Redox Reactions and Electrochemistry — NEET Chemistry

Every battery, every rusting nail, every metabolic energy step is electron transfer — one species loses electrons while another gains them. Redox is the bookkeeping of those electrons; electrochemistry turns that transfer into (or draws it from) electricity. NEET pulls 2–3 questions from this block — oxidation numbers, cell EMF, the Nernst equation and Faraday's laws recur every year. This chapter fixes redox balancing, builds the galvanic cell and the electrochemical series, derives the Nernst equation and its link to free energy, and works through electrolysis and the real devices that use it.

Part A — Redox Reactions

1. Oxidation, reduction and oxidation number

The modern definition is in terms of electrons:

• Oxidation = loss of electrons (oxidation number increases).
• Reduction = gain of electrons (oxidation number decreases).
• OIL RIG — Oxidation Is Loss, Reduction Is Gain.

An oxidising agent accepts electrons (is itself reduced); a reducing agent donates electrons (is itself oxidised). The two always occur together — you cannot have one without the other.

Oxidation number (ON) rules (apply in order):

1. Free element = 0 (O₂, Na, P₄).
2. Monatomic ion = its charge (Na⁺ = +1, S²⁻ = −2).
3. O = −2 (except peroxides −1, superoxides −½, OF₂ +2).
4. H = +1 (except metal hydrides −1, e.g. NaH).
5. F = −1 always; group 1 = +1, group 2 = +2.
6. Sum of ON = 0 for a neutral molecule, = charge for an ion.

Worked example 1.1. Oxidation number of Mn in KMnO₄? K = +1, O = −2 (×4 = −8). Let Mn = $x$: $+1+x-8=0\Rightarrow x=+7$. Manganese is in its highest oxidation state — a strong oxidiser.

Worked example 1.2. Oxidation number of S in H₂SO₄? H = +1 (×2 = +2), O = −2 (×4 = −8): $2+x-8=0\Rightarrow x=+6$.

2. Types of redox reactions and balancing

Four types NEET names:

• Combination: C + O₂ → CO₂.
• Decomposition: 2KClO₃ → 2KCl + 3O₂.
• Displacement: Zn + CuSO₄ → ZnSO₄ + Cu (a more reactive metal displaces a less reactive one).
• Disproportionation: the same element is simultaneously oxidised and reduced, e.g. $2H_2{O}_2\to 2H_{2}O+O_2$ (O goes −1 → −2 and −1 → 0); Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O.

Balancing by the ion–electron (half-reaction) method: split into oxidation and reduction half-reactions, balance atoms then charge (add electrons), balance O with H₂O, H with H⁺ (acidic) or OH⁻ (basic), then add so electrons cancel.

Worked example 2.1. Which is the oxidising agent in Zn + CuSO₄ → ZnSO₄ + Cu? Zn goes 0 → +2 (oxidised, reducing agent); Cu goes +2 → 0 (reduced). The species reduced is the oxidising agent → Cu²⁺ (CuSO₄).

Part B — Electrochemistry

3. Galvanic (voltaic) cells

A galvanic cell converts the energy of a spontaneous redox reaction into electricity. In the Daniell cell (Zn | Zn²⁺ ‖ Cu²⁺ | Cu):

• Anode (Zn) — oxidation, the negative terminal: $Zn\to Z{n}^{2+}+2e^{-}$.
• Cathode (Cu) — reduction, the positive terminal: $C{u}^{2+}+2e^{-}\to Cu$.
• Electrons flow through the external wire from anode to cathode; the salt bridge completes the circuit and maintains electrical neutrality.

Mnemonic: An Ox (anode oxidation), Red Cat (reduction cathode). In a galvanic cell anode is −, cathode is + (this reverses in electrolysis — a classic trap).

Cell notation: anode on the left, cathode on the right, ‖ for the salt bridge: $Zn|Z{n}^{2+}\Vert C{u}^{2+}|Cu$.

4. Electrode potential and the electrochemical series

Every electrode has a tendency to be reduced, measured by its standard reduction potential $E^{\ominus }$ (at 298 K, 1 M, 1 bar), referenced to the standard hydrogen electrode (SHE) defined as $E^{\ominus }=0$.

The electrochemical series ranks electrodes by $E^{\ominus }$:

• More positive $E^{\ominus }$ → stronger tendency to be reduced → better oxidising agent (F₂ at the top, +2.87 V).
• More negative $E^{\ominus }$ → stronger tendency to be oxidised → better reducing agent (Li at the bottom, −3.05 V).

