ChemistryCBSE Class 12 · Electrochemistry

Electrolysis and Faraday's Laws: Class 12 Chemistry

Updated 15 October 2026 · Fact-checked

In electrolysis an external current drives a non-spontaneous reaction. Faraday's first law says the mass deposited is proportional to the charge passed; the second says equal charges deposit masses in the ratio of the substances' equivalent masses. One faraday (96487 C, about 96500 C) is the charge on one mole of electrons.

Understand Electrolysis and Faraday's Laws

In a galvanic cell a spontaneous reaction produces electricity. An electrolytic cell does the opposite: an external source pushes electrons through and forces a non-spontaneous reaction. Electrons enter at the cathode (negative), where reduction takes place, and leave from the anode (positive), where oxidation takes place. In molten NaCl, for example, $N a^{+} + e^{-} \to N a$ at the cathode and $2 C l^{-} \to C l_{2} + 2 e^{-}$ at the anode.

Faraday's laws. Michael Faraday measured how much substance is produced by a given charge.

*First law:* the amount of chemical reaction at an electrode is proportional to the quantity of electricity passed: $m = Z Q = Z I t$ , where $Z$ is the electrochemical equivalent.

*Second law:* when the same quantity of electricity passes through different electrolytes, the amounts of substances liberated are proportional to their chemical equivalent weights (atomic mass ÷ number of electrons needed per atom).

The modern way to use both laws is through moles of electrons. The charge on one electron is $1.6021 \times 1 0^{- 19}$ C, so one mole of electrons carries $6.02 \times 1 0^{23} \times 1.6021 \times 1 0^{- 19} = 96487$ C. This is 1 faraday (F). The charge passed is $Q = I \times t$ (A × s). So moles of electrons $= Q / F$ . From the electrode equation you know how many electrons each atom or molecule needs: $A g^{+} + e^{-} \to A g$ needs 1 F per mole, $C u^{2 +} + 2 e^{-} \to C u$ needs 2 F, $A l^{3 +} + 3 e^{-} \to A l$ needs 3 F. Hence $m = \frac{Q}{n F} \times M$ .

Products of electrolysis. What forms depends on the electrolyte, its state and concentration, and the electrodes. With inert electrodes (Pt, graphite) in aqueous solution, water competes with the ions:

At the cathode, the species with