Electrochemistry — Cells, Nernst Equation, Electrolysis, Corrosion

Where Chemistry Meets Electricity

Electrochemistry is about the interconversion of chemical and electrical energy. Galvanic cells convert chemical energy to electrical energy (batteries). Electrolytic cells do the reverse — using electricity to drive non-spontaneous reactions. This chapter is one of the highest-scoring in Class 12 chemistry for both boards and JEE.

The chapter has three big ideas: (1) cell potential and the Nernst equation, (2) Faraday’s laws of electrolysis, (3) conductance and Kohlrausch’s law. Master the formulas and their applications and this becomes a reliable 8-10 mark scorer.

Core Concepts

Galvanic (Voltaic) cells

A spontaneous chemical reaction generates electricity. The classic example is the Daniell cell: Zn in ZnSO $_4$ solution (anode) and Cu in CuSO $_4$ solution (cathode), connected by a salt bridge.

Anode (oxidation): $\text{Zn} \to \text{Zn}^{2+} + 2e^-$ (negative terminal in galvanic cell) Cathode (reduction): $\text{Cu}^{2+} + 2e^- \to \text{Cu}$ (positive terminal in galvanic cell)

Salt bridge: KCl or KNO $_3$ in agar. Maintains electrical neutrality by allowing ion flow between half-cells. Without it, charge buildup stops the reaction.

Cell notation: $\text{Zn}(s) | \text{Zn}^{2+}(aq) || \text{Cu}^{2+}(aq) | \text{Cu}(s)$ . Single vertical line = phase boundary. Double line = salt bridge. Anode on left, cathode on right.

Electrolytic cells

An external power source drives a non-spontaneous reaction. The polarity is reversed compared to galvanic cells:

Anode is positive (connected to + terminal of battery)
Cathode is negative (connected to - terminal)
But oxidation still occurs at the anode and reduction at the cathode

Examples: electrolysis of molten NaCl, electroplating, aluminium extraction (Hall-Heroult process).

Essential Formulas

E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}}

For a spontaneous reaction: $E°_{\text{cell}} > 0$ . Higher reduction potential = stronger oxidising agent. More negative reduction potential = stronger reducing agent.

E_{\text{cell}} = E°_{\text{cell}} - \frac{RT}{nF}\ln Q = E°_{\text{cell}} - \frac{0.0591}{n}\log Q \quad \text{(at 25°C)}

$Q$ = reaction quotient = $[\text{products}]/[\text{reactants}]$ (same form as equilibrium expression).

At equilibrium: $E_{\text{cell}} = 0$ and $\ln K = \frac{nFE°}{RT}$ , so $E° = \frac{0.0591}{n}\log K$ .

\Delta G° = -nFE°_{\text{cell}}

$n$ = moles of electrons transferred, $F = 96485$ C/mol. If $E°_{\text{cell}} > 0$ , then $\Delta G° < 0$ (spontaneous). This links thermodynamics to electrochemistry.

First law: Mass deposited $\propto$ charge passed: $m = \frac{MIt}{nF}$

where $M$ = molar mass, $I$ = current (A), $t$ = time (s), $n$ = electrons per ion, $F$ = 96485 C/mol.

Second law: For the same charge, mass deposited $\propto$ equivalent weight ( $M/n$ ).

Specific conductance: $\kappa = \frac{1}{\rho}$ (S/cm)

Molar conductivity: $\Lambda_m = \frac{\kappa \times 1000}{M}$ (S cm $^2$ mol $^{-1}$ ), where $M$ = molarity.

Kohlrausch’s Law: $\Lambda°_m = \nu_+ \lambda°_+ + \nu_- \lambda°_-$

Limiting molar conductivity = sum of ionic contributions. Used to calculate $\Lambda°_m$ for weak electrolytes (which cannot be measured directly by extrapolation).

Conductance — variation with dilution

Strong electrolytes (NaCl, KCl, HCl): $\Lambda_m$ increases slightly with dilution (less interionic attraction). The graph of $\Lambda_m$ vs $\sqrt{c}$ is a straight line (Debye-Huckel-Onsager equation). Extrapolation to $c = 0$ gives $\Lambda°_m$ .

