Electrochemistry Class 12: Concepts, Formulas and Board Numericals

What Electrochemistry is about

Weightage: This chapter is Unit II, Electrochemistry, which carries 9 of the 70 theory marks.Source: CBSE Chemistry (043) syllabus 2026-27, course structure, Unit II Electrochemistry

In real board papers: asked in 8 of the 8 board papers we hold (2023 to 2026), on average 9 marks per paper. See the questions

Key concepts

Galvanic cells and how to write them

NCERT section 2.1 Electrochemical Cells; 2.2 Galvanic Cells

A galvanic (voltaic) cell turns the Gibbs energy of a spontaneous redox reaction into electrical work. The Daniell cell uses $Zn(s)+C{u}^{2+}(aq)\to Z{n}^{2+}(aq)+Cu(s)$, with a potential of 1.1 V when both ion concentrations are 1 mol dm⁻³.

Each half of the cell is a half-cell (or redox couple): the metal dipped in its ions. The two half-cells are joined by a wire and a salt bridge (an inert electrolyte such as KCl or NH₄NO₃ in agar), which completes the circuit and keeps both solutions electrically neutral.

• Anode: oxidation happens; in a galvanic cell it is the negative electrode, because electrons are released there.
• Cathode: reduction happens; it is the positive electrode.
• Electrons flow from anode to cathode in the external wire; current flows the other way.

Cell notation: anode on the left, cathode on the right, a single vertical line for a phase boundary and a double line for the salt bridge:
$Zn(s)|Z{n}^{2+}(aq)||C{u}^{2+}(aq)|Cu(s)$

If an opposing external potential greater than 1.1 V is applied to the Daniell cell, the reaction reverses and the cell behaves as an electrolytic cell.

Worked example

A galvanic cell is made from a magnesium electrode in 1 M Mg²⁺ solution and a silver electrode in 1 M Ag⁺ solution. (a) Which electrode is the anode? (b) Write the half-reactions and the overall reaction. (c) Write the cell notation and calculate $E_{cell}^{\circ }$. ($E_{M{g}^{2+}/Mg}^{\circ }$ = −2.37 V, $E_{A{g}^{+}/Ag}^{\circ }$ = +0.80 V) (3 marks)

Try it first, then show the solutionSolution

1. (a) The lower reduction potential is oxidised: magnesium is the anode. (½)
2. (b) Anode: $Mg(s)\to M{g}^{2+}(aq)+2e^{-}$; cathode: $A{g}^{+}(aq)+e^{-}\to Ag(s)$, doubled to match electrons. (1)
3. Overall: $Mg(s)+2A{g}^{+}(aq)\to M{g}^{2+}(aq)+2Ag(s)$. (½)
4. (c) $Mg(s)|M{g}^{2+}(1M)||A{g}^{+}(1M)|Ag(s)$; $E_{cell}^{\circ }=0.80-(-2.37)=3.17$ V. (1)

Answer: Mg is the anode; Mg + 2Ag⁺ → Mg²⁺ + 2Ag; E°cell = 3.17 V.

Electrode potential, the SHE and the electrochemical series

NCERT section 2.2.1 Measurement of Electrode Potential

An electrode potential cannot be measured alone; only the difference between two electrodes can. So every electrode is compared with the standard hydrogen electrode (SHE): a platinum electrode coated with platinum black, dipped in an acid solution with H⁺ at unit activity (about 1 M), with pure hydrogen gas at 1 bar bubbled through it. Its potential is taken as zero at all temperatures:
$H^{+}(aq)+e^{-}\to \frac{1}{2}{H}_2(g)$, $E^{\circ }=0.00$ V

By convention, potentials are written as standard reduction potentials. Then:
$E_{cell}^{\circ }=E_{cathode}^{\circ }-E_{anode}^{\circ }=E_{right}^{\circ }-E_{left}^{\circ }$

Using the series:

• A more negative $E^{\circ }$ means a stronger reducing agent (Li, K, Mg, Zn are high reducers). Lithium has the most negative value.
• A more positive $E^{\circ }$ means a stronger oxidising agent; fluorine has the highest.
• A metal with lower $E^{\circ }$ displaces the ions of a metal with higher $E^{\circ }$ from solution.
• A positive $E_{cell}^{\circ }$ means the reaction as written is spontaneous.

Copper ($E^{\circ }$ = +0.34 V) does not dissolve in HCl, because its ions are more easily reduced than H⁺; zinc (−0.76 V) does.

Worked example

Can a solution of copper sulphate be stored (a) in a zinc pot, (b) in a silver vessel? Justify with $E_{cell}^{\circ }$. ($E^{\circ }$: Zn²⁺/Zn = −0.76 V, Cu²⁺/Cu = +0.34 V, Ag⁺/Ag = +0.80 V) (2 marks)

Try it first, then show the solutionSolution

1. (a) Possible reaction $Zn+C{u}^{2+}\to Z{n}^{2+}+Cu$: $E_{cell}^{\circ }=0.34-(-0.76)=+1.10$ V, positive, so zinc reacts. It cannot be stored in a zinc pot. (1)
2. (b) Possible reaction $2Ag+C{u}^{2+}\to 2A{g}^{+}+Cu$: $E_{cell}^{\circ }=0.34-0.80=-0.46$ V, negative, so no reaction. It can be stored in a silver vessel. (1)

Answer: Not in zinc (E°cell = +1.10 V, reaction occurs); yes in silver (E°cell = −0.46 V, no reaction).

