Class 12 Chemistry · Original EZT practice
Electrochemistry (Class 12 Chemistry Chapter 2): turning redox reactions into electric current
Every battery is a controlled redox reaction: oxidation at one electrode, reduction at the other, and electrons forced through a wire between them. This chapter gives you the numbers behind that - electrode potentials, the Nernst equation, conductance and Faraday's laws. Below: the ideas in plain words, then original EZT practice with worked answers. (Note: this is original practice, not a textbook exercise key.)
10 original worked examples, 10 fresh self-check questions, with explained answers. These are not copied textbook exercises, an official chapter index or a full answer key. No affiliation with NCERT is claimed.
Official sources for topic and edition context, checked 17 October 2026:
1. The galvanic cell: chemistry makes electricity
In a Daniell cell, a zinc rod in zinc sulphate and a copper rod in copper sulphate are joined by a wire and a salt bridge. Zinc is oxidised at the anode (Zn to Zn2+ + 2e-), copper ions are reduced at the cathode (Cu2+ + 2e- to Cu), and electrons flow from zinc to copper through the wire. The salt bridge completes the circuit and keeps the solutions electrically neutral.
2. Electrode potential and EMF
Each half-cell has an electrode potential measured against the standard hydrogen electrode (SHE), fixed at 0 volts. The cell's EMF: E cell = E cathode - E anode. A more negative standard potential means a stronger reducing agent.
3. EMF, free energy and equilibrium
The cell's driving force links to thermodynamics: delta G = -nFE, where n is electrons transferred and F = 96500 C/mol. A positive E means a negative delta G - a spontaneous reaction. At 298 K, E0 = (0.059 / n) log K, tying EMF to the equilibrium constant.
4. Conductance of solutions
Electrolyte solutions conduct through their ions. Molar conductivity: Lm = kappa x 1000 / M, where kappa is conductivity in S/cm and M is molar concentration, giving S cm2/mol. For weak electrolytes, molar conductivity rises sharply on dilution; Kohlrausch's law says each ion contributes its own fixed share at infinite dilution.
5. Faraday's laws and electrolysis
Faraday's first law: the mass deposited at an electrode is proportional to the charge passed. One mole of electrons (96500 C, one faraday) deposits one gram-equivalent - so 96500 C deposits 31.75 g of copper from Cu2+, half of its 63.5 g molar mass, because Cu2+ takes two electrons.
Choose, solve, check, then try a fresh question.
Before calculating, read the method prompt and name the condition that makes it useful. Write your own attempt. Open the steps only when you need them, and use the check to catch a wrong sign, unit or interpretation. Finish with the fresh question before reading its answer.
Identify the anode reaction
Choose the method: Find the electrode where oxidation loses electrons.
EZT practice question
In a Daniell cell, which electrode is the anode and what happens there?
Show the worked solution
- Zinc has the stronger tendency to give electrons.
- Zinc atoms lose two electrons and enter solution as Zn2+.
Answer: The zinc rod is the anode; oxidation occurs there: Zn loses two electrons and goes into solution as Zn2+.
Check: Electrons then travel through the wire from zinc to copper.
Watch for this: The cathode is copper, where reduction occurs; do not swap the electrodes.
Now try it yourself
What happens at the copper cathode?
Check your answer
Cu2+ gains two electrons and deposits as Cu.
Reduction is gain of electrons at the cathode.
Explain the salt bridge
Choose the method: Connect ion flow, circuit completion and charge balance.
EZT practice question
What does the salt bridge do?
Show the worked solution
- The wire carries electrons but not ions.
- The salt bridge lets ions move between the half-cells.
Answer: It lets ions flow between the two half-cells, completing the circuit and preventing charge build-up; without it the cell stops in seconds.
Check: The solutions remain electrically neutral as the cell runs.
Watch for this: The salt bridge does not carry the main electron current; the wire does that.
Now try it yourself
Why does a cell stop without a salt bridge?
Check your answer
Charge builds up in the half-cells and stops the redox flow.
Ion movement is needed to keep each solution neutral.
Calculate standard EMF
Choose the method: Subtract the anode potential from the cathode potential.
EZT practice question
Given E0(Zn2+/Zn) = -0.76 V and E0(Cu2+/Cu) = +0.34 V, find the standard EMF of the Daniell cell.
Show the worked solution
- Copper is the cathode and zinc is the anode.
- E0cell = E0cathode - E0anode.
Answer: E0cell = 0.34 - (-0.76) = 1.10 V.
Check: A positive EMF matches a spontaneous cell reaction.
Watch for this: Subtracting a negative potential adds 0.76 V; do not report -0.42 V.
Now try it yourself
Find E0cell for E0(Ag+/Ag) = +0.80 V with zinc.
Check your answer
1.56 V.
0.80 - (-0.76) = 1.56.
Compare reducing agents
Choose the method: Read the more negative standard potential as the stronger electron giver.
EZT practice question
Between zinc (E0 = -0.76 V) and copper (E0 = +0.34 V), which is the stronger reducing agent?
Show the worked solution
- A reducing agent gives electrons.
- Zinc has the more negative standard potential.
Answer: Zinc is the stronger reducing agent; the more negative the standard potential, the stronger the tendency to give electrons.
Check: Zinc is oxidised in the Daniell cell.
Watch for this: A positive potential does not mean stronger reduction power for the metal.
Now try it yourself
Which is stronger as a reducing agent, zinc or silver (E0 = +0.80 V)?
Check your answer
Zinc.
Zinc's standard potential is much more negative.
Recall the hydrogen electrode reference
Choose the method: Treat the standard hydrogen electrode as the zero reference.
EZT practice question
What is the potential of the standard hydrogen electrode, and why?