Weak electrolytes (CH $_3$ COOH, NH $_4$ OH): $\Lambda_m$ increases sharply with dilution because the degree of dissociation increases. The graph rises steeply at low concentrations. $\Lambda°_m$ cannot be determined by extrapolation — Kohlrausch’s law is needed.

Electrochemical series (standard electrode potentials)

A table of half-reactions arranged by $E°$ value (most negative at the top, most positive at the bottom).

Key entries to memorise:

Half-reaction	$E°$ (V)
Li $^+$ /Li	-3.04
K $^+$ /K	-2.93
Na $^+$ /Na	-2.71
Zn $^{2+}$ /Zn	-0.76
Fe $^{2+}$ /Fe	-0.44
H $^+$ /H $_2$	0.00 (reference)
Cu $^{2+}$ /Cu	+0.34
Ag $^+$ /Ag	+0.80
Au $^{3+}$ /Au	+1.50
F $_2$ /F $^-$	+2.87

The species at the top (more negative $E°$ ) are strong reducing agents. The species at the bottom (more positive $E°$ ) are strong oxidising agents.

Corrosion

Electrochemical oxidation of metals in the presence of moisture and oxygen. Rusting of iron:

Anode (iron surface): $\text{Fe} \to \text{Fe}^{2+} + 2e^-$ Cathode (less active area): $\text{O}_2 + 4\text{H}^+ + 4e^- \to 2\text{H}_2\text{O}$

Fe $^{2+}$ is further oxidised to Fe $^{3+}$ , which forms hydrated iron(III) oxide — rust ( $\text{Fe}_2\text{O}_3 \cdot x\text{H}_2\text{O}$ ).

Prevention methods: galvanisation (zinc coating), cathodic protection (sacrificial anode), painting, oiling, alloying (stainless steel with Cr and Ni), electroplating.

Corrosion as an electrochemical process is a 2-3 mark question in CBSE boards. Remember: the iron itself is the anode (oxidised), and a less reactive area with moisture acts as the cathode. Galvanisation works because zinc ( $E° = -0.76$ V) is more reactive than iron ( $E° = -0.44$ V) and is oxidised preferentially.

Solved Examples

Example 1 (Easy — CBSE)

$E°_{\text{Cu}^{2+}/\text{Cu}} = +0.34$ V, $E°_{\text{Zn}^{2+}/\text{Zn}} = -0.76$ V.

Cathode = Cu (higher $E°$ ). $E°_{\text{cell}} = 0.34 - (-0.76) = \mathbf{1.10}$ V.

Example 2 (Medium — JEE Main)

For the Daniell cell, find $E_{\text{cell}}$ when $[\text{Zn}^{2+}] = 0.1$ M and $[\text{Cu}^{2+}] = 1$ M at 25°C.

$Q = [\text{Zn}^{2+}]/[\text{Cu}^{2+}] = 0.1/1 = 0.1$

$E = 1.10 - \frac{0.0591}{2}\log(0.1) = 1.10 - \frac{0.0591}{2}(-1) = 1.10 + 0.0296 = \mathbf{1.13}$ V.

Note: reducing $[\text{Zn}^{2+}]$ (product) increases the cell potential — Le Chatelier applied to electrochemistry.

Example 3 (Medium — CBSE)

How much copper is deposited by passing 2 A for 1 hour through CuSO $_4$ solution?

$m = \frac{MIt}{nF} = \frac{63.5 \times 2 \times 3600}{2 \times 96485} = \frac{457200}{192970} = \mathbf{2.37}$ g.

Note: $n = 2$ because Cu $^{2+}$ gains 2 electrons. Time must be in seconds.

Example 4 (Hard — JEE Main)

For a cell with $E° = 0.295$ V and $n = 2$ :

$\log K = \frac{nE°}{0.0591} = \frac{2 \times 0.295}{0.0591} = 9.98$

$K = 10^{9.98} \approx 10^{10}$ .

A large $K$ confirms the reaction is strongly spontaneous (goes nearly to completion).