Nernst equation

NCERT section 2.3 Nernst Equation

Standard potentials hold only when all ions are at 1 M. For any concentration, the Nernst equation for a reduction $M^{n+}(aq)+n{e}^{-}\to M(s)$ is:
$E=E^{\circ }-\frac{RT}{nF}\ln \frac{1}{[M^{n+}]}$
The concentration of a pure solid is taken as unity.

For a whole cell with reaction $aA+bB\to cC+dD$:
$E_{cell}=E_{cell}^{\circ }-\frac{2.303RT}{nF}\log \frac{[C{]}^c[D{]}^d}{[A{]}^a[B{]}^b}$
At 298 K, $\frac{2.303RT}{F}=0.059$ V, so
$E_{cell}=E_{cell}^{\circ }-\frac{0.059}{n}\log Q$

For the Daniell cell: $E_{cell}=1.10-\frac{0.059}{2}\log \frac{[Z{n}^{2+}]}{[C{u}^{2+}]}$. Raising [Cu²⁺] or lowering [Zn²⁺] raises the cell potential.

Steps for any Nernst sum: (1) write the balanced cell reaction and find $n$; (2) find $E_{cell}^{\circ }$; (3) write $Q$ with products on top, leaving out solids; (4) substitute and simplify the logarithm.

Worked example

Calculate the emf of the cell at 298 K: $Zn(s)|Z{n}^{2+}(0.001M)||C{u}^{2+}(0.1M)|Cu(s)$. ($E_{cell}^{\circ }$ = 1.10 V) (3 marks)

Try it first, then show the solutionSolution

1. Cell reaction: $Zn(s)+C{u}^{2+}(aq)\to Z{n}^{2+}(aq)+Cu(s)$, so $n=2$. (½)
2. $E_{cell}=E_{cell}^{\circ }-\frac{0.059}{2}\log \frac{[Z{n}^{2+}]}{[C{u}^{2+}]}$ (1)
3. $=1.10-0.0295\log \frac{0.001}{0.1}=1.10-0.0295\times (-2)$ (1)
4. $=1.10+0.059=1.159$ V. (½)

Answer: E_cell ≈ 1.16 V (1.159 V).

Equilibrium constant and Gibbs energy from cell potential

NCERT section 2.3.1 Equilibrium Constant from Nernst Equation; 2.3.2 Electrochemical Cell and Gibbs Energy of the Reaction

At equilibrium the cell can do no more work, so $E_{cell}=0$ and $Q$ becomes $K_c$. Putting this into the Nernst equation gives
$E_{cell}^{\circ }=\frac{2.303RT}{nF}\log K_c$, i.e. at 298 K: $\log K_c=\frac{n{E}_{cell}^{\circ }}{0.059}$

Notice that $K_c$ is linked to $E_{cell}^{\circ }$, not to $E_{cell}$, because $E_{cell}$ changes as the reaction proceeds while $E_{cell}^{\circ }$ is fixed.

The electrical work done by a cell equals the decrease in Gibbs energy:
${\Delta }_{r}G=-nF{E}_{cell}$ and ${\Delta }_r{G}^{\circ }=-nF{E}_{cell}^{\circ }$
${\Delta }_r{G}^{\circ }=-RT\ln K=-2.303RT\log K$

Signs to remember: a positive $E_{cell}^{\circ }$ gives a negative ${\Delta }_r{G}^{\circ }$ and $K>1$ (spontaneous). $E_{cell}$ is an intensive property, but ${\Delta }_{r}G$ depends on how the reaction is written: doubling the equation doubles $n$ and doubles ${\Delta }_{r}G$, but leaves $E_{cell}$ unchanged.

Use F = 96500 C mol⁻¹ and give ${\Delta }_{r}G$ in kJ mol⁻¹.

Worked example

The standard emf of a cell in which the reaction involves the transfer of 2 electrons is 0.236 V at 298 K. Calculate (a) the standard Gibbs energy change and (b) the equilibrium constant of the cell reaction. (F = 96500 C mol⁻¹) (3 marks)

Try it first, then show the solutionSolution

1. (a) ${\Delta }_r{G}^{\circ }=-nF{E}_{cell}^{\circ }=-2\times 96500\times 0.236$ (1)
2. $=-45548$ J mol⁻¹ $=-45.55$ kJ mol⁻¹. (½)
3. (b) $\log K=\frac{n{E}^{\circ }}{0.059}=\frac{2\times 0.236}{0.059}=8$ (1)
4. $K=1\times 10^8$. (½)

Answer: (a) ΔrG° ≈ −45.5 kJ mol⁻¹ (b) K = 1 × 10⁸

Conductivity, cell constant and molar conductivity

NCERT section 2.4 Conductance of Electrolytic Solutions; 2.4.1 Measurement of the Conductivity of Ionic Solutions

Metals conduct by moving electrons (electronic conductance). Solutions of electrolytes conduct by moving ions (electrolytic or ionic conductance). Ionic conductance depends on the nature of the electrolyte, the size and solvation of the ions, the solvent and its viscosity, concentration and temperature (it increases with temperature).

