Oxidation was originally defined as addition of oxygen or removal of hydrogen from a substance; reduction as removal of oxygen or addition of hydrogen. E.g., $2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$ — Mg is oxidised (gains oxygen).
Oxidation is the loss of electrons by a species; reduction is the gain of electrons. A redox reaction always involves both processes occurring simultaneously — one species is oxidised while another is reduced, since electrons lost by one must be gained by another.
The species that gets oxidised (loses electrons) is called the reducing agent (it causes reduction of the other species); the species that gets reduced (gains electrons) is the oxidising agent.
In the reaction $\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}$, identify the species oxidised, reduced, and the oxidising and reducing agents.
Zn loses 2 electrons to become Zn²⁺ — Zn is oxidised, and acts as the reducing agent.
Cu²⁺ gains 2 electrons to become Cu — Cu²⁺ is reduced, and acts as the oxidising agent.
The oxidation number of an atom is the hypothetical charge it would have if all bonds to atoms of different elements were considered fully ionic, with electrons assigned entirely to the more electronegative atom.
Calculate the oxidation number of manganese in $\text{KMnO}_4$.
K = +1, O = −2 (×4 = −8). Let Mn = $x$. $1 + x + (-8) = 0 \Rightarrow x = +7$
Calculate the oxidation number of sulphur in $\text{Na}_2\text{S}_2\text{O}_3$.
Na = +1 (×2 = +2), O = −2 (×3 = −6). Let S (×2) = $x$. $2 + x + (-6) = 0 \Rightarrow x = +4$, so average oxidation number of S = $+4/2 = +2$
| Type | Description | Example |
|---|---|---|
| Combination reaction | Two elements/compounds combine to form one product | $\text{C} + \text{O}_2 \rightarrow \text{CO}_2$ |
| Decomposition reaction | A compound breaks down into two or more simpler substances | $2\text{KClO}_3 \rightarrow 2\text{KCl} + 3\text{O}_2$ |
| Displacement reaction | One element displaces another from its compound | $\text{Fe} + \text{CuSO}_4 \rightarrow \text{FeSO}_4 + \text{Cu}$ |
| Disproportionation | The same element is simultaneously oxidised and reduced | $2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2$ |
In a disproportionation reaction, one species in an intermediate oxidation state is converted into both a higher and a lower oxidation state simultaneously. In $2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2$, oxygen goes from −1 (in H₂O₂) to −2 (in H₂O, reduced) and to 0 (in O₂, oxidised) at the same time.
Balance $\text{MnO}_4^- + \text{Fe}^{2+} \rightarrow \text{Mn}^{2+} + \text{Fe}^{3+}$ in acidic medium using the ion-electron method.
Oxidation half: $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$
Reduction half: $\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$
Multiply oxidation half by 5 and add: $\text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}$
A device that converts the chemical energy of a spontaneous redox reaction directly into electrical energy. It consists of two half-cells (electrodes dipped in electrolyte solutions), connected externally by a wire and internally by a salt bridge that maintains electrical neutrality by allowing ion flow.
By convention: anode | anode solution || cathode solution | cathode. E.g., the Daniell cell: $\text{Zn}(s) | \text{Zn}^{2+}(aq) \,||\, \text{Cu}^{2+}(aq) | \text{Cu}(s)$, where the single line represents a phase boundary and the double line represents the salt bridge.
The potential of an electrode measured relative to a Standard Hydrogen Electrode (SHE), assigned a reference potential of exactly 0 V, under standard conditions (1 M concentration, 1 atm pressure, 298 K).
A positive $E^\circ_{cell}$ indicates a spontaneous reaction as written.
Elements arranged in order of increasing standard reduction potential. Elements higher up (more negative E°) are stronger reducing agents and more easily oxidised (more reactive metals, e.g., Li, K, Na); elements lower down (more positive E°) are stronger oxidising agents (e.g., F₂, easily reduced).
Given $E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34\ \text{V}$ and $E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76\ \text{V}$, calculate the standard cell potential for the Daniell cell (Zn anode, Cu cathode).
$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.34 - (-0.76) = 1.10\ \text{V}$
Standard electrode potentials apply only under standard conditions (1 M, 1 atm). The Nernst equation calculates the actual electrode/cell potential under any non-standard concentration or pressure conditions.
At 298 K, this simplifies to: $E_{cell} = E^\circ_{cell} - \dfrac{0.0591}{n}\log Q$, where $n$ = number of electrons transferred, $F$ = Faraday constant, and $Q$ = reaction quotient.
Calculate the cell potential for $\text{Zn}(s) | \text{Zn}^{2+}(0.1\ M) \,||\, \text{Cu}^{2+}(1\ M) | \text{Cu}(s)$ at 298 K, given $E^\circ_{cell} = 1.10\ \text{V}$.
$n = 2$. $Q = \dfrac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = \dfrac{0.1}{1} = 0.1$
$E_{cell} = 1.10 - \dfrac{0.0591}{2}\log(0.1) = 1.10 - \dfrac{0.0591}{2}(-1) = 1.10 + 0.0296 = 1.13\ \text{V}$
where $F$ = Faraday constant $= 96{,}500\ \text{C/mol}$. A positive $E_{cell}$ gives a negative $\Delta G$, confirming the reaction is spontaneous — consistent with thermodynamic criteria studied earlier.