The standard EMF of a cell:

$E_{\text{cell}}^{\ominus }=E_{\text{cathode}}^{\ominus }-E_{\text{anode}}^{\ominus }$

A positive $E_{\text{cell}}^{\ominus }$ means a spontaneous cell reaction (and a working galvanic cell). A metal higher in the reactivity series (more negative $E^{\ominus }$) displaces one lower down from solution.

Worked example 4.1. For the Daniell cell, $E^{\ominus }(C{u}^{2+}/Cu)=+0.34$ V, $E^{\ominus }(Z{n}^{2+}/Zn)=-0.76$ V. Find $E_{\text{cell}}^{\ominus }$. Cu²⁺ is reduced (cathode), Zn is oxidised (anode): $E_{\text{cell}}^{\ominus }=E_{\text{cathode}}^{\ominus }-E_{\text{anode}}^{\ominus }=0.34-(-0.76)=+1.10$ V. Positive → spontaneous.

Worked example 4.2. Can Cu displace Zn from ZnSO₄? No. Cu (E° = +0.34) is below Zn (E° = −0.76) in reactivity; only a metal with a more negative E° can displace one with a more positive E°. Zn displaces Cu, not the reverse.

5. The Nernst equation

Standard potentials assume 1 M concentrations. The Nernst equation gives the potential at any concentration:

$E_{\text{cell}}=E_{\text{cell}}^{\ominus }-\frac{0.0591}{n}\log Q(\text{at 298 K})$

where $n$ = electrons transferred and $Q$ = reaction quotient. For a single electrode $M^{n+}+n{e}^{-}\to M$:

$E=E^{\ominus }+\frac{0.0591}{n}\log [M^{n+}]$

• Increasing product concentration ($Q$ up) lowers the cell potential.
• At equilibrium $E_{\text{cell}}=0$ and $Q=K$.

Worked example 5.1. A Daniell cell has $[Z{n}^{2+}]=0.1$ M and $[C{u}^{2+}]=0.1$ M. Find $E_{\text{cell}}$ ($E^{\ominus }=1.10$ V, $n=2$). $Q=\frac{[Z{n}^{2+}]}{[C{u}^{2+}]}=\frac{0.1}{0.1}=1$, so $\log Q=0$ and $E_{\text{cell}}=1.10-0=1.10$ V (unchanged, since the ratio is 1).

6. EMF, free energy and the equilibrium constant

The cell potential links directly to thermodynamics:

$\Delta G=-nF{E}_{\text{cell}},\Delta G^{\ominus }=-nF{E}_{\text{cell}}^{\ominus }=-RT\ln K$

where $F=96500$ C/mol (Faraday constant). Consequences NEET tests:

• $E_{\text{cell}}^{\ominus }>0\iff \Delta G^{\ominus }<0\iff K>1$ → spontaneous.
• Combining the last two: $\log K=\frac{n{E}_{\text{cell}}^{\ominus }}{0.0591}$ at 298 K.

Worked example 6.1. Find $\Delta G^{\ominus }$ for the Daniell cell ($n=2$, $E^{\ominus }=1.10$ V). $\Delta G^{\ominus }=-nF{E}^{\ominus }=-2\times 96500\times 1.10=-212300$ J $=-212.3$ kJ/mol. Large negative → strongly spontaneous.

7. Conductance and Kohlrausch's law

Electrolyte solutions conduct via ion movement. Key quantities:

• Conductivity $\kappa$ (S/cm) — conductance of a unit cube of solution; decreases on dilution (fewer ions per volume).
• Molar conductivity ${\Lambda }_m=\frac{\kappa \times 1000}{C}$ (S cm² mol⁻¹) — increases on dilution (each ion moves more freely). For strong electrolytes it rises slowly toward ${\Lambda }_m^0$; for weak electrolytes it rises steeply near infinite dilution (more dissociation).

Kohlrausch's law of independent migration: at infinite dilution each ion contributes independently:

${\Lambda }_m^0={\nu }_{+}{\lambda }_{+}^0+{\nu }_{-}{\lambda }_{-}^0$

This lets you find ${\Lambda }_m^0$ for a weak electrolyte (e.g. acetic acid) from strong-electrolyte values, and hence its degree of dissociation $\alpha ={\Lambda }_m/{\Lambda }_m^0$.

Worked example 7.1. Why does molar conductivity of a weak acid rise sharply on dilution while its conductivity falls? Conductivity depends on ion concentration, which dilution lowers → $\kappa$ falls. Molar conductivity accounts for concentration; dilution greatly increases the degree of dissociation of a weak acid, releasing far more ions per mole → ${\Lambda }_m$ rises steeply.

8. Electrolysis and Faraday's laws

Electrolysis uses external electricity to drive a non-spontaneous redox reaction. Here polarity reverses: cathode is negative (reduction), anode is positive (oxidation).