Resistance $R=\rho \frac{l}{A}$. Its inverse is conductance $G$ (unit siemens, S). The inverse of resistivity is conductivity, $\kappa$ (kappa), the conductance of a solution between electrodes of 1 cm² area placed 1 cm apart (unit S cm⁻¹ or S m⁻¹).

The cell constant $G^{\ast }=\frac{l}{A}$ is found by measuring the resistance of a KCl solution of known conductivity: $G^{\ast }=\kappa \times R$. Resistance is measured with alternating current (a Wheatstone bridge with a conductivity cell), because direct current would electrolyse the solution.

Molar conductivity, ${\Lambda }_m$, is the conductivity per unit molar concentration. With $\kappa$ in S cm⁻¹ and $c$ in mol L⁻¹:
${\Lambda }_m=\frac{\kappa \times 1000}{c}$ (S cm² mol⁻¹)
It is the conductance of the volume of solution that contains one mole of electrolyte, placed between electrodes 1 cm apart.

Worked example

A conductivity cell filled with 0.1 M KCl (κ = 0.0129 S cm⁻¹) has a resistance of 100 Ω. The same cell filled with a 0.05 M solution of an electrolyte has a resistance of 260 Ω. Find the cell constant and the molar conductivity of the second solution. (3 marks)

Try it first, then show the solutionSolution

1. $G^{\ast }=\kappa \times R=0.0129\times 100=1.29$ cm⁻¹. (1)
2. $\kappa =\frac{G^{\ast }}{R}=\frac{1.29}{260}=4.96\times 10^{-3}$ S cm⁻¹. (1)
3. ${\Lambda }_m=\frac{4.96\times 10^{-3}\times 1000}{0.05}=99.2$ S cm² mol⁻¹. (1)

Answer: Cell constant = 1.29 cm⁻¹; Λm ≈ 99.2 S cm² mol⁻¹.

Variation with concentration and Kohlrausch's law

NCERT section 2.4.2 Variation of Conductivity and Molar Conductivity with Concentration

On dilution, conductivity falls for every electrolyte, because there are fewer ions per cm³. But molar conductivity rises, because it refers to one mole of electrolyte, whatever volume it occupies.

• Strong electrolytes (KCl, NaCl) are fully ionised. ${\Lambda }_m$ rises slowly on dilution because interionic attractions weaken: ${\Lambda }_m={\Lambda }_m^{\circ }-A\sqrt{c}$. A plot of ${\Lambda }_m$ against $\sqrt{c}$ is a straight line whose intercept gives the limiting molar conductivity ${\Lambda }_m^{\circ }$ and slope $-A$.
• Weak electrolytes (acetic acid) ionise only partly. On dilution the degree of dissociation increases sharply, so ${\Lambda }_m$ rises steeply at low concentration. The curve cannot be extrapolated to zero concentration, so ${\Lambda }_m^{\circ }$ is found another way.

Kohlrausch's law of independent migration of ions: the limiting molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its cation and anion, each multiplied by the number of that ion in the formula:
${\Lambda }_m^{\circ }={\nu }_{+}{\lambda }_{+}^{\circ }+{\nu }_{-}{\lambda }_{-}^{\circ }$

Applications:

• ${\Lambda }_m^{\circ }$ of a weak electrolyte from strong ones, e.g. ${\Lambda }^{\circ }(C{H}_{3}COOH)={\Lambda }^{\circ }(HCl)+{\Lambda }^{\circ }(C{H}_{3}COONa)-{\Lambda }^{\circ }(NaCl)$.
• Degree of dissociation $\alpha =\frac{{\Lambda }_m}{{\Lambda }_m^{\circ }}$ and dissociation constant $K_a=\frac{c{\alpha }^2}{1-\alpha }$.

NCERT values at 298 K (S cm² mol⁻¹): H⁺ 349.6, Na⁺ 50.1, K⁺ 73.5, Cl⁻ 76.3, CH₃COO⁻ 40.9.

Worked example

The molar conductivity of 0.01 M acetic acid is 16.4 S cm² mol⁻¹. Using ${\lambda }^{\circ }(H^{+})$ = 349.6 and ${\lambda }^{\circ }(C{H}_{3}CO{O}^{-})$ = 40.9 S cm² mol⁻¹, calculate the degree of dissociation and the dissociation constant of acetic acid. (3 marks)

Try it first, then show the solutionSolution

1. Kohlrausch's law: ${\Lambda }_m^{\circ }=349.6+40.9=390.5$ S cm² mol⁻¹. (1)
2. $\alpha =\frac{16.4}{390.5}=0.042$. (1)
3. $K_a=\frac{c{\alpha }^2}{1-\alpha }=\frac{0.01\times (0.042{)}^2}{0.958}=1.84\times 10^{-5}$. (1)

Answer: α ≈ 0.042; Ka ≈ 1.8 × 10⁻⁵

Electrolysis: Faraday's laws and products

NCERT section 2.5 Electrolytic Cells and Electrolysis; 2.5.1 Quantitative Aspects of Electrolysis; 2.5.2 Products of Electrolysis

In an electrolytic cell, a direct current drives a non-spontaneous reaction. Copper refining and electroplating are examples: in refining, impure copper is the anode, pure copper is the cathode and CuSO₄ solution is the electrolyte.

