6 H$^+$ and 3 H$_2$O):

$$\mathrm{Cr}_2\mathrm{O}_7^{2-} + 8\mathrm{H}^+ + 3\mathrm{SO}_3^{2-} \to 2\mathrm{Cr}^{3+} + 3\mathrm{SO}_4^{2-} + 4\mathrm{H}_2\mathrm{O}$$

Question 3: Cell EMF and Spontaneity

Given the following standard reduction potentials:

Half-Reaction	$E^\circ$ (V)
$\mathrm{Ag}^+ + e^- \to \mathrm{Ag}$	+0.80
$\mathrm{Fe}^{3+} + e^- \to \mathrm{Fe}^{2+}$	+0.77
$\mathrm{Zn}^{2+} + 2e^- \to \mathrm{Zn}$	-0.76
$\mathrm{Ni}^{2+} + 2e^- \to \mathrm{Ni}$	-0.25

(a) Calculate $E^\circ_{\mathrm{cell}}$ for the reaction: $\mathrm{Zn}(s) + 2\mathrm{Ag}^+(aq) \to \mathrm{Zn}^{2+}(aq) + 2\mathrm{Ag}(s)$

(b) Write the cell notation for this cell.

(c) Will Ni spontaneously reduce Fe$^{3+}$? Justify with a calculation.

Answer:

(a) Cathode: Ag$^+$/Ag, $E^\circ = +0.80\mathrm{ V}$. Anode: Zn$^{2+}$/Zn, $E^\circ = -0.76\mathrm{ V}$.

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

(b) $\mathrm{Zn}(s) \mid \mathrm{Zn}^{2+}(aq) \parallel \mathrm{Ag}^+(aq) \mid \mathrm{Ag}(s)$

(c) $\mathrm{Ni}(s) + 2\mathrm{Fe}^{3+}(aq) \to \mathrm{Ni}^{2+}(aq) + 2\mathrm{Fe}^{2+}(aq)$

Cathode: Fe$^{3+}$/Fe$^{2+}$, $E^\circ = +0.77\mathrm{ V}$. Anode: Ni$^{2+}$/Ni, $E^\circ = -0.25\mathrm{ V}$.

$$E^\circ_{\mathrm{cell}} = 0.77 - (-0.25) = +1.02\mathrm{ V}$$

Since $E^\circ_{\mathrm{cell}} = +1.02\mathrm{ V} \gt 0$, Ni will spontaneously reduce Fe$^{3+}$.

Question 4: Electrolysis Quantitative

A $3.00\mathrm{ A}$ current is passed through aqueous CuSO$_4$ using copper electrodes for 1.50 hours.

(a) Calculate the mass of copper deposited at the cathode.

(b) Explain why the mass of the anode decreases during electrolysis.

(c) Calculate the volume of oxygen gas produced at the anode at $25\degree\mathrm{C}$ and $100\mathrm{ kPa}$. (Use $R = 8.314\mathrm{ J/(mol}\cdot\mathrm{K)}$)

Answer:

(a)

$$Q = 3.00 \times 1.50 \times 3600 = 16200\mathrm{ C}$$$$m(\mathrm{Cu}) = \frac{Q \times M}{n \times F} = \frac{16200 \times 63.5}{2 \times 96500} = \frac{1028700}{193000} = 5.33\mathrm{ g}$$

(b) In electrolysis of CuSO$_4$ with copper electrodes (electrorefining), the anode is made of impure copper. At the anode, copper oxidises: $\mathrm{Cu}(s) \to \mathrm{Cu}^{2+}(aq) + 2e^-$. This means the anode loses mass as copper atoms dissolve into solution. The Cu$^{2+}$ ions then travel to the cathode where they are deposited as pure copper.

Note: With inert electrodes (e.g., graphite), oxygen is produced at the anode from water oxidation, not from copper dissolution. The question specifies copper electrodes, so copper oxidation occurs.

(c) With copper electrodes, the anode reaction is $\mathrm{Cu} \to \mathrm{Cu}^{2+} + 2e^-$, and no O$_2$ is produced. If the electrodes were inert (graphite), then:

At the anode: $2\mathrm{H}_2\mathrm{O} \to \mathrm{O}_2 + 4\mathrm{H}^+ + 4e^-$

$$n(\mathrm{O}_2) = \frac{Q}{4 \times F} = \frac{16200}{4 \times 96500} = 0.04197\mathrm{ mol}$$

Using ideal gas law at $T = 298\mathrm{ K}$ and $P = 100\mathrm{ kPa}$:

V = \frac`\{nRT}`{P} = \frac{0.04197 \times 8.314 \times 298}{100} = \frac{103.9}{100} = 1.04\mathrm{ L}

Question 5: Fuel Cell Calculations

A hydrogen-oxygen fuel cell (alkaline) operates at a constant current of $10.0\mathrm{ A}$.

(a) Write the overall reaction.

(b) Calculate the mass of hydrogen consumed per hour.

(c) If the fuel cell has an efficiency of 65% and $\Delta H$ for the reaction is $-286\mathrm{ kJ/mol}$, calculate the electrical energy output per hour.

