Electrochemistry Strategy — Concepts, Formulas & Examples

Electrochemistry is one of the most problem-heavy chapters in physical chemistry. NEET and JEE both test Nernst equation, electrolysis, EMF and conductance. This topic is a strategy guide, not a content dump.

Core Concepts

Know the standard electrode potentials

Memorise the electrochemical series — top is most reactive (K, Na, Ca), bottom is least (Au, Pt). The difference determines cell direction and EMF.

Key standard reduction potentials to memorise:

Half-reaction	$E^\circ$ (V)
$\text{Li}^+ + e^- \rightarrow \text{Li}$	-3.04
$\text{K}^+ + e^- \rightarrow \text{K}$	-2.93
$\text{Zn}^{2+} + 2e^- \rightarrow \text{Zn}$	-0.76
$\text{Fe}^{2+} + 2e^- \rightarrow \text{Fe}$	-0.44
$2\text{H}^+ + 2e^- \rightarrow \text{H}_2$	0.00 (reference)
$\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$	+0.34
$\text{Ag}^+ + e^- \rightarrow \text{Ag}$	+0.80
$\text{Au}^{3+} + 3e^- \rightarrow \text{Au}$	+1.50

The rule: The species with higher (more positive) reduction potential gets reduced. The other gets oxidised. The cell EMF = $E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$ .

Nernst equation

At non-standard conditions — $E = E^\circ - (0.0591/n)\log Q$ at 25°C. Always write Q as products over reactants, including activities of species involved.

E = E^\circ - \frac{0.0591}{n}\log Q

Where $E$ = cell potential, $E^\circ$ = standard cell potential, $n$ = number of electrons transferred, $Q$ = reaction quotient.

E = E^\circ - \frac{RT}{nF}\ln Q = E^\circ - \frac{2.303RT}{nF}\log Q

At 25°C (298 K): $\frac{2.303RT}{F} = \frac{2.303 \times 8.314 \times 298}{96500} = 0.0591$ V.

At equilibrium: $E = 0$ and $Q = K$ (equilibrium constant). So:

E^\circ = \frac{0.0591}{n}\log K

This connects thermodynamics ( $\Delta G$ ), equilibrium ( $K$ ), and electrochemistry ( $E^\circ$ ) — a favourite JEE question type.

Cell notation

Anode on left, cathode on right, with double vertical line for salt bridge. Oxidation happens at anode; reduction at cathode. Memorise ‘An Ox’ and ‘Red Cat’.

Example — Daniel cell:

\text{Zn(s)} | \text{Zn}^{2+}\text{(aq)} || \text{Cu}^{2+}\text{(aq)} | \text{Cu(s)}

Anode (left): $\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-$ (oxidation)
Cathode (right): $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$ (reduction)
$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.34 - (-0.76) = 1.10$ V

Gibbs energy and EMF

\Delta G = -nFE

Where $n$ = moles of electrons, $F$ = Faraday constant (96500 C/mol), $E$ = cell potential.

At standard conditions: $\Delta G^\circ = -nFE^\circ$

If $E > 0$ , $\Delta G < 0$ → spontaneous. If $E < 0$ , $\Delta G > 0$ → non-spontaneous (needs external power = electrolysis).

Electrolysis calculations

Use Faraday’s law — $m = (ZeIt)/F$ where Z is equivalent weight, I is current, t is time, F is Faraday constant (96500 C). Common mistake is forgetting to convert time to seconds.

First law: Mass deposited $\propto$ charge passed: $m = \frac{M \times I \times t}{n \times F}$

Where $M$ = molar mass, $I$ = current (A), $t$ = time (s), $n$ = electrons per ion, $F$ = 96500 C/mol.

Second law: When same charge passes through different electrolytes, masses deposited are proportional to their equivalent weights.

Conductance and conductivity

Conductance ( $G$ ) = $1/R$ (measured in siemens, S).

Conductivity ( $\kappa$ ) = $G \times (l/A)$ = conductance × cell constant. Units: S/m or S/cm.

Molar conductivity ( $\Lambda_m$ ) = $\kappa/c$ where $c$ is concentration in mol/L. Units: S cm $^2$ mol $^{-1}$ .

\Lambda_m^\circ = \nu_+ \lambda_+^\circ + \nu_- \lambda_-^\circ

The limiting molar conductivity of an electrolyte equals the sum of limiting ionic conductivities of its constituent ions, each multiplied by its stoichiometric coefficient.