Faraday's first law: the mass deposited is proportional to charge passed:

$m=Zq=\frac{M}{nF}It$

where $I$ = current, $t$ = time, $M$ = molar mass, $n$ = electrons per ion, $F=96500$ C/mol. One faraday (96500 C) deposits one gram-equivalent.

Faraday's second law: the same charge deposits masses in the ratio of their equivalent masses.

Worked example 8.1. What mass of copper is deposited by 2 A for 965 s from CuSO₄ solution? (Cu = 63.5, $n=2$.) Charge $q=It=2\times 965=1930$ C. $m=\frac{M}{nF}q=\frac{63.5}{2\times 96500}\times 1930=\frac{63.5}{100}\times 1=0.635$ g.

Worked example 8.2. How many faradays are needed to deposit 1 mole of Al from Al³⁺? Al³⁺ + 3e⁻ → Al needs 3 electrons per atom, so 3 faradays (3 × 96500 C) per mole.

9. Batteries, fuel cells and corrosion

• Primary cells (non-rechargeable): dry cell (Leclanché, Zn anode, MnO₂), mercury cell (constant voltage).
• Secondary cells (rechargeable): lead–acid battery (Pb / PbO₂ in H₂SO₄, used in cars), nickel–cadmium.
• Fuel cells convert the energy of a fuel directly to electricity — the H₂–O₂ fuel cell (used in spacecraft) is efficient and pollution-free, producing only water.
• Corrosion (rusting) is an electrochemical process: iron is oxidised at anodic spots, O₂ reduced at cathodic spots, forming hydrated iron(III) oxide. Prevented by barrier coating (paint, galvanising with zinc) or cathodic (sacrificial) protection using a more reactive metal (Mg, Zn) that corrodes instead.

Worked example 9.1. Why does galvanising (zinc coating) protect iron even if the coating is scratched? Zinc is more reactive (more negative $E^{\ominus }$) than iron, so it acts as the sacrificial anode — it is oxidised in preference to iron, protecting the iron even where the coating is broken. This is cathodic protection.

10. Common traps NEET sets here

• OIL RIG — oxidation is loss, reduction is gain of electrons; the oxidising agent is reduced.
• Peroxide/hydride exceptions — O = −1 in peroxides, H = −1 in metal hydrides.
• Galvanic vs electrolytic polarity: galvanic anode −, electrolytic anode +. Oxidation is always at the anode, though.
• $E_{\text{cell}}^{\ominus }=E_{\text{cathode}}^{\ominus }-E_{\text{anode}}^{\ominus }$ — always cathode minus anode, using reduction potentials.
• Nernst sign: $E$ decreases as $\log Q$ increases; $-0.0591/n$, not $+$.
• $\Delta G=-nFE$ — a positive $E$ gives negative $\Delta G$ (spontaneous).
• Conductivity falls, molar conductivity rises on dilution — opposite directions.
• One faraday = one gram-equivalent; count electrons $n$ correctly (Al needs 3).

11. Memory aids

• "OIL RIG" — Oxidation Is Loss, Reduction Is Gain.
• "An Ox, Red Cat" — Anode Oxidation, Reduction Cathode (always true).
• "Cathode minus anode" — $E_{\text{cell}}^{\ominus }$ from reduction potentials.
• "Higher $E^{\ominus }$ = better oxidiser" — top of the electrochemical series.
• "$-nFE$ = go" — negative $\Delta G$ from positive EMF.
• "$m=MIt/nF$" — Faraday's first law in one line.
• "Sacrifice the reactive metal" — cathodic protection.

12. Exam protocol

1. Assign oxidation numbers by the rules; identify what is oxidised/reduced and the agents (OIL RIG).
2. Balance redox by half-reactions; recognise disproportionation (same element up and down).
3. Galvanic cell: anode oxidation (−), cathode reduction (+); electrons flow anode → cathode.
4. $E_{\text{cell}}^{\ominus }=E_{\text{cathode}}^{\ominus }-E_{\text{anode}}^{\ominus }$; positive means spontaneous.
5. Non-standard conditions: apply Nernst $E=E^{\ominus }-\frac{0.0591}{n}\log Q$.
6. Link EMF to thermodynamics: $\Delta G^{\ominus }=-nF{E}^{\ominus }=-RT\ln K$.
7. Conductance: ${\Lambda }_m=\kappa \times 1000/C$; use Kohlrausch for weak-electrolyte ${\Lambda }_m^0$ and $\alpha$.
8. Electrolysis: reverse the polarity; apply $m=MIt/nF$ (one faraday = one equivalent).