Why the bookkeeping is the difficulty
Consider a question giving two standard electrode potentials and asking for the cell EMF. The formula is one subtraction. The work is deciding which electrode is the cathode - the one with the higher reduction potential - and then remembering that the formula subtracts the anode value, which may itself be negative, producing a double negative.
That pattern repeats. The Nernst equation is one substitution, provided the reaction quotient is written the right way up. Faraday's law is one multiplication, provided the electron count per ion is right. Getting the bookkeeping automatic is what makes this chapter fast.
How JEE actually asks Electrochemistry
NTA does not publish chapter-wise weightage, so figures online are coaching estimates from past papers; check the official NTA information bulletin for your session's syllabus and pattern.
The recurring types are numerically dense and repeat closely: compute a cell EMF from standard potentials, apply the Nernst equation at non-standard concentrations, relate EMF to ΔG or to the equilibrium constant, calculate a deposited mass by Faraday's laws, and distinguish conductivity from molar conductivity. Preparing those five covers most of what is asked.
Key concepts, compressed
- ✓A galvanic cell converts chemical energy into electrical energy spontaneously; an electrolytic cell uses electrical energy to drive a non-spontaneous reaction.
- ✓Standard electrode potentials are measured relative to the standard hydrogen electrode, defined as exactly 0.00 V.
- ✓A more positive reduction potential means a stronger tendency to be reduced, so that electrode becomes the cathode.
- ✓The Nernst equation corrects the standard potential for non-standard concentrations.
- ✓Faraday's constant, 96500 C per mole of electrons, converts between charge and moles of electrons.
- ✓Conductivity is per unit volume and falls on dilution; molar conductivity is per mole and rises, because the ions present dissociate more completely.
Relations you need before attempting the questions
| Quantity | Formula | Note |
|---|---|---|
| Cell potential | E°(cell) = E°(cathode) − E°(anode) | both as reduction potentials |
| Nernst equation at 298 K | E = E° − (0.059/n) log Q | 298 K only |
| Free energy and EMF | ΔG = −nFE | F = 96500 C/mol |
| EMF and equilibrium | E° = (0.059/n) log K | at 298 K |
| At equilibrium | E = 0 and ΔG = 0 | |
| Charge | Q = It | coulombs |
| Moles of electrons | Q / F | |
| Faraday's first law | m = ZIt | Z is the electrochemical equivalent |
| Molar conductivity | Λm = κ × 1000 / molarity | |
| Kohlrausch's law | Λm° = λ°(cation) + λ°(anion) |
The five mistakes that cost the most marks
- ✓Mixing up anode polarity between cell types. Oxidation is at the anode in both, but the anode is negative in a galvanic cell and positive in an electrolytic one.
- ✓Using 0.059/n away from 298 K. That coefficient already contains the temperature; at any other value you need 2.303RT/nF.
- ✓Writing the reaction quotient upside down in the Nernst equation, which flips the sign of the correction.
- ✓Confusing conductivity with molar conductivity. On dilution the first falls and the second rises.
- ✓Forgetting the electron count in Faraday calculations. One mole of Al³⁺ requires three faradays.
Practice set 1: galvanic cells
1. Where does oxidation occur and where does reduction occur?
Oxidation at the anode and reduction at the cathode, in every electrochemical cell without exception. A useful mnemonic is that the vowels go together - anode with oxidation, cathode with reduction. What changes between cell types is the polarity, not the chemistry.
2. What is the polarity of the anode in a galvanic cell and in an electrolytic cell?
Negative in a galvanic cell, positive in an electrolytic one. In a galvanic cell electrons are produced at the anode by spontaneous oxidation, making it the negative terminal. In an electrolytic cell an external supply forces oxidation at the electrode connected to its positive terminal. The site of oxidation is unchanged; only the sign differs.
3. How is a galvanic cell written in cell notation?
Anode on the left and cathode on the right, with a double vertical line for the salt bridge: Zn | Zn²⁺ || Cu²⁺ | Cu. Single lines mark phase boundaries. The convention is fixed, so reading a given notation immediately tells you which half-cell is being oxidised.
4. What is the function of a salt bridge?
To maintain electrical neutrality in both half-cells by allowing ions to migrate, and to complete the circuit. Without it, positive charge accumulates in the anode compartment and negative charge in the cathode compartment, and current stops almost immediately. It also prevents the two solutions from mixing directly.
5. Calculate E° for the cell Zn | Zn²⁺ || Cu²⁺ | Cu.
1.10 V. Copper has the higher reduction potential so it is the cathode: E°(cell) = E°(cathode) − E°(anode) = 0.34 − (−0.76) = 1.10 V. Watch the double negative - subtracting a negative adds. A positive result confirms the reaction as written is spontaneous.