Faraday's first law: the mass of substance deposited or liberated at an electrode is proportional to the charge passed: $m=Z\times I\times t$, where $Z$ is the electrochemical equivalent.
Faraday's second law: when the same charge passes through different electrolytes, the masses deposited are proportional to their chemical equivalent weights (atomic mass ÷ number of electrons needed).

One mole of electrons carries 1 faraday, F = 96487 C mol⁻¹ (use 96500 for calculations). So:
$Q=I\times t$; moles of electrons $=\frac{Q}{F}$; moles of product $=\frac{Q}{nF}$
For example, depositing 1 mol of Cu from Cu²⁺ needs 2F, and of Al from Al³⁺ needs 3F.

Products depend on competition. In molten NaCl, Na is formed at the cathode and Cl₂ at the anode. In aqueous NaCl, water is reduced in preference to Na⁺ (H₂ at the cathode, OH⁻ left in solution). At the anode, water should be oxidised (lower $E^{\circ }$ = 1.23 V than Cl⁻ at 1.36 V), but oxygen has a large overpotential, so Cl₂ is formed. The products are NaOH, Cl₂ and H₂. Dilute sulphuric acid gives O₂ at the anode; concentrated sulphuric acid gives peroxodisulphate ions.

Worked example

A current of 1.5 A is passed for 30 minutes through copper sulphate solution. Calculate the mass of copper deposited at the cathode. (Cu = 63.5 g mol⁻¹, F = 96500 C mol⁻¹) (2 marks)

Try it first, then show the solutionSolution

1. $Q=I\times t=1.5\times 30\times 60=2700$ C. (½)
2. $C{u}^{2+}+2e^{-}\to Cu$: 2 × 96500 C deposit 63.5 g. (½)
3. Mass $=\frac{63.5\times 2700}{2\times 96500}=0.888$ g. (1)

Answer: About 0.89 g of copper.

Batteries, fuel cells and corrosion

NCERT section 2.6 Batteries; 2.7 Fuel Cells; 2.8 Corrosion

Primary batteries work once. The dry (Leclanché) cell has a zinc container as anode and a graphite rod surrounded by MnO₂ and carbon as cathode, in a moist paste of NH₄Cl and ZnCl₂; it gives about 1.5 V. The mercury cell (hearing aids, watches) has a zinc–mercury amalgam anode, HgO and carbon paste as cathode, and KOH and ZnO paste as electrolyte. Its overall reaction $Zn+HgO(s)\to ZnO(s)+Hg(l)$ (the zinc is present as an amalgam) involves no ion in solution, so its potential stays constant at about 1.35 V.

Secondary batteries can be recharged. The lead storage battery (cars and inverters) has a lead anode and a lead grid packed with PbO₂ as cathode, in 38% sulphuric acid. On discharge, PbSO₄ forms at both electrodes and sulphuric acid is used up; on charging the reaction reverses. The nickel-cadmium cell lasts longer but costs more.

Fuel cells convert the energy of combustion of fuels such as hydrogen directly into electricity. In the H₂–O₂ cell, the gases bubble through porous carbon electrodes (with Pt or Pd catalyst) into concentrated NaOH. It was used in the Apollo space programme, and the water formed was used for drinking. Efficiency is about 70%, against about 40% for thermal power plants, and it is pollution-free.

Corrosion is an electrochemical process. On iron, one spot acts as anode ($Fe\to F{e}^{2+}+2e^{-}$, $E^{\circ }$ = −0.44 V) and another as cathode, where oxygen is reduced with H⁺ from dissolved CO₂ ($E^{\circ }$ = 1.23 V). $E_{cell}^{\circ }$ = 1.67 V. Fe²⁺ is further oxidised to hydrated ferric oxide, rust ($F{e}_2{O}_3\cdot x{H}_{2}O$). Prevention: paint or chemicals (such as bisphenol) as a barrier, coating with other metals (Sn, Zn), or a sacrificial anode of a more reactive metal (Mg, Zn).

Worked example

(a) Why does the mercury cell give a constant voltage throughout its life? (b) Write the overall reaction of the lead storage battery during discharge, and state what happens to the density of the acid. (3 marks)

Try it first, then show the solutionSolution

1. (a) Its overall reaction, $Zn+HgO(s)\to ZnO(s)+Hg(l)$ (the zinc is present as an amalgam), has no ion in solution whose concentration changes, so by the Nernst equation its potential stays constant (about 1.35 V). (1)
2. (b) $Pb(s)+Pb{O}_2(s)+2H_{2}S{O}_4(aq)\to 2PbS{O}_4(s)+2H_{2}O(l)$ (1)
3. Sulphuric acid is used up and water is formed, so the density of the electrolyte falls during discharge. (1)

Answer: (a) No ion whose concentration changes takes part in the overall reaction. (b) Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O; acid density decreases.

