Electrochemistry ISC Class 12 Chemistry: Notes and Numericals

What Electrochemistry is about

Weightage: Electrochemistry is part of the Physical Chemistry group, which carries 25 of the 70 marks in the ISC Chemistry Paper 1 (Theory), shared with 2 other chapters. Questions on it can appear in any section of the paper.Source: CISCE ISC Chemistry syllabus, examination year 2027, distribution of marks

Key concepts

Galvanic cells and cell notation

In the Daniell cell a zinc rod dips in ZnSO₄ solution and a copper rod in CuSO₄ solution, the two joined by a wire and a salt bridge. Zinc is oxidised at the anode ($Zn\to Z{n}^{2+}+2e^{-}$) and the electrons flow through the wire to copper, where Cu²⁺ is reduced at the cathode ($C{u}^{2+}+2e^{-}\to Cu$). Overall: $Zn(s)+C{u}^{2+}(aq)\to Z{n}^{2+}(aq)+Cu(s)$, $E^{\circ }$ = 1.10 V. Salt bridge (KCl or KNO₃ in agar): completes the circuit by letting ions migrate and keeps both solutions electrically neutral; without it the current stops at once. Cell notation: anode on the left, cathode on the right, single line for a phase boundary, double line for the salt bridge: Zn(s) | Zn²⁺(aq) ‖ Cu²⁺(aq) | Cu(s). Electrolytic vs galvanic: a galvanic cell converts chemical to electrical energy, has a negative anode and a spontaneous reaction (ΔG < 0); an electrolytic cell converts electrical to chemical energy, has a positive anode and needs an external source.

Worked example

Write the cell notation and the electrode reactions for a cell in which the reaction $Mg(s)+2A{g}^{+}(aq)\to M{g}^{2+}(aq)+2Ag(s)$ takes place. Which electrode is negative?

Try it first, then show the solutionSolution

1. Oxidation (anode): $Mg\to M{g}^{2+}+2e^{-}$; reduction (cathode): $A{g}^{+}+e^{-}\to Ag$ (×2)
2. Notation: Mg(s) | Mg²⁺(aq) ‖ Ag⁺(aq) | Ag(s)
3. In a galvanic cell the anode, where electrons are released, is negative: the magnesium electrode.

Answer: Mg | Mg²⁺ ‖ Ag⁺ | Ag; magnesium is the negative electrode

Standard hydrogen electrode and measuring electrode potentials

A single electrode potential cannot be measured on its own, only a difference. The standard hydrogen electrode is the reference: a platinum foil coated with platinum black, dipped in 1 M H⁺ (HCl), with pure H₂ gas bubbled at 1 bar and 298 K. Its potential is taken as 0.00 V: $2H^{+}(aq)+2e^{-}\rightleftharpoons H_2(g)$. **Measuring $E^{\circ }$ of Zn²⁺/Zn:** the cell Zn | Zn²⁺(1 M) ‖ H⁺(1 M) | H₂(1 bar), Pt reads 0.76 V with zinc as the anode, so $E_{Z{n}^{2+}/Zn}^{\circ }$ = −0.76 V. Cu²⁺/Cu: Pt, H₂ | H⁺ ‖ Cu²⁺ | Cu reads 0.34 V with copper as the cathode, so $E^{\circ }$ = +0.34 V. Limitations of SHE: it is hard to keep H₂ at exactly 1 bar and H⁺ at exactly 1 M; platinum is easily poisoned by impurities; it cannot be used in solutions with oxidising agents. Factors affecting electrode potential: nature of the metal (its tendency to lose electrons), concentration of ions (Nernst equation) and temperature.

Worked example

A cell made from SHE and a Ni²⁺(1 M)/Ni half-cell gives an emf of 0.25 V, with electrons flowing from nickel to the hydrogen electrode. Find $E_{N{i}^{2+}/Ni}^{\circ }$.