Calculate ΔG° for a cell reaction with n = 2 and E°cell = 1.10 V.
$\Delta G^\circ = -nFE^\circ_{cell} = -(2)(96500)(1.10) = -212{,}300\ \text{J/mol} = -212.3\ \text{kJ/mol}$
Conductivity (κ, specific conductance) is the conductance of a solution of unit length and unit cross-sectional area — it decreases with dilution for strong electrolytes because the number of ions per unit volume decreases. Molar conductivity (Λm) is the conductivity contributed by one mole of electrolyte, and it increases with dilution for both strong and weak electrolytes.
where $\kappa$ = conductivity (S/cm), $C$ = molar concentration (mol/L). Units of $\Lambda_m$: S cm² mol⁻¹.
For strong electrolytes, ionic interactions decrease on dilution, allowing ions to move more freely, so Λm rises modestly and reaches a limiting value ($\Lambda_m^0$) at infinite dilution. For weak electrolytes, dilution increases the degree of dissociation substantially — producing far more ions — so Λm rises sharply, often without a clear plateau, as concentration approaches zero.
The conductivity of a 0.01 M KCl solution is $0.0014\ \text{S/cm}$. Calculate its molar conductivity.
$\Lambda_m = \dfrac{\kappa \times 1000}{C} = \dfrac{0.0014 \times 1000}{0.01} = 140\ \text{S cm}^2\text{mol}^{-1}$
The limiting molar conductivity ($\Lambda_m^0$) of an electrolyte at infinite dilution equals the sum of the individual contributions of its constituent ions, regardless of the nature of the other ion present — this is the law of independent migration of ions.
where $\nu_+, \nu_-$ = number of cations/anions per formula unit, and $\lambda^0_+, \lambda^0_-$ = limiting molar conductivities of the individual ions.
Weak electrolytes (like acetic acid) never reach a clear plateau even at very high dilution, so their $\Lambda_m^0$ can't be measured directly by extrapolation. Kohlrausch's law solves this: $\Lambda_m^0(\text{CH}_3\text{COOH}) = \Lambda_m^0(\text{HCl}) + \Lambda_m^0(\text{CH}_3\text{COONa}) - \Lambda_m^0(\text{NaCl})$ — all values on the right are for strong electrolytes, which are easy to measure directly.
Given $\Lambda_m^0(\text{HCl}) = 426.2$, $\Lambda_m^0(\text{CH}_3\text{COONa}) = 91.0$, $\Lambda_m^0(\text{NaCl}) = 126.5$ (all in S cm² mol⁻¹), calculate $\Lambda_m^0(\text{CH}_3\text{COOH})$.
$\Lambda_m^0(\text{CH}_3\text{COOH}) = 426.2 + 91.0 - 126.5 = 390.7\ \text{S cm}^2\text{mol}^{-1}$
Unlike a galvanic cell, an electrolytic cell uses electrical energy from an external source to drive a non-spontaneous redox reaction (electrolysis). Here, the anode is connected to the positive terminal of the external source (oxidation still occurs at the anode), and the cathode to the negative terminal (reduction still occurs at the cathode) — but the anode is now positive and cathode negative, opposite to a galvanic cell.
The mass ($m$) of a substance deposited/liberated at an electrode is directly proportional to the quantity of charge ($Q = It$) passed through the electrolyte, where $Z$ = electrochemical equivalent.
When the same quantity of charge is passed through different electrolytes connected in series, the masses of substances deposited are proportional to their chemical equivalent weights.
$$\dfrac{m_1}{m_2} = \dfrac{E_1}{E_2}$$Calculate the mass of copper deposited when a current of 2 A is passed through CuSO₄ solution for 1 hour. (Molar mass of Cu = 63.5, n = 2)
$Q = It = 2 \times 3600 = 7200\ \text{C}$
Moles of electrons $= Q/F = 7200/96500 = 0.0746\ \text{mol}$
Cu²⁺ + 2e⁻ → Cu, so moles of Cu deposited $= 0.0746/2 = 0.0373\ \text{mol}$
Mass $= 0.0373 \times 63.5 = 2.37\ \text{g}$
A primary battery (e.g., dry cell) cannot be recharged — its electrode reactions are not reversible, so once the reactants are consumed, the battery is dead. A secondary battery (e.g., lead-acid battery, used in vehicles) can be recharged by passing current in the reverse direction, reversing the cell reaction and regenerating the original reactants.
A fuel cell (e.g., the hydrogen-oxygen fuel cell) generates electricity directly from the continuous supply of a fuel (like H₂) and an oxidant (like O₂), rather than from a fixed, sealed amount of reactant like ordinary batteries — making it more efficient and eco-friendly, since the product is simply water.
Corrosion (e.g., rusting of iron) is an electrochemical process where a metal is gradually oxidised by oxygen (and moisture) in the environment, forming an oxide/hydroxide layer. It essentially involves the metal acting as an anode (oxidised to metal ions), atmospheric oxygen acting as the cathode (reduced, in presence of water, to hydroxide ions), which then combine to form rust.