6. Why is the standard hydrogen electrode assigned exactly 0.00 V?
Because only potential differences can be measured, never absolute electrode potentials. The SHE is a defined reference point, and every other standard potential is quoted relative to it. The choice is a convention, not a measurement - any other reference would shift all values by a constant without changing any cell EMF.
7. What does a positive cell potential indicate?
That the cell reaction as written is spontaneous, since ΔG = −nFE makes ΔG negative when E is positive. A negative cell potential means the reverse reaction is the spontaneous one, and driving the written reaction would require an external supply - which is exactly what an electrolytic cell does.
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Practice set 2: the Nernst equation
8. Write the Nernst equation at 298 K.
E = E° − (0.059/n) log Q, where n is the number of electrons transferred and Q the reaction quotient. It corrects the standard potential for concentrations away from 1 M. The full form is E = E° − (2.303RT/nF) log Q, and substituting 298 K produces the 0.059 coefficient.
9. For the Zn-Cu cell with [Zn²⁺] = 0.1 M and [Cu²⁺] = 0.01 M, find E.
About 1.07 V. The cell reaction is Zn + Cu²⁺ → Zn²⁺ + Cu with n = 2, so Q = [Zn²⁺]/[Cu²⁺] = 0.1/0.01 = 10 and log Q = 1. Then E = 1.10 − (0.059/2)(1) = 1.10 − 0.0295 = 1.0705 V. The potential falls slightly because product concentration exceeds reactant concentration.
10. How does increasing the product ion concentration affect the cell potential?
It decreases it. A larger Q makes log Q more positive, so more is subtracted from E°. This is Le Chatelier's principle expressed electrically - accumulating products reduces the driving force, and the cell potential falls steadily as a battery discharges.
11. What is the cell potential at equilibrium?
Zero. At equilibrium there is no net driving force, so no work can be extracted - which is exactly what a dead battery is. Setting E = 0 in the Nernst equation gives E° = (0.059/n) log K, which is how equilibrium constants are determined electrochemically.
12. How is the standard cell potential related to the equilibrium constant?
E° = (0.059/n) log K at 298 K, obtained by setting E = 0 in the Nernst equation, at which point Q equals K. A large positive E° therefore corresponds to a very large K - a strongly spontaneous reaction that proceeds nearly to completion.
13. Why can the 0.059 coefficient not be used at other temperatures?
Because it is 2.303RT/F evaluated at 298 K. At any other temperature the coefficient differs, and the full expression 2.303RT/nF must be used. Questions specifying a temperature other than 25 °C are testing exactly this, and using 0.059 anyway gives a plausible but wrong answer.
Practice set 3: EMF and thermodynamics
14. How are cell potential and Gibbs free energy related?
ΔG = −nFE, where n is the moles of electrons transferred and F is Faraday's constant. The negative sign means a positive EMF corresponds to a negative ΔG, that is a spontaneous reaction. This relation is the bridge between electrochemistry and thermodynamics.
15. Calculate ΔG° for the Zn-Cu cell.
About −212.3 kJ/mol. Using ΔG° = −nFE° = −(2)(96500)(1.10) = −212300 J/mol, which is −212.3 kJ/mol. The large negative value confirms a strongly spontaneous reaction. Note that n = 2 because two electrons transfer per zinc atom oxidised.
16. What is Faraday's constant and what does it represent?
96500 coulombs per mole of electrons - the charge carried by one mole of electrons. It is Avogadro's number multiplied by the elementary charge, 6.022 × 10²³ × 1.6 × 10⁻¹⁹, which gives approximately 96500 C. It converts between electrical charge and moles in every electrochemical calculation.
17. What combination of ΔG and E indicates a spontaneous cell reaction?
Negative ΔG and positive E, which are equivalent statements given ΔG = −nFE. A spontaneous reaction can deliver electrical work, which is what a galvanic cell does. If E is negative the reaction runs the other way, and driving it forward requires electrolysis.
18. What are ΔG and E at equilibrium?
Both zero. No further net reaction occurs, so no work can be extracted. This is the same condition from question 11 seen thermodynamically, and it is what makes the relation E° = (0.059/n) log K work - the standard values are related through the equilibrium point.
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Practice set 4: electrolysis and Faraday's laws
19. State Faraday's first law of electrolysis.
The mass of substance deposited or liberated at an electrode is proportional to the quantity of electricity passed: m = ZIt, where Z is the electrochemical equivalent. In practice it is easier to work through moles of electrons - charge divided by Faraday's constant - than to look up Z values.