Example 5 (Application)

Given: $\Lambda°_m(\text{NaCl}) = 126.5$ , $\Lambda°_m(\text{NaOH}) = 248.1$ , $\Lambda°_m(\text{HCl}) = 426.2$ S cm $^2$ /mol.

$\Lambda°_m(\text{H}_2\text{O}) = \Lambda°_m(\text{HCl}) + \Lambda°_m(\text{NaOH}) - \Lambda°_m(\text{NaCl})$ $= 426.2 + 248.1 - 126.5 = \mathbf{547.8}$ S cm $^2$ /mol.

This works because: $\lambda°(\text{H}^+) + \lambda°(\text{OH}^-) = [\lambda°(\text{H}^+) + \lambda°(\text{Cl}^-)] + [\lambda°(\text{Na}^+) + \lambda°(\text{OH}^-)] - [\lambda°(\text{Na}^+) + \lambda°(\text{Cl}^-)]$ .

Common Mistakes to Avoid

Wrong sign for anode potential. $E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}}$ . Students sometimes add both values or subtract in the wrong order. Remember: cathode minus anode, using the standard reduction potentials as given.

Wrong Q in the Nernst equation. $Q = [\text{products}]/[\text{reactants}]$ — the same reaction quotient as in equilibrium. For the Daniell cell: $Q = [\text{Zn}^{2+}]/[\text{Cu}^{2+}]$ , not the reverse.

Forgetting $n$ in Faraday’s law. For Cu $^{2+}$ deposition, $n = 2$ (each Cu $^{2+}$ gains 2 electrons). For Ag $^+$ , $n = 1$ . Using the wrong $n$ doubles or halves your answer.

Mixing galvanic and electrolytic conventions. In galvanic cells: anode is negative, cathode is positive. In electrolytic cells: anode is positive, cathode is negative. The reactions (oxidation at anode, reduction at cathode) are the same — only the polarity labels differ.

Thinking $E°$ changes with stoichiometry. Standard electrode potential is an intensive property — it does not change when you multiply the half-reaction by a coefficient. But $n$ (number of electrons) does change, and you must use the correct $n$ in the Nernst equation and $\Delta G° = -nFE°$ .

Exam Weightage and Strategy

Electrochemistry carries 5-8 marks in CBSE Class 12 boards and 2-3 JEE Main questions per year. NEET asks 1-2. This is one of the most numerically intensive chapters in chemistry — practice the formulas with different types of problems. The Nernst equation and Faraday’s law are the two most tested formulas.

Memorise five formulas: $E°_{\text{cell}}$ , Nernst equation, $\Delta G° = -nFE°$ , Faraday’s law ( $m = MIt/nF$ ), and Kohlrausch’s law. Know 5-6 standard electrode potentials (Li, Zn, Fe, H, Cu, Ag). Practice 10-15 numericals. That covers the entire chapter for any exam.

Practice Questions

Q1. Find $\Delta G°$ for a cell with $E° = 1.5$ V and $n = 2$ .

$\Delta G° = -nFE° = -2 \times 96485 \times 1.5 = -289,455$ J $\approx -289.5$ kJ. The large negative value confirms the reaction is highly spontaneous.

Q2. Using Kohlrausch’s law, find $\Lambda°_m$ for H $_2$ O given: $\Lambda°_m$ (HCl) = 426.2, $\Lambda°_m$ (NaOH) = 248.1, $\Lambda°_m$ (NaCl) = 126.5 S cm $^2$ /mol.

$\Lambda°_m$ (H $_2$ O) = $\Lambda°_m$ (HCl) + $\Lambda°_m$ (NaOH) - $\Lambda°_m$ (NaCl) = 426.2 + 248.1 - 126.5 = 547.8 S cm $^2$ /mol.

Q3. Current of 5 A is passed through molten NaCl for 965 s. Find mass of Na deposited.

$m = MIt/(nF) = 23 \times 5 \times 965/(1 \times 96485) = 111,475/96,485 = \mathbf{1.155}$ g. ( $n = 1$ because Na $^+$ gains 1 electron.)

Q4. At what $[\text{Zn}^{2+}]/[\text{Cu}^{2+}]$ ratio does the Daniell cell potential drop to 1.0 V?