Chemistry in real life: Electrochemistry

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Nernst Equation, Gibbs Energy and Equilibrium Constant
• Conductivity, Molar Conductivity and Kohlrausch's Law
• Electrolysis and Faraday's Laws

Electrochemistry formula sheet

Standard cell potential

$E_{cell}^{\circ }=E_{cathode}^{\circ }-E_{anode}^{\circ }=E_{right}^{\circ }-E_{left}^{\circ }$

Use reduction potentials for both electrodes; never change the sign of a table value.

Nernst equation (electrode)

$E_{M^{n+}/M}=E_{M^{n+}/M}^{\circ }-\frac{0.059}{n}\log \frac{1}{[M^{n+}]}$

At 298 K. A pure solid has activity 1.

Nernst equation (cell)

$E_{cell}=E_{cell}^{\circ }-\frac{0.059}{n}\log \frac{[\text{products}]}{[\text{reactants}]}$

General form: $E_{cell}=E_{cell}^{\circ }-\frac{2.303RT}{nF}\log Q$.

Hydrogen electrode in a solution of given pH

$E_{H^{+}/H_2}=-0.059\text{pH}$

For hydrogen gas at 1 bar and 298 K.

Equilibrium constant

$\log K_c=\frac{n{E}_{cell}^{\circ }}{0.059}$

At 298 K; from $E_{cell}=0$ at equilibrium.

Gibbs energy

${\Delta }_{r}G=-nF{E}_{cell},{\Delta }_r{G}^{\circ }=-nF{E}_{cell}^{\circ }=-2.303RT\log K$

F = 96500 C mol⁻¹; convert J to kJ.

Resistance, conductance and conductivity

$R=\rho \frac{l}{A},G=\frac{1}{R},\kappa =\frac{1}{\rho }$

Units: Ω, S and S cm⁻¹ (or S m⁻¹).

Cell constant

$G^{\ast }=\frac{l}{A}=\kappa \times R$

Unit cm⁻¹ (or m⁻¹).

Molar conductivity

${\Lambda }_m=\frac{\kappa \times 1000}{c}$

κ in S cm⁻¹, c in mol L⁻¹, Λm in S cm² mol⁻¹. 1 S cm² mol⁻¹ = 10⁻⁴ S m² mol⁻¹.

Strong electrolytes (Debye–Hückel–Onsager form)

${\Lambda }_m={\Lambda }_m^{\circ }-A\sqrt{c}$

Plot Λm against √c: intercept Λ°m, slope −A.

Kohlrausch's law

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

For example ${\Lambda }^{\circ }(CaC{l}_2)={\lambda }^{\circ }(C{a}^{2+})+2{\lambda }^{\circ }(C{l}^{-})$.

Weak electrolytes

$\alpha =\frac{{\Lambda }_m}{{\Lambda }_m^{\circ }},K_a=\frac{c{\alpha }^2}{1-\alpha }$

Λ°m of the weak electrolyte comes from Kohlrausch's law.

Faraday's laws

$Q=It,m=ZIt=\frac{M}{nF}It,\frac{m_1}{m_2}=\frac{M_1/n_1}{M_2/n_2}$

1 F = 96500 C mol⁻¹ = charge on 1 mol of electrons.

Daniell cell reaction

$Zn(s)+C{u}^{2+}(aq)\to Z{n}^{2+}(aq)+Cu(s)$

E°cell = 1.10 V.

Lead storage battery (discharge)

$Pb(s)+Pb{O}_2(s)+2H_{2}S{O}_4(aq)\to 2PbS{O}_4(s)+2H_{2}O(l)$

Reversed on charging.

Mercury cell

$Zn+HgO(s)\to ZnO(s)+Hg(l)$

Zinc is used as a zinc–mercury amalgam. About 1.35 V, constant.

Hydrogen–oxygen fuel cell

$2H_2(g)+O_2(g)\to 2H_{2}O(l)$

Cathode: $O_2+2H_{2}O+4e^{-}\to 4O{H}^{-}$; anode: $2H_2+4O{H}^{-}\to 4H_{2}O+4e^{-}$.

Corrosion of iron

$2Fe(s)+O_2(g)+4H^{+}(aq)\to 2F{e}^{2+}(aq)+2H_{2}O(l)$

E°cell = 1.23 − (−0.44) = 1.67 V; Fe²⁺ then forms rust, Fe₂O₃·xH₂O.

Electrolysis of aqueous NaCl

$2NaCl(aq)+2H_{2}O(l)\to 2NaOH(aq)+H_2(g)+C{l}_2(g)$

H₂ at the cathode, Cl₂ at the anode (overpotential of O₂).

Electrochemistry: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the board asks Electrochemistry

The question patterns to expect, and the method that works for each.

Tricks to remember Electrochemistry

Short memory aids. Say each one aloud once, then use it on a question the same day.

• An Ox, Red Cat

  ANode = OXidation, REDuction = CAThode. In a galvanic cell the anode is negative; in an electrolytic cell it is positive, but oxidation is always at the anode.