20. What charge passes when a current of 2 A flows for 965 seconds?
1930 coulombs. Charge is current times time: Q = It = 2 × 965 = 1930 C. Time must be in seconds, so a question quoting minutes or hours needs conversion first - and that conversion is a frequent source of order-of-magnitude errors.
21. How many moles of electrons does that charge represent?
0.02 moles. Divide by Faraday's constant: 1930/96500 = 0.02 mol of electrons. This intermediate step is worth doing explicitly, because everything that follows depends on the electron count of the particular ion being deposited.
22. What mass of copper is deposited by that charge from a Cu²⁺ solution?
0.635 g. The half-reaction Cu²⁺ + 2e⁻ → Cu needs two electrons per copper atom, so 0.02 mol of electrons deposits 0.01 mol of copper. With a molar mass of 63.5 g/mol, that is 0.635 g. Omitting the division by two would give double the correct answer.
23. State Faraday's second law of electrolysis.
When the same quantity of electricity passes through different electrolytes, the masses deposited are proportional to their chemical equivalent weights. In practice this means that comparing two metals requires comparing molar mass divided by charge number - which is why the same charge deposits very different masses of silver and aluminium.
24. How many faradays are needed to deposit one mole of aluminium from Al³⁺?
Three. The half-reaction Al³⁺ + 3e⁻ → Al requires three electrons per aluminium atom, so one mole of aluminium needs three moles of electrons, that is 3 × 96500 C. Assuming one faraday per mole regardless of the ion is the most common error in this section.
25. In the electrolysis of aqueous sodium chloride, what is produced at each electrode?
Hydrogen at the cathode and chlorine at the anode, leaving sodium hydroxide in solution. Water is reduced in preference to sodium ions, because sodium has a far more negative reduction potential. This is why aqueous and molten electrolysis of the same salt give different products - molten sodium chloride gives sodium metal.
Practice set 5: conductance
26. What is the difference between conductivity and molar conductivity?
Conductivity κ is the conductance of a unit volume of solution and depends on how many ions are present per unit volume. Molar conductivity Λm is the conductance due to all the ions from one mole of electrolyte. The distinction matters because they respond oppositely to dilution.
27. How is molar conductivity calculated from conductivity?
Λm = κ × 1000 / molarity, with the factor of 1000 converting between cubic centimetres and litres. Unit consistency is the main hazard here - conductivity is usually quoted per centimetre and concentration per litre, and the 1000 exists precisely to reconcile them.
28. What happens to conductivity and molar conductivity on dilution?
Conductivity decreases and molar conductivity increases. Diluting reduces the number of ions per unit volume, so κ falls. But molar conductivity counts the ions from a fixed amount of electrolyte, and dilution promotes fuller dissociation, so Λm rises. They genuinely move in opposite directions, which is what makes this a favourite exam question.
29. State Kohlrausch's law.
At infinite dilution, the molar conductivity of an electrolyte is the sum of the independent contributions of its cation and anion: Λm° = λ°(cation) + λ°(anion). Its practical value is that it allows the limiting molar conductivity of a weak electrolyte to be obtained indirectly, by combining values from strong electrolytes.
30. How do strong and weak electrolytes differ in their conductivity behaviour on dilution?
A strong electrolyte's molar conductivity rises only slightly and levels off, since it is already fully dissociated and the small increase comes from reduced ionic interference. A weak electrolyte's rises sharply and does not level off in the measurable range, because dilution genuinely increases its degree of dissociation - which is why its limiting value must be found by Kohlrausch's law rather than by extrapolation.
How to study this chapter efficiently
- ✓Write the two half-reactions before anything else. They identify the anode, the cathode and the electron count n in one step.
- ✓Check the sign twice when subtracting a negative anode potential. That double negative causes more errors here than any concept.
- ✓For Nernst problems, write Q explicitly as products over reactants for the cell reaction as written.
- ✓For Faraday problems, always compute moles of electrons as an intermediate, then divide by the charge number of the ion.
- ✓Learn the conductivity and molar conductivity behaviours as an opposing pair - one falls on dilution, the other rises.
- ✓Study this chapter after Mole Concept and Thermodynamics. It uses both, and the ΔG link is only meaningful if thermodynamics is secure.
Turn this into active practice
This chapter is procedural enough that speed matters more than insight. The concepts are few and the calculations short, so the difference between students is largely how quickly the setup happens - which half-cell is the cathode, what n is, which way up Q goes.
The JEE Electrochemistry quiz on QUFF generates fresh questions across cells, the Nernst equation, Faraday's laws and conductance, marks them instantly and explains each answer. Do timed sets and watch specifically for sign errors and missing electron counts - between them they account for most of what goes wrong here.
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