$1.0 = 1.10 - \frac{0.0591}{2}\log Q$ . $\log Q = \frac{0.10}{0.0296} = 3.38$ . $Q = 10^{3.38} \approx 2400$ . So $[\text{Zn}^{2+}]/[\text{Cu}^{2+}] \approx 2400$ . The Cu $^{2+}$ concentration must be very low relative to Zn $^{2+}$ for the potential to drop significantly.

Q5. Why does iron rust faster near the sea?

Seawater contains dissolved NaCl — a strong electrolyte. The ions increase the conductivity of the moisture film on the iron surface, accelerating the electrochemical corrosion process. Na $^+$ and Cl $^-$ ions facilitate faster electron transfer between the anodic (iron) and cathodic (moisture + O $_2$ ) areas. Salt spray also prevents protective oxide layers from forming properly.

FAQs

What is the difference between galvanic and electrolytic cells?

Galvanic: spontaneous reaction generates electricity ( $\Delta G < 0$ , $E°_{\text{cell}} > 0$ ). Electrolytic: external electricity drives a non-spontaneous reaction ( $\Delta G > 0$ ). In both, oxidation occurs at the anode and reduction at the cathode. The key difference is the energy direction — galvanic outputs energy, electrolytic inputs energy.

What is corrosion and how to prevent it?

Corrosion is electrochemical oxidation of metals (e.g., Fe → Fe $_2$ O $_3$ · xH $_2$ O = rust). Prevention: galvanisation (zinc coating — zinc corrodes preferentially), cathodic protection (connecting to a more reactive metal), painting (physical barrier), alloying (stainless steel with Cr forms protective Cr $_2$ O $_3$ layer).

Why does molar conductivity increase on dilution?

For strong electrolytes, dilution reduces interionic attractions, allowing ions to move more freely — the increase is slight and linear with $\sqrt{c}$ (Debye-Huckel-Onsager). For weak electrolytes, dilution shifts the dissociation equilibrium to the right (Le Chatelier), producing many more ions — the increase is sharp and nonlinear. Both effects increase $\Lambda_m$ , but for fundamentally different reasons.

What is a fuel cell?

A galvanic cell that converts the chemical energy of a fuel (usually H $_2$ ) directly into electricity without combustion. At the anode: $\text{H}_2 \to 2\text{H}^+ + 2e^-$ . At the cathode: $\frac{1}{2}\text{O}_2 + 2\text{H}^+ + 2e^- \to \text{H}_2\text{O}$ . The product is water — zero pollution. Fuel cells are more efficient than heat engines (no Carnot limitation) and are used in spacecraft, buses and increasingly in cars. The hydrogen-oxygen fuel cell has $E° \approx 1.23$ V.

What is the difference between EMF and terminal voltage?

EMF ( $E$ ) is the maximum potential difference a cell can deliver when no current flows (open circuit). Terminal voltage ( $V$ ) is the actual voltage when current flows — it is always less than EMF due to internal resistance: $V = E - Ir$ , where $r$ is the internal resistance. The greater the current drawn, the larger the voltage drop across the internal resistance.

Why is $E°$ an intensive property?

Standard electrode potential measures the tendency of a species to be reduced — it is a potential (energy per unit charge), not a total energy. Multiplying a half-reaction by 2 doubles the number of electrons but does not change the potential at which the reaction occurs. This is why we do not multiply $E°$ by stoichiometric coefficients. However, $\Delta G° = -nFE°$ does depend on $n$ because it is an extensive property (total energy).

Why does molar conductivity increase on dilution?

For strong electrolytes, dilution reduces interionic attractions, allowing ions to move more freely — slight increase. For weak electrolytes, dilution shifts the dissociation equilibrium to the right (Le Chatelier), producing more ions — sharp increase. Both effects increase $\Lambda_m$ , but for different reasons.

What is a fuel cell?

A galvanic cell that converts the chemical energy of a fuel (usually H $_2$ ) directly into electricity. H $_2$ is oxidised at the anode, O $_2$ is reduced at the cathode, and the product is water. Fuel cells are more efficient than combustion engines (no Carnot limit) and produce no pollution. Used in spacecraft and increasingly in vehicles.