Try it first, then show the solutionSolution

1. Electrons leave nickel, so nickel is the anode and SHE is the cathode.
2. $E_{cell}^{\circ }=E_{SHE}^{\circ }-E_{Ni}^{\circ }=0-E_{Ni}^{\circ }=0.25$ V

Answer: $E_{N{i}^{2+}/Ni}^{\circ }$ = −0.25 V

Electrochemical series and feasibility

Half-cells listed in order of standard reduction potential: Li⁺/Li −3.05 V, K⁺/K −2.93, Mg²⁺/Mg −2.37, Al³⁺/Al −1.66, Zn²⁺/Zn −0.76, Fe²⁺/Fe −0.44, Ni²⁺/Ni −0.25, H⁺/H₂ 0.00, Cu²⁺/Cu +0.34, Ag⁺/Ag +0.80, F₂/F⁻ +2.87. Uses: (1) the more negative $E^{\circ }$, the stronger the reducing agent (Li strongest); the more positive, the stronger the oxidising agent (F₂ strongest). (2) A metal above hydrogen (negative $E^{\circ }$) displaces H₂ from dilute acids; metals below cannot. (3) A metal displaces any metal below it from a solution of its salt. (4) In a galvanic cell, the half-cell with the lower $E^{\circ }$ is the anode; $E_{cell}^{\circ }=E_{cathode}^{\circ }-E_{anode}^{\circ }$. (5) A reaction is feasible if the $E_{cell}^{\circ }$ calculated for it is positive.

Worked example

Can a copper vessel be used to store 1 M silver nitrate solution? Can an iron spoon be used to stir 1 M copper sulphate? ($E^{\circ }$: Ag⁺/Ag +0.80, Cu²⁺/Cu +0.34, Fe²⁺/Fe −0.44 V)

Try it first, then show the solutionSolution

1. Cu + 2Ag⁺ → Cu²⁺ + 2Ag: $E^{\circ }=0.80-0.34=+0.46$ V, feasible. Copper would dissolve and silver deposit.
2. Fe + Cu²⁺ → Fe²⁺ + Cu: $E^{\circ }=0.34-(-0.44)=+0.78$ V, feasible. Iron would dissolve.

Answer: Neither: both reactions are feasible.

Nernst equation, ΔG and equilibrium constant

For $M^{n+}(aq)+n{e}^{-}\to M(s)$: $E=E^{\circ }-\frac{2.303RT}{nF}\log \frac{1}{[M^{n+}]}$, i.e. $E=E^{\circ }+\frac{0.0591}{n}\log [M^{n+}]$ at 298 K. For a cell $aA+bB\to cC+dD$: $E_{cell}=E_{cell}^{\circ }-\frac{0.0591}{n}\log \frac{[C{]}^c[D{]}^d}{[A{]}^a[B{]}^b}$ (pure solids have activity 1). Free energy: $\Delta G=-nF{E}_{cell}$ and $\Delta G^{\circ }=-nF{E}_{cell}^{\circ }$. A positive emf gives negative ΔG, so the reaction is spontaneous. The maximum electrical work obtainable is $-\Delta G=nFE$. Equilibrium constant: at equilibrium $E_{cell}=0$ and Q = K, so $E_{cell}^{\circ }=\frac{0.0591}{n}\log K$ and $\Delta G^{\circ }=-2.303RT\log K$.

Worked example

For Zn | Zn²⁺(0.1 M) ‖ Cu²⁺(0.01 M) | Cu at 298 K, $E_{cell}^{\circ }$ = 1.10 V. Find $E_{cell}$ and the standard Gibbs energy change.