  Use it when: Labelling electrodes and writing half-reactions.

• Right minus Left

  E°cell = E°(right) − E°(left), where right is the cathode as written in cell notation. Never flip the sign of a table value.

  Use it when: Every E°cell calculation.

• Products over reactants, solids out

  In the Nernst log term, put product ion concentrations on top and reactant ion concentrations below, each raised to its coefficient; leave solids and pure liquids out.

  Use it when: Nernst equation numericals.

• 0.059 is the magic number

  At 298 K, 2.303RT/F = 0.059 V. So log K = nE°/0.059: if E° = 0.059 V and n = 1, K = 10.

  Use it when: Equilibrium constant without a calculator.

• Kappa falls, Lambda climbs

  On dilution conductivity (κ) falls but molar conductivity (Λm) climbs; for weak electrolytes it climbs steeply.

  Use it when: Assertion-reason and graph questions on dilution.

Electrochemistry: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Electrochemistry

The slips examiners see most often, and what to write instead.

• ✗Mistake: Wrong n in the Nernst equation

  Students use the number of electrons in one half-reaction before balancing the two.

  Wrong working

  Mg + 2Ag⁺ → Mg²⁺ + 2Ag: n = 1, because Ag⁺ + e⁻ → Ag

  Right working

  Mg + 2Ag⁺ → Mg²⁺ + 2Ag: n = 2, the electrons transferred in the balanced overall reaction

  ✓Do this: Balance the overall reaction first and count electrons lost by the anode.

  Usually costs: 1 to 2 marks in a 3-mark numerical.

• ✗Mistake: Forgetting to square concentrations in Q

  Coefficients are ignored when writing Q.

  Wrong working

  Q = [Mg²⁺] ÷ [Ag⁺]

  Right working

  Q = [Mg²⁺] ÷ [Ag⁺]²

  ✓Do this: Raise each concentration to its coefficient in the balanced reaction.

  Usually costs: The final answer mark.

• ✗Mistake: Doubling E° when doubling a half-reaction

  Students treat E° like ΔG.

  Wrong working

  2Ag⁺ + 2e⁻ → 2Ag, E° = 2 × 0.80 = 1.60 V

  Right working

  2Ag⁺ + 2e⁻ → 2Ag, E° = 0.80 V (intensive); only ΔG doubles

  ✓Do this: Remember that E is intensive; n carries the multiplication into ΔG = −nFE.

  Usually costs: 1 mark, and every later step that uses E°cell.

• ✗Mistake: Missing the 1000 factor in Λm

  κ in S cm⁻¹ is divided directly by c in mol L⁻¹.

  Wrong working

  Λm = 4.96 × 10⁻³ ÷ 0.05 = 0.0992 S cm² mol⁻¹

  Right working

  Λm = 4.96 × 10⁻³ × 1000 ÷ 0.05 = 99.2 S cm² mol⁻¹

  ✓Do this: With κ in S cm⁻¹ and c in mol L⁻¹, always use Λm = κ × 1000 ÷ c.

  Usually costs: The answer mark.

• ✗Mistake: Saying O₂ forms at the anode in aqueous NaCl

  Students compare only E° values (1.23 V for water, 1.36 V for Cl⁻).

  ✓Do this: Add the reason: oxygen has a high overpotential, so Cl⁻ is oxidised and Cl₂ forms.

  Usually costs: 1 mark in a products-of-electrolysis question.

• ✗Mistake: Giving ΔG in joules without a sign or unit

  ✓Do this: Write ΔrG° = −nFE°, keep the minus sign, convert to kJ mol⁻¹ and state whether the reaction is spontaneous.

  Usually costs: ½ to 1 mark.

NCERT exercise guide: Electrochemistry

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Electrochemistry: previous year questions

• Asked in 8 of the 8 board papers we hold (2023 to 2026), on average 9 marks per paper; 37 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Electrochemistry

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

Every Electrochemistry question, by year

Practice questions on Electrochemistry

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQBoard level

   The unit of molar conductivity is:
   (a) S cm⁻¹ (b) S cm² mol⁻¹ (c) S⁻¹ cm² mol (d) Ω cm

   Hint

   Λm = κ × 1000 ÷ c.

   Answer

   (b) S cm² mol⁻¹

2. Q2 · 1 mark · MCQBoard level

   Which metal can act as a sacrificial anode to protect an iron pipe? (E° values: Mg²⁺/Mg −2.37 V, Fe²⁺/Fe −0.44 V, Sn²⁺/Sn −0.14 V, Cu²⁺/Cu +0.34 V)
   (a) Cu (b) Sn (c) Mg (d) None of these

   Hint

   It must be oxidised more easily than iron.

   Answer

   (c) Mg, because its reduction potential is more negative than that of iron.

3. Q3 · 1 mark · Assertion-reasonBoard level

   Assertion (A): Λ°m of acetic acid cannot be obtained by extrapolating the graph of Λm against √c.
   Reason (R): Acetic acid is a strong electrolyte.
   (a) Both A and R are true and R is the correct explanation of A.
   (b) Both A and R are true but R is not the correct explanation of A.
   (c) A is true but R is false.
   (d) A is false but R is true.