Show the worked solution
- Electrode potentials need a reference point.
- The standard hydrogen electrode is assigned the reference value.
Answer: Exactly 0 V by international convention; it is the reference every other electrode is measured against.
Check: Other standard potentials are relative to this zero.
Watch for this: The value is assigned, not measured as an absolute potential.
Now try it yourself
What does SHE stand for?
Check your answer
Standard hydrogen electrode.
It is the conventional reference half-cell.
Calculate free energy from EMF
Choose the method: Use delta G = -nFE with the charge in coulombs per mole.
EZT practice question
For the Daniell cell (E0 = 1.10 V, n = 2), find delta G0.
Show the worked solution
- Multiply n, F and E.
- Attach the negative sign from the formula.
Answer: delta G0 = -2 x 96500 x 1.10 = -212300 J = -212.3 kJ/mol, so the reaction is strongly spontaneous.
Check: Positive E gives negative delta G.
Watch for this: Do not drop the negative sign in delta G = -nFE.
Now try it yourself
Find delta G0 for a cell with E0 = 0.50 V and n = 2.
Check your answer
-96.5 kJ/mol.
-2 x 96500 x 0.50 = -96500 J.
Estimate the equilibrium constant
Choose the method: Rearrange E0 = (0.059 / n) log K at 298 K.
EZT practice question
Estimate the equilibrium constant K for the Daniell cell at 298 K.
Show the worked solution
- log K = nE0 / 0.059.
- Substitute n = 2 and E0 = 1.10.
Answer: log K = 2 x 1.10 / 0.059 = 37.3, so K is about 2 x 10^37; the reaction goes essentially to completion.
Check: A large positive EMF gives an enormous K.
Watch for this: Do not report log K as K; take 10 to the power.
Now try it yourself
For E0 = 0.59 V and n = 2, what is log K?
Check your answer
20.
2 x 0.59 / 0.059 = 20.
Calculate molar conductivity
Choose the method: Use Lm = kappa x 1000 / M.
EZT practice question
A 0.1 M solution has kappa = 0.02 S/cm. Find its molar conductivity.
Show the worked solution
- Multiply conductivity by 1000.
- Divide by the molar concentration.
Answer: Lm = 0.02 x 1000 / 0.1 = 200 S cm2/mol.
Check: The units reduce to S cm2/mol.
Watch for this: Do not divide by 1000; the formula multiplies kappa by 1000.
Now try it yourself
A 0.5 M solution has kappa = 0.01 S/cm. Find Lm.
Check your answer
20 S cm2/mol.
0.01 x 1000 / 0.5 = 20.
Explain dilution in weak electrolytes
Choose the method: Link dilution to greater ionisation.
EZT practice question
Why does the molar conductivity of acetic acid jump on dilution while KCl's changes little?
Show the worked solution
- Acetic acid is weak and only partly ionised.
- Dilution shifts it towards fuller ionisation; KCl is already fully ionised.
Answer: Acetic acid is a weak electrolyte - dilution shifts it towards full ionisation, so far more ions carry current; KCl is fully ionised at all dilutions.
Check: The change is in the number of charge carriers, not only the volume.
Watch for this: Do not treat all electrolytes as weak.
Now try it yourself
Which law gives each ion's contribution at infinite dilution?
Check your answer
Kohlrausch's law.
It says each ion contributes its own fixed share at infinite dilution.
Use one faraday of charge
Choose the method: Convert electron moles into metal moles using the half-reaction.
EZT practice question
How much copper does 96500 C deposit from a Cu2+ solution?
Show the worked solution
- Cu2+ + 2e- to Cu needs 2 F per mole of copper.
- One faraday therefore deposits half a mole.
Answer: Cu2+ + 2e- to Cu needs 2 F per mole, so 1 F deposits half a mole = 31.75 g of copper.
Check: Half of the 63.5 g molar mass is 31.75 g.
Watch for this: Do not deposit one full mole for one faraday when the ion needs two electrons.
Now try it yourself
How much copper would 2 F deposit?
Check your answer
63.5 g.
Two faradays deposit one mole of copper from Cu2+.
Try it yourself
Work these on paper before checking any notes. Answers are deliberately not shown here.
- Write the two half-reactions in a Daniell cell.
- Find E0cell for E0(Ag+/Ag) = +0.80 V with zinc.
- Find delta G0 for a cell with E0 = 0.50 V and n = 2.
- A 0.5 M solution has kappa = 0.01 S/cm - find Lm.
Know the rule but lose track of the steps?
Bring your book edition, your school topic list and a question you attempted. Ask a tutor to explain the choice of method and test it with a different problem. Agree the lesson topics, schedule and fee before starting.
For cost context, see home tuition fees in Delhi NCR. Agree the topics, schedule and fee with your chosen tutor.
Questions about this practice page
- What is a galvanic (voltaic) cell?
- A device that converts a spontaneous redox reaction into electric current - oxidation at the anode, reduction at the cathode, electrons flowing through the external wire.
- How do you calculate a cell's standard EMF?
- E0cell = E0cathode - E0anode using standard electrode potentials measured against the hydrogen electrode (0 V).
- What is the relation between EMF and Gibbs free energy?
- delta G = -nFE - a positive EMF means a negative free-energy change, so the reaction is spontaneous.
- What does Faraday's first law of electrolysis say?
- The mass of substance deposited or dissolved at an electrode is directly proportional to the total charge passed; 96500 C (one faraday) deposits one gram-equivalent.
- Are these the NCERT exercise questions for Electrochemistry?
- No. EZT pages are original practice on this chapter - use them to prepare for the textbook exercises, not as a substitute for your book.