Try it first, then show the solutionSolution

1. Reaction: $Zn+C{u}^{2+}\to Z{n}^{2+}+Cu$, n = 2, $Q=\frac{[Z{n}^{2+}]}{[C{u}^{2+}]}=\frac{0.1}{0.01}=10$
2. $E_{cell}=1.10-\frac{0.0591}{2}\log 10=1.10-0.0296=1.070$ V
3. $\Delta G^{\circ }=-nF{E}^{\circ }=-2\times 96500\times 1.10=-212300$ J = −212.3 kJ mol⁻¹

Answer: 1.070 V; −212.3 kJ mol⁻¹

Conductance, cell constant and molar and equivalent conductivity

Metallic conductance is by electrons, involves no chemical change and decreases with rising temperature. Electrolytic conductance is by ions, involves chemical change at the electrodes and increases with rising temperature (ions move faster, viscosity falls). Conductance G = 1/R (siemens, S). $R=\rho \frac{l}{A}$: ρ is specific resistance (resistivity, Ω cm). Specific conductance (conductivity) $\kappa =\frac{1}{\rho }=G\times \frac{l}{A}$, in S cm⁻¹. Cell constant $=\frac{l}{A}=\kappa R$, found using a KCl solution of known κ. Molar conductivity ${\Lambda }_m=\frac{1000\kappa }{C}$ (S cm² mol⁻¹, C in mol L⁻¹): conductance of all ions from one mole of electrolyte. Equivalent conductivity ${\Lambda }_{eq}=\frac{1000\kappa }{N}$ (S cm² eq⁻¹), so ${\Lambda }_m=n\times {\Lambda }_{eq}$ where n is the number of equivalents per mole. Effect of dilution: κ *decreases* (fewer ions per unit volume); ${\Lambda }_m$ *increases* (the whole mole is still present; ions are freer for strong electrolytes, and more ionised for weak ones). For strong electrolytes ${\Lambda }_m={\Lambda }_m^{\circ }-A\sqrt{C}$ (a straight line against $\sqrt{C}$); for weak electrolytes ${\Lambda }_m$ rises steeply at low concentration and cannot be extrapolated.

Worked example

A conductivity cell with cell constant 1.30 cm⁻¹ filled with 0.02 M solution of an electrolyte has a resistance of 520 Ω. Find κ and ${\Lambda }_m$.

Try it first, then show the solutionSolution

1. $\kappa =\frac{\text{cell constant}}{R}=\frac{1.30}{520}=2.50\times 10^{-3}$ S cm⁻¹
2. ${\Lambda }_m=\frac{1000\times 2.50\times 10^{-3}}{0.02}=125$ S cm² mol⁻¹

Answer: κ = $2.50\times 10^{-3}$ S cm⁻¹; ${\Lambda }_m$ = 125 S cm² mol⁻¹

Kohlrausch's law and its applications

Kohlrausch's law of independent migration of ions: at infinite dilution, every ion makes a definite contribution to the molar conductivity of an electrolyte, whatever the other ion: ${\Lambda }_m^{\circ }={\nu }_{+}{\lambda }_{+}^{\circ }+{\nu }_{-}{\lambda }_{-}^{\circ }$ (ν = number of each ion in the formula). Applications: (1) ${\Lambda }_m^{\circ }$ of a weak electrolyte, which cannot be found by extrapolation, from strong electrolytes: ${\Lambda }^{\circ }(C{H}_{3}COOH)={\Lambda }^{\circ }(HCl)+{\Lambda }^{\circ }(C{H}_{3}COONa)-{\Lambda }^{\circ }(NaCl)$. (2) Degree of dissociation $\alpha =\frac{{\Lambda }_m^c}{{\Lambda }_m^{\circ }}$. (3) Dissociation constant of a weak acid $K_a=\frac{C{\alpha }^2}{1-\alpha }$. (4) Solubility of sparingly soluble salts: $S=\frac{1000\kappa }{{\Lambda }_m^{\circ }}$.

Worked example

${\Lambda }_m^{\circ }$ values (S cm² mol⁻¹): HCl 426.2, CH₃COONa 91.0, NaCl 126.5. A 0.001 M ethanoic acid solution has ${\Lambda }_m$ = 48.15 S cm² mol⁻¹. Find α and $K_a$.