   Hint

   Is acetic acid fully ionised?

   Answer

   (c) A is true, but acetic acid is a weak electrolyte, so R is false.

4. Q4 · 2 marks · Very short answerBoard level

   Write the reactions at the cathode and anode of the hydrogen–oxygen fuel cell and give one advantage of fuel cells over thermal power plants. (2 marks)

   Hint

   The electrolyte is concentrated NaOH.

   Answer

   Cathode: O₂(g) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq); anode: 2H₂(g) + 4OH⁻(aq) → 4H₂O(l) + 4e⁻ (1). Advantage: about 70% efficient compared with about 40%, or pollution-free (1).

5. Q5 · 2 marks · Very short answerHarder board level

   A current of 2.0 A is passed through acidified water for 16 minutes 5 seconds. Calculate (a) the number of electrons that pass and (b) the volume of hydrogen collected at the cathode at 273 K and 1 bar (molar volume 22.7 L). (F = 96500 C mol⁻¹, N_A = 6.022 × 10²³ mol⁻¹) (2 marks)

   Hint

   Charge → moles of electrons; 2 electrons make one H₂.

   Answer

   t = 965 s; Q = 2.0 × 965 = 1930 C; moles of electrons = 1930 ÷ 96500 = 0.02, i.e. 1.2 × 10²² electrons (1). 2H⁺ + 2e⁻ → H₂, so 0.01 mol H₂ = 0.01 × 22.7 = 0.227 L (227 mL) (1).

6. Q6 · 3 marks · Short answerHarder board level

   The cell Zn(s) | Zn²⁺(0.1 M) || Cu²⁺(c) | Cu(s) shows an emf of 1.0115 V at 298 K. Find c. If [Cu²⁺] is now raised to 0.1 M, will the emf rise or fall, and to what value? (E°cell = 1.10 V) (3 marks)

   Hint

   Write E = 1.10 − 0.0295 log([Zn²⁺]/[Cu²⁺]) and solve for the log term.

   Answer

   1.0115 = 1.10 − 0.0295 log(0.1/c), so log(0.1/c) = 0.0885 ÷ 0.0295 = 3 (1); 0.1/c = 10³, c = 1 × 10⁻⁴ M (1). With [Cu²⁺] = 0.1 M, Q = 1 and log Q = 0, so the emf rises to 1.10 V (1).

7. Q7 · 3 marks · Short answerHarder board level

   For the cell reaction Zn(s) + Fe²⁺(aq) → Zn²⁺(aq) + Fe(s) at 298 K, calculate (a) E°cell, (b) ΔrG°, (c) the equilibrium constant and (d) E_cell when [Zn²⁺] = 0.1 M and [Fe²⁺] = 0.001 M. (E°: Zn²⁺/Zn −0.76 V, Fe²⁺/Fe −0.44 V; F = 96500 C mol⁻¹; antilog 0.85 = 7.08) (3 marks)

   Hint

   Iron ions are reduced, so Fe is the cathode.

   Answer

   (a) E°cell = −0.44 − (−0.76) = 0.32 V (½). (b) ΔrG° = −2 × 96500 × 0.32 = −61 760 J mol⁻¹ ≈ −61.8 kJ mol⁻¹ (1). (c) log K = 2 × 0.32 ÷ 0.059 = 10.85, so K = 7.08 × 10¹⁰ (1). (d) E = 0.32 − 0.0295 log(0.1/0.001) = 0.32 − 0.059 = 0.261 V (½).

8. Q8 · 3 marks · Short answerHOTS / competency

   A hydrogen electrode (H₂ at 1 bar) placed in a solution shows a potential of −0.236 V against the SHE at 298 K. Find the pH of the solution. What would the potential be in a solution of pH 2? (3 marks)

   Hint

   E = −0.059 pH.

   Answer

   For H⁺ + e⁻ → ½H₂, E = 0 − 0.059 log(1/[H⁺]) = −0.059 pH (1). pH = 0.236 ÷ 0.059 = 4 (1). At pH 2, E = −0.118 V (1).

9. Q9 · 5 marks · Long answerHarder board level

   (a) State Faraday's first and second laws of electrolysis. (b) A silver nitrate cell and a copper sulphate cell are joined in series. When a steady current of 0.4 A passes, 2.16 g of silver is deposited. Calculate the time taken and the mass of copper deposited. (Ag = 108, Cu = 63.5 g mol⁻¹, F = 96500 C mol⁻¹) (5 marks)

   Hint

   The same charge passes through both cells.

   Answer

   (a) First law: mass deposited is proportional to the charge passed (m = ZIt) (1). Second law: for the same charge, masses deposited are proportional to chemical equivalent weights (1). (b) Ag: 2.16 ÷ 108 = 0.02 mol, so 0.02 mol electrons, Q = 1930 C (1); t = 1930 ÷ 0.4 = 4825 s (about 80 min) (1). Cu²⁺ needs 2 electrons: 0.01 mol Cu = 0.01 × 63.5 = 0.635 g (1).