Answer:

(a) $2\mathrm{H}_2(g) + \mathrm{O}_2(g) \to 2\mathrm{H}_2\mathrm{O}(l)$

(b) From the anode half-equation: $2\mathrm{H}_2 + 4\mathrm{OH}^- \to 4\mathrm{H}_2\mathrm{O} + 4e^-$

4 moles of electrons are produced per 2 moles of H$_2$ consumed. So 2 moles of $e^-$ per mole of H$_2$.

$$Q = 10.0 \times 3600 = 36000\mathrm{ C}$$$$n(\mathrm{H}_2) = \frac{Q}{2 \times F} = \frac{36000}{2 \times 96500} = 0.1865\mathrm{ mol}$$$$m(\mathrm{H}_2) = 0.1865 \times 2.016 = 0.376\mathrm{ g}$$

(c) Energy released by the reaction per hour:

$$\mathrm{Energy}_{\mathrm{chemical}} = n \times |\Delta H| = 0.1865 \times 286 = 53.3\mathrm{ kJ}$$$$\mathrm{Energy}_{\mathrm{electrical}} = 0.65 \times 53.3 = 34.7\mathrm{ kJ}$$

Question 6: Corrosion Prevention Analysis (Paper 2 Style)

An iron water pipe is to be protected from corrosion. Two methods are proposed: (i) connecting it to a block of magnesium, and (ii) coating it with tin.

(a) For each method, explain the electrochemical principle involved.

(b) If the tin coating is scratched, explain whether the pipe would still be protected. Use oxidation states and electrode potential reasoning.

(c) Write the half-equations for the corrosion of iron if neither protection method is applied.

Answer:

(a) (i) Sacrificial anode: Mg is more reactive than Fe (Mg has a more negative $E^\circ$, $-2.37\mathrm{ V}$ vs Fe's $-0.44\mathrm{ V}$). Mg acts as the anode and is oxidised: $\mathrm{Mg} \to \mathrm{Mg}^{2+} + 2e^-$. Fe becomes the cathode and is protected from oxidation. Mg is "sacrificed" instead of Fe.

(ii) Tin plating: Tin provides a physical barrier between Fe and the environment (water/oxygen). As long as the coating is intact, Fe cannot come into contact with H$_2$O and O$_2$, preventing corrosion.

(b) If the tin coating is scratched, the pipe would not be protected and would actually corrode faster. Sn has a less negative $E^\circ$ ($-0.14\mathrm{ V}$) than Fe ($-0.44\mathrm{ V}$), so at the scratch, Fe becomes the anode and Sn becomes the cathode:

• Anode: $\mathrm{Fe} \to \mathrm{Fe}^{2+} + 2e^-$ (oxidation of Fe)
• Cathode: $\mathrm{O}_2 + 2\mathrm{H}_2\mathrm{O} + 4e^- \to 4\mathrm{OH}^-$ (reduction at Sn surface)

Fe is preferentially oxidised, accelerating corrosion at the scratch.

(c) Without protection:

• Anode: $\mathrm{Fe}(s) \to \mathrm{Fe}^{2+}(aq) + 2e^-$
• Cathode: $\mathrm{O}_2(g) + 2\mathrm{H}_2\mathrm{O}(l) + 4e^- \to 4\mathrm{OH}^-(aq)$

Question 7: Predicting Redox Reactions (Paper 1 Style)

Which of the following species is the strongest oxidising agent?

A. $\mathrm{Na}^+$ B. $\mathrm{Zn}^{2+}$ C. $\mathrm{Cu}^{2+}$ D. $\mathrm{Ag}^+$

Answer: D. Ag$^+$

The strongest oxidising agent has the most positive $E^\circ$ value (most easily reduced).

• Na$^+$: $E^\circ = -2.71\mathrm{ V}$
• Zn$^{2+}$: $E^\circ = -0.76\mathrm{ V}$
• Cu$^{2+}$: $E^\circ = +0.34\mathrm{ V}$
• Ag$^+$: $E^\circ = +0.80\mathrm{ V}$

Ag$^+$ has the most positive $E^\circ$ and is therefore the strongest oxidising agent.

Question 8: Electrolysis with Variable Products

Aqueous NaCl is electrolysed using inert graphite electrodes.

(a) Write the half-equations at each electrode.

(b) What observation would be made at each electrode?

(c) If the electrolysis is continued for a long time, the solution becomes more acidic. Explain why.

Answer:

(a) Cathode: $2\mathrm{H}_2\mathrm{O} + 2e^- \to \mathrm{H}_2(g) + 2\mathrm{OH}^-(aq)$

Na$^+$ is not discharged because it is too reactive (above H in the reactivity series).

Anode: $2\mathrm{Cl}^- \to \mathrm{Cl}_2(g) + 2e^-$

Cl$^-$ is a halide and is discharged in preference to OH$^-$/water.

(b) Cathode: Bubbles of colourless gas (H$_2$) and the solution around the cathode turns basic (due to OH$^-$ production, detectable with universal indicator or phenolphthalein).

Anode: Bubbles of greenish-yellow gas with a pungent smell (Cl$_2$).

(c) If Cl$_2$ dissolves in the alkaline solution near the cathode, it disproportionates:

$\mathrm{Cl}_2 + 2\mathrm{OH}^- \to \mathrm{Cl}^- + \mathrm{ClO}^- + \mathrm{H}_2\mathrm{O}$

However, if Cl$_2$ escapes (which it tends to do), the remaining solution contains OH$^-$ and unreacted Na$^+$, making it basic, not acidic. The question implies that Cl$_2$ reacts with water:

$\mathrm{Cl}_2 + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{HCl} + \mathrm{HClO}$

This produces HCl (a strong acid), making the solution acidic. In practice, with thorough mixing, the solution becomes more acidic because Cl$_2$ dissolves and hydrolyses to form HCl and HClO, while the OH$^-$ at the cathode is consumed by this acid.

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