Try it first, then show the solutionSolution

1. ${\Lambda }^{\circ }(C{H}_{3}COOH)=426.2+91.0-126.5=390.7$ S cm² mol⁻¹
2. $\alpha =\frac{48.15}{390.7}=0.1232$
3. $K_a=\frac{C{\alpha }^2}{1-\alpha }=\frac{0.001\times 0.1232^2}{0.8768}=1.73\times 10^{-5}$

Answer: α = 0.123; $K_a$ = $1.73\times 10^{-5}$

Faraday's laws of electrolysis

First law: the mass of a substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed: $w=ZQ=ZIt$. Z is the electrochemical equivalent, the mass deposited by 1 coulomb. Second law: when the same quantity of electricity is passed through different electrolytes in series, the masses deposited are proportional to their chemical equivalent masses: $\frac{w_1}{w_2}=\frac{E_1}{E_2}$. The faraday: the charge on one mole of electrons, $F=N_{A}e=6.023\times 10^{23}\times 1.602\times 10^{-19}\approx 96500$ C mol⁻¹; one faraday deposits one gram-equivalent of any substance, so $Z=\frac{E}{96500}$. Useful form: $w=\frac{MIt}{nF}$, where n is the number of electrons per ion discharged.

Worked example

A current of 2.0 A is passed through CuSO₄ solution for 30 minutes. Find the mass of copper deposited. (Cu = 63.5)

Try it first, then show the solutionSolution

1. Q = It = 2.0 × 1800 = 3600 C
2. $C{u}^{2+}+2e^{-}\to Cu$: 2 × 96 500 C deposit 63.5 g
3. $w=\frac{63.5\times 3600}{2\times 96500}=1.18$ g

Answer: 1.18 g of copper

Batteries and fuel cells

Primary cells cannot be recharged (the reaction is not reversible); secondary cells can be recharged by passing current in the opposite direction. Leclanché (dry) cell: zinc container is the anode; a graphite (carbon) rod surrounded by MnO₂ and carbon powder is the cathode; electrolyte is a moist paste of NH₄Cl and ZnCl₂. Anode: $Zn\to Z{n}^{2+}+2e^{-}$; cathode: $2Mn{O}_2+2N{H}_4^{+}+2e^{-}\to M{n}_2{O}_3+2N{H}_3+H_{2}O$; about 1.5 V; primary. Mercury cell: Zn-Hg amalgam anode, paste of HgO and carbon cathode, KOH-ZnO electrolyte; overall (zinc as amalgam) $Zn(s)+HgO(s)\to ZnO(s)+Hg(l)$; constant 1.35 V throughout its life, so used in hearing aids and watches. Lead storage battery: anode Pb, cathode PbO₂ (lead grid packed with PbO₂), electrolyte 38% H₂SO₄. Discharge: anode $Pb+S{O}_4^{2-}\to PbS{O}_4+2e^{-}$; cathode $Pb{O}_2+S{O}_4^{2-}+4H^{+}+2e^{-}\to PbS{O}_4+2H_{2}O$; overall $Pb+Pb{O}_2+2H_{2}S{O}_4\to 2PbS{O}_4+2H_{2}O$. On charging the reactions reverse; the acid's density falls on discharge. H₂-O₂ fuel cell: H₂ and O₂ bubbled through porous carbon electrodes (with Pt/Pd catalyst) into hot aqueous KOH. Anode $2H_2+4O{H}^{-}\to 4H_{2}O+4e^{-}$; cathode $O_2+2H_{2}O+4e^{-}\to 4O{H}^{-}$; overall $2H_2+O_2\to 2H_{2}O$. Advantages: high efficiency (about 70%), continuous supply as long as fuel is fed, pollution-free, the water is drinkable (used in spacecraft). Disadvantages: costly catalysts, storing hydrogen is hazardous and difficult.

Worked example

Why does the voltage of a mercury cell stay constant during use, while that of a dry cell falls?