10. Q10 · 3 marks · Short answerHOTS / competency

    Explain why, on dilution, the conductivity of a solution decreases while its molar conductivity increases. Why is the increase much larger for a weak electrolyte? (3 marks)

    Hint

    Think 'per cm³' versus 'per mole'.

    Answer

    Conductivity counts ions per unit volume, which falls on dilution (1). Molar conductivity refers to one mole of electrolyte in whatever volume it occupies; that volume grows and, for strong electrolytes, interionic attractions weaken, so Λm rises (1). For a weak electrolyte the degree of dissociation rises sharply on dilution, adding many new ions, so Λm rises steeply (1).

11. Q11 · 5 marks · Long answerHOTS / competency

    Suppose an iron gate in a coastal town rusts fast where its paint is scratched. (a) Explain rusting as an electrochemical cell, giving the anode and cathode reactions and E°cell. (b) Why does rusting speed up in salty, humid air? (c) Why does a zinc-coated (galvanised) gate keep protecting the iron even after the coating is scratched, while a tin-coated one rusts faster? (E°: Fe²⁺/Fe −0.44 V, Zn²⁺/Zn −0.76 V, Sn²⁺/Sn −0.14 V, O₂/H₂O 1.23 V) (5 marks)

    Hint

    Compare each coating metal's E° with that of iron.

    Answer

    (a) Anode: 2Fe → 2Fe²⁺ + 4e⁻; cathode: O₂ + 4H⁺ + 4e⁻ → 2H₂O (H⁺ from dissolved CO₂ and other acidic oxides) (1); overall 2Fe + O₂ + 4H⁺ → 2Fe²⁺ + 2H₂O, E°cell = 1.23 − (−0.44) = 1.67 V; Fe²⁺ is oxidised further to rust, Fe₂O₃·xH₂O (1). (b) Dissolved salts make the moisture a better electrolyte, so the corrosion cells work faster (1). (c) Zinc has a more negative E° than iron, so at a scratch zinc is oxidised as a sacrificial anode and iron stays the cathode (1). Tin has a less negative E° than iron, so at a scratch iron becomes the anode and corrodes faster (1).

12. Q12 · 3 marks · Short answerHarder board level

    Λ°m values for HCl, HCOONa and NaCl are 426, 105 and 126 S cm² mol⁻¹. (a) Find Λ°m for formic acid, HCOOH. (b) A 0.025 M formic acid solution has Λm = 33.6 S cm² mol⁻¹. Find its degree of dissociation and dissociation constant. (3 marks)

    Hint

    Add the two salts that contain H⁺ and HCOO⁻, remove the one that adds Na⁺ and Cl⁻.

    Answer

    (a) Λ°(HCOOH) = Λ°(HCl) + Λ°(HCOONa) − Λ°(NaCl) = 426 + 105 − 126 = 405 S cm² mol⁻¹ (1). (b) α = 33.6 ÷ 405 = 0.083 (1); Ka = cα² ÷ (1 − α) = 0.025 × (0.083)² ÷ 0.917 ≈ 1.9 × 10⁻⁴ (1).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.What is the role of the salt bridge?Show answer

   It completes the electrical circuit and keeps the two solutions electrically neutral, allowing ions to move.

2. 2.What is the potential of the standard hydrogen electrode?Show answer

   0.00 V at all temperatures, by convention.

3. 3.Write the Nernst equation for a cell at 298 K.Show answer

   E_cell = E°cell − (0.059/n) log Q.

4. 4.Why is Kc related to E°cell and not to E_cell?Show answer

   At equilibrium E_cell = 0; E°cell is fixed for a reaction, whereas E_cell keeps changing as concentrations change.

5. 5.State Kohlrausch's law.Show answer

   The limiting molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its cation and anion, each multiplied by its number in the formula.

6. 6.What happens to conductivity and molar conductivity on dilution?Show answer

   Conductivity decreases (fewer ions per unit volume); molar conductivity increases.

7. 7.How many coulombs are needed to reduce 1 mol of Cr₂O₇²⁻ to Cr³⁺?Show answer

   6 mol of electrons: 6 × 96500 = 579 000 C.

8. 8.Why is the cell potential of a mercury cell constant?Show answer

   The overall reaction involves no ion in solution whose concentration changes.

9. 9.Give two advantages of fuel cells.Show answer

   High efficiency (about 70%) and no pollution.

10. 10.What is the formula of rust?Show answer

    Hydrated ferric oxide, Fe₂O₃·xH₂O.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you write the cell notation for a cell made from any two given half-cells and find E°cell?
• ☐Can you decide whether a metal will displace another from solution using E° values?
• ☐Can you calculate E_cell for given concentrations, including squared terms in Q?
• ☐Can you find ΔrG° and K from E°cell without a calculator?
• ☐Can you find G*, κ and Λm from resistance data?
• ☐Can you sketch Λm against √c for KCl and acetic acid and explain the difference?
• ☐Can you use Kohlrausch's law to get Λ° of acetic acid, then α and Ka?
• ☐Can you predict the products of electrolysis of aqueous NaCl and explain the overpotential point?
• ☐Can you write the reactions of the lead storage battery, mercury cell, fuel cell and rusting?

Electrochemistry: frequently asked questions