Application: Kohlrausch’s law can calculate $\Lambda_m^\circ$ for weak electrolytes (like acetic acid) that cannot be measured directly:

\Lambda_m^\circ(\text{CH}_3\text{COOH}) = \Lambda_m^\circ(\text{CH}_3\text{COONa}) + \Lambda_m^\circ(\text{HCl}) - \Lambda_m^\circ(\text{NaCl})

Common problem types

Calculate EMF from standard potentials. 2. Apply Nernst to non-standard conditions. 3. Predict spontaneity ( $\Delta G = -nFE$ ). 4. Electrolysis — mass deposited. 5. Conductance and molar conductivity — dilution effects.

Problem-solving framework:

Always start by identifying what gets oxidised and what gets reduced.

The species with lower $E^\circ$ gets oxidised (anode). Higher $E^\circ$ gets reduced (cathode).

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

Plug in concentrations into $Q$ and use $E = E^\circ - (0.0591/n)\log Q$ .

Key Formulas

E = E^\circ - \frac{0.0591}{n}\log Q

Standard form at 25°C. Use ln form with RT/nF for other temperatures.

Worked Examples

If E is positive, the reaction is spontaneous as written. If E is negative, it is spontaneous in the reverse direction. Equal to $\Delta G = -nFE$ .

Higher T generally lowers EMF for reactions with negative entropy change. Nernst equation’s RT/nF term gets larger, affecting Q contribution.

How much copper is deposited when 2 A of current passes through CuSO $_4$ solution for 30 minutes?

$m = \frac{M \times I \times t}{n \times F} = \frac{63.5 \times 2 \times 1800}{2 \times 96500} = \frac{228600}{193000} = 1.18\;\text{g}$

Note: Cu $^{2+}$ requires 2 electrons per ion ( $n = 2$ ), and time must be in seconds (30 min = 1800 s).

For the Daniel cell: $E^\circ = 1.10$ V, $n = 2$ .

\log K = \frac{nE^\circ}{0.0591} = \frac{2 \times 1.10}{0.0591} = 37.2

K = 10^{37.2} \approx 1.6 \times 10^{37}

This enormous K confirms the reaction is essentially irreversible under standard conditions.

Solved Problems (Exam Style)

Problem 1 (JEE Main): The EMF of a cell is 1.10 V at 25°C. If [Zn $^{2+}$ ] = 0.1 M and [Cu $^{2+}$ ] = 1.0 M, find the new EMF.

E = E^\circ - \frac{0.0591}{n}\log\frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = 1.10 - \frac{0.0591}{2}\log\frac{0.1}{1.0}

E = 1.10 - 0.02955 \times (-1) = 1.10 + 0.03 = 1.13\;\text{V}

Lower product concentration (Zn $^{2+}$ ) and higher reactant concentration (Cu $^{2+}$ ) shift the reaction forward, increasing EMF.

Problem 2 (NEET): How long will it take to deposit 10.8 g of silver from AgNO $_3$ solution using 5 A current?

$m = \frac{M \times I \times t}{n \times F}$ . For Ag: $M = 108$ , $n = 1$ (Ag $^+ + e^- \rightarrow$ Ag).

$10.8 = \frac{108 \times 5 \times t}{1 \times 96500}$

$t = \frac{10.8 \times 96500}{108 \times 5} = \frac{1042200}{540} = 1930\;\text{s} \approx 32.2\;\text{min}$

Common Mistakes

Forgetting the sign convention. Reduction potentials are reported as written; for oxidation at the anode, you may need to flip the sign.

Mixing up Q direction. Q is always products over reactants.

Using natural log instead of log $_{10}$ in the Nernst equation at 25°C. The 0.0591 factor assumes log $_{10}$ .

Forgetting to convert time to seconds in Faraday’s law. 30 minutes = 1800 seconds, not 30.

Not multiplying by the stoichiometric coefficient in Faraday’s law. For Cu $^{2+}$ , $n = 2$ (needs 2 electrons). For Al $^{3+}$ , $n = 3$ . Getting $n$ wrong halves or thirds your answer.