Try it first, then show the solutionSolution

1. In the mercury cell, the overall reaction $Zn+HgO\to ZnO+Hg$ involves only solids and a liquid: no ion concentration appears in the Nernst equation, so E stays constant.
2. In the dry cell, NH₄⁺ and Zn²⁺ concentrations change as it discharges, and NH₃ builds up, so the voltage drops.

Answer: No dissolved species in the overall mercury-cell reaction

Corrosion: mechanism and prevention

Corrosion is the slow destruction of a metal by the action of air, moisture and other substances in its surroundings; rusting of iron gives hydrated ferric oxide $F{e}_2{O}_3\cdot x{H}_{2}O$. Electrochemical mechanism: a water drop containing dissolved CO₂ (and O₂) on iron forms a tiny galvanic cell. At the anodic spot iron is oxidised: $Fe\to F{e}^{2+}+2e^{-}$. Electrons flow through the metal to a cathodic spot, where oxygen is reduced in acid: $O_2+4H^{+}+4e^{-}\to 2H_{2}O$. Overall: $2Fe+O_2+4H^{+}\to 2F{e}^{2+}+2H_{2}O$. Fe²⁺ is then oxidised further by air to rust. Factors: moisture, oxygen, electrolytes (salt water speeds it), acidic gases (CO₂, SO₂), contact with a less active metal, strain in the metal, temperature. Prevention: barrier protection (paint, oil, grease, electroplating with tin or chromium); galvanising (coating with zinc, which is more reactive and corrodes in preference even when scratched); sacrificial protection/cathodic protection (a block of Mg or Zn joined to iron pipes and ship hulls corrodes instead of the iron); alloying (stainless steel); anti-rust solutions (alkaline phosphates).

Worked example

A tin-plated iron can and a galvanised (zinc-coated) iron bucket are both scratched. Which rusts faster at the scratch? Explain with electrode potentials (Zn²⁺/Zn −0.76, Fe²⁺/Fe −0.44, Sn²⁺/Sn −0.14 V).

Try it first, then show the solutionSolution

1. At a scratch the two metals form a cell. With zinc, zinc ($E^{\circ }$ lower) is the anode and is oxidised; iron is the cathode and is protected.
2. With tin, iron ($E^{\circ }$ lower than tin) becomes the anode, so iron corrodes faster than it would alone.

Answer: The tin-plated can rusts faster once scratched; zinc protects iron sacrificially.

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 emf

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

Use reduction potentials for both; positive means feasible.

Daniell cell reaction

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

$E^{\circ }$ = 1.10 V.

Nernst equation (298 K)

$E_{cell}=E_{cell}^{\circ }-\frac{0.0591}{n}\log Q$

Products over reactants; omit solids.

Gibbs energy and emf

$\Delta G^{\circ }=-nF{E}_{cell}^{\circ }$

Maximum work = nFE; keep the negative sign for ΔG.

Equilibrium constant

$\log K=\frac{n{E}_{cell}^{\circ }}{0.0591},\Delta G^{\circ }=-2.303RT\log K$

At equilibrium E = 0.

Conductivity and cell constant

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

κ in S cm⁻¹.

Molar and equivalent conductivity

${\Lambda }_m=\frac{1000\kappa }{C},{\Lambda }_{eq}=\frac{1000\kappa }{N}$

S cm² mol⁻¹ and S cm² eq⁻¹.

Kohlrausch's law

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

For weak electrolytes, combine three strong ones.

Degree of dissociation and Ka

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

Weak electrolytes only.

Faraday's first law

$w=ZIt=\frac{MIt}{nF}$

Z = E/96 500 g C⁻¹.

Faraday's second law

$\frac{w_1}{w_2}=\frac{E_1}{E_2}$

Cells in series, same charge.

Faraday constant

$F=N_{A}e\approx 96500Cmo{l}^{-1}$

Charge on one mole of electrons.