Exam Weightage and Revision

This topic is a repeat performer in board papers and entrance exams. NEET typically asks one to two questions on the core mechanisms, CBSE boards give three to six marks, and state PMT papers often include a diagram-based long answer. The PYQs cluster around a small set of facts — lock those and you clear the topic.

JEE Main 2024 had a Nernst equation numerical. JEE Main 2023 tested Faraday’s law. NEET 2023 asked about the relationship between $\Delta G$ and $E$ . CBSE boards ask for cell diagram + EMF calculation as a five-mark question. Electrochemistry gives 2-3 questions across JEE + NEET every year.

When a question gives a scenario, identify the core mechanism first, then match it to the concepts above. Most wrong answers come from reading the scenario too quickly.

Before starting any problem, write the two half-reactions and identify anode and cathode. Half the mistakes come from skipping this step.

Practice Questions

Q1. Calculate $\Delta G^\circ$ for the Daniel cell ( $E^\circ = 1.10$ V, $n = 2$ ).

$\Delta G^\circ = -nFE^\circ = -2 \times 96500 \times 1.10 = -212300$ J/mol = -212.3 kJ/mol. The large negative value confirms the reaction is highly spontaneous.

Q2. Using Kohlrausch’s law, calculate $\Lambda_m^\circ$ for AgCl given: $\Lambda_m^\circ$ (AgNO $_3$ ) = 133.4, $\Lambda_m^\circ$ (NaCl) = 126.5, $\Lambda_m^\circ$ (NaNO $_3$ ) = 121.6 (all in S cm $^2$ mol $^{-1}$ ).

$\Lambda_m^\circ(\text{AgCl}) = \Lambda_m^\circ(\text{AgNO}_3) + \Lambda_m^\circ(\text{NaCl}) - \Lambda_m^\circ(\text{NaNO}_3)$ $= 133.4 + 126.5 - 121.6 = **138.3$ S cm $^2$ mol $^{-1}$ **

Q3. A cell has $E^\circ = 0.46$ V. What is the equilibrium constant if $n = 2$ ?

$\log K = nE^\circ/0.0591 = 2 \times 0.46/0.0591 = 15.6$ . $K = 10^{15.6} \approx 4 \times 10^{15}$ . A very large K — the reaction goes essentially to completion.

Q4. Why does molar conductivity increase on dilution for both strong and weak electrolytes?

For strong electrolytes: All ions are already dissociated. Dilution reduces interionic attractions (Debye-Huckel effects), allowing ions to move more freely. The increase is small and linear. For weak electrolytes: Dilution increases the degree of dissociation ( $\alpha$ ), producing more ions. The increase is dramatic. At infinite dilution, dissociation is complete.

Q5. In the electrolysis of water, what volume of H $_2$ is produced at STP when 96500 C of charge is passed?

At the cathode: $2\text{H}^+ + 2e^- \rightarrow \text{H}_2$ . 96500 C = 1 Faraday = 1 mole of electrons. Since 2 moles of electrons produce 1 mole of H $_2$ : 1 mole of electrons produces 0.5 mole of H $_2$ . Volume at STP = $0.5 \times 22.4 = **11.2$ L**.

FAQs

What is the difference between a galvanic cell and an electrolytic cell? A galvanic (voltaic) cell converts chemical energy to electrical energy spontaneously ( $E > 0$ , $\Delta G < 0$ ). An electrolytic cell uses external electrical energy to drive a non-spontaneous reaction ( $E < 0$ , $\Delta G > 0$ ). In a galvanic cell, the anode is negative; in electrolytic, the anode is positive.

Why is the Faraday constant 96500 C/mol? It is the charge of one mole of electrons: $F = N_A \times e = 6.022 \times 10^{23} \times 1.602 \times 10^{-19} = 96485 \approx 96500$ C/mol.

Can the EMF of a cell be negative? If you write the cell reaction in the non-spontaneous direction, $E$ will be negative. This means the reverse reaction is spontaneous. In practice, we arrange the cell so the anode (oxidation) is on the left and the cathode (reduction) is on the right, giving a positive EMF for the spontaneous direction.

Why do we use a salt bridge? The salt bridge completes the circuit by allowing ion flow between the two half-cells. Without it, charge would accumulate (positive in the anode compartment, negative in the cathode), and the cell would stop working. The salt bridge maintains electrical neutrality.

Electrochemistry rewards discipline. Write every step, label every species, and the Nernst equation gives up its answer without a fight.