Lead storage battery (discharge)

$Pb+Pb{O}_2+2H_{2}S{O}_4\to 2PbS{O}_4+2H_{2}O$

Reverses on charging.

Fuel cell

$2H_2+O_2\to 2H_{2}O$

KOH electrolyte, porous carbon electrodes.

Rusting (electrochemical)

$2Fe+O_2+4H^{+}\to 2F{e}^{2+}+2H_{2}O$

Fe²⁺ then oxidised to $F{e}_2{O}_3\cdot x{H}_{2}O$.

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 and RED CAT

  ANode = OXidation; REDuction = CAThode.

  Use it when: Every cell question.

• Right minus left

  $E_{cell}^{\circ }=E_{right}^{\circ }-E_{left}^{\circ }$ = cathode − anode (both as reduction potentials).

  Use it when: Emf from cell notation.

• Dilution splits them

  On dilution κ goes Down, ${\Lambda }_m$ goes Up.

  Use it when: Conductance one-markers.

• LiKe My Aunt's Zippy Fe Nickel Hat Costs Silver

  Li, K, Mg, Al, Zn, Fe, Ni, H, Cu, Ag: reducing power falls along the line.

  Use it when: Electrochemical series order.

• 0.0591 over n

  At 298 K, 2.303RT/F = 0.0591 V; divide by electrons transferred.

  Use it when: Nernst and log K.

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 5-mark question split into 1 + 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: Stopping at E° or dropping the sign of ΔG

  CISCE examiners' comments, 2024: asked for maximum work from Zn | Zn²⁺ ‖ Ag⁺ | Ag, candidates calculated only $E_{cell}^{\circ }$, or found ΔG° but omitted the negative sign or used a wrong formula; in the Cu-Ag long question some calculated ΔG° instead of ΔG.

  Wrong working

  $E^{\circ }$ = 1.56 V (answer ends here), or ΔG = nFE.

  Right working

  $\Delta G^{\circ }=-nF{E}^{\circ }$; maximum work = $nF{E}^{\circ }$, stated with units.

  ✓Do this: Underline what is asked (work, ΔG or ΔG°) and finish with that quantity.

  Usually costs: 1 mark.

• ✗Mistake: Wrong substitution in the Nernst equation

  CISCE examiners' comments, 2024 and 2025: candidates wrote the Nernst equation wrongly, substituted incorrect values, wrote a negative $E_{cell}^{\circ }$, wrote unbalanced cell reactions or only half-reactions, and calculated $E_{cell}$ when $E_{cell}^{\circ }$ was asked.

  Wrong working

  $E=E^{\circ }-\frac{0.0591}{n}\log \frac{[C{u}^{2+}]}{[N{i}^{2+}]}$ (reactant over product).

  Right working

  Q = [products]/[reactants] from the balanced cell reaction: $\frac{[N{i}^{2+}]}{[C{u}^{2+}]}$.

  ✓Do this: Always write the overall balanced reaction before the Nernst equation.

  Usually costs: 1-2 marks.

• ✗Mistake: Incomplete reason for conductivity falling on dilution

  CISCE examiners' comments, 2024 and 2025: candidates rejected the correct reason in an assertion-reason item, and wrote 'the number of ions decreases' without 'per unit volume'.

  Wrong working

  On dilution the number of ions decreases.

  Right working

  On dilution the number of ions per unit volume (per cm³) decreases, so κ falls.

  ✓Do this: Include 'per unit volume' every time.

  Usually costs: 1 mark.

• ✗Mistake: Confusing primary, secondary and fuel cells

  CISCE examiners' comments, 2024 and 2025: candidates said a dry cell can be easily recharged, interchanged examples of primary and secondary cells, named nickel-cadmium or fuel cells as primary cells, and gave 'cannot be recharged' as the fuel cell's disadvantage.

  Wrong working

  Dry cell: secondary. Fuel cell disadvantage: cannot be recharged.

  Right working

  Dry and mercury cells are primary; lead storage and Ni-Cd are secondary; fuel cell drawbacks are cost of catalysts and hydrogen storage.

  ✓Do this: Learn a three-row table: primary, secondary, fuel cell.

  Usually costs: 1 mark each.

• ✗Mistake: Answering Faraday questions in the wrong unit or with approximate values

  CISCE examiners' comments, 2024 and 2025: candidates substituted wrongly in Faraday's law, gave the answer in faradays when coulombs were asked, used approximate values that gave an incorrect molar conductance, and used a wrong formula for degree of dissociation.

  Wrong working

  Charge = 0.53 F (coulombs were asked); α = Λ°/Λ.

  Right working

  Convert to coulombs (× 96 500); $\alpha ={\Lambda }_m/{\Lambda }_m^{\circ }$; keep four decimal places.

  ✓Do this: Re-read the unit asked and recheck the arithmetic.

  Usually costs: 1 mark.

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

   In a galvanic cell, the salt bridge:
   (a) supplies electrons
   (b) keeps the two solutions electrically neutral
   (c) is the site of oxidation
   (d) increases the emf

   Hint

   Ion migration.

   Answer

   (b)

2. Q2 · 1 mark · MCQBoard level

   The unit of molar conductivity is:
   (a) S cm⁻¹
   (b) S cm² mol⁻¹
   (c) Ω cm
   (d) S mol cm⁻²

   Hint

   1000κ/C.

   Answer

   (b)

3. Q3 · 1 mark · Very short answerBoard level

   Fill in the blanks: The electrochemical equivalent is the mass deposited by ______ coulomb, while one gram-equivalent is deposited by ______ coulombs.

   Hint

   Faraday's laws.

   Answer

   one; 96 500

4. Q4 · 1 mark · Very short answerBoard level

   Why can the emf of a working galvanic cell not be negative?

   Hint

   ΔG = −nFE.

   Answer

   A cell works only if its reaction is spontaneous, ΔG < 0; since ΔG = −nFE, E must be positive.

5. Q5 · 2 marks · Short answerHarder board level

   How long must a current of 1.5 A flow to deposit 2.0 g of zinc from ZnSO₄? (Zn = 65.4)

   Hint

   n = 2.

   Answer

   Q = 2.0 × 2 × 96 500/65.4 = 5902 C; t = 5902/1.5 = 3935 s (about 65.6 min).

6. Q6 · 2 marks · Short answerBoard level

   Show that $F=N_{A}e$ gives about 96 500 C mol⁻¹. ($N_A=6.023\times 10^{23}$, e = $1.602\times 10^{-19}$ C)

   Hint

   Multiply.

   Answer

   $6.023\times 10^{23}\times 1.602\times 10^{-19}=96488\approx 96500$ C mol⁻¹: the charge on one mole of electrons.

7. Q7 · 2 marks · Short answerHarder board level

   Calculate the potential of a hydrogen electrode in a solution of pH 3 at 298 K.

   Hint

   E = −0.0591 pH.

   Answer

   $E=0-\frac{0.0591}{1}\log \frac{1}{[H^{+}]}=-0.0591\times 3=-0.177$ V.

8. Q8 · 3 marks · Short answerHarder board level

   Compare metallic and electrolytic conduction in three ways.

   Hint

   Carriers, chemical change, temperature.

   Answer

   Metallic: by electrons, no chemical change, conductance decreases with temperature. Electrolytic: by ions, chemical change at electrodes, conductance increases with temperature.

9. Q9 · 3 marks · Short answerHOTS / competency

   For the Daniell cell, find the ratio $[Z{n}^{2+}]/[C{u}^{2+}]$ at which the emf becomes zero ($E^{\circ }$ = 1.10 V). Comment on the answer.

   Hint

   E = 0 at equilibrium.

   Answer

   $\log Q=\frac{2\times 1.10}{0.0591}=37.2$, so the ratio is about $1.7\times 10^{37}$. The reaction goes essentially to completion: the cell keeps working until Cu²⁺ is almost totally used up.

10. Q10 · 3 marks · Short answerHarder board level

    Write the electrode reactions in the Leclanché cell and explain why its voltage falls with use.

    Hint

    Ion concentrations change.

    Answer

    Anode: $Zn\to Z{n}^{2+}+2e^{-}$; cathode: $2Mn{O}_2+2N{H}_4^{+}+2e^{-}\to M{n}_2{O}_3+2N{H}_3+H_{2}O$. As it discharges, ion concentrations in the paste change and NH₃ accumulates around the cathode, so the voltage falls.

11. Q11 · 5 marks · Long answerHarder board level

    (a) State Kohlrausch's law. [1]
    (b) ${\lambda }^{\circ }$ of Ca²⁺ and Cl⁻ are 119.0 and 76.3 S cm² mol⁻¹. Find ${\Lambda }_m^{\circ }$ of CaCl₂. [1]
    (c) A 0.05 M NH₄OH solution has resistance 30.0 Ω in a cell of constant 0.30 cm⁻¹. If ${\Lambda }_m^{\circ }$(NH₄OH) = 271.0 S cm² mol⁻¹, find κ, ${\Lambda }_m$ and α. [3]

    Hint

    ${\Lambda }^{\circ }={\lambda }_{+}+2{\lambda }_{-}$.

    Answer

    (a) At infinite dilution each ion contributes a definite amount to the molar conductivity of the electrolyte, independent of the other ion. (b) 119.0 + 2 × 76.3 = 271.6 S cm² mol⁻¹. (c) κ = 0.30/30.0 = 0.0100 S cm⁻¹; ${\Lambda }_m$ = 1000 × 0.0100/0.05 = 200 S cm² mol⁻¹; α = 200/271.0 = 0.738.

12. Q12 · 2 marks · Short answerHOTS / competency

    An iron pipe and a copper pipe are joined in a domestic water system. Predict which corrodes near the joint and suggest a cheap remedy.

    Hint

    Lower $E^{\circ }$ = anode.

    Answer

    Iron ($E^{\circ }$ = −0.44 V) is the anode relative to copper (+0.34 V), so iron corrodes faster near the joint. Insert an insulating (dielectric) union between them, or attach a zinc or magnesium sacrificial anode.

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.Where does oxidation occur in any cell?Show answer

   At the anode.

2. 2.What is the potential of SHE?Show answer

   0.00 V by convention.

3. 3.Write the Nernst equation for $M^{n+}/M$ at 298 K.Show answer

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

4. 4.Relation between ΔG° and $E_{cell}^{\circ }$?Show answer

   $\Delta G^{\circ }=-nF{E}_{cell}^{\circ }$

5. 5.Define cell constant.Show answer

   The ratio l/A of the distance between electrodes to their area, equal to κR.

6. 6.State Faraday's second law.Show answer

   For the same charge, masses deposited are proportional to equivalent masses.

7. 7.Electrolyte in a lead storage battery?Show answer

   About 38% sulphuric acid.

8. 8.What is galvanising?Show answer

   Coating iron with zinc to protect it from rusting.

Quick revision checklist

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

• ☐Can you write cell notation and both half-reactions for any pair of half-cells?
• ☐Can you describe SHE and give two limitations?
• ☐Can you use the Nernst equation to find E, $E^{\circ }$ or an unknown concentration?
• ☐Can you convert $E^{\circ }$ into ΔG° and K?
• ☐Can you define κ, ${\Lambda }_m$, ${\Lambda }_{eq}$ and cell constant with units?
• ☐Can you apply Kohlrausch's law to a weak acid?
• ☐Can you solve Faraday first- and second-law problems?
• ☐Can you write the reactions of the dry, mercury, lead storage and fuel cells and of rusting?

Electrochemistry: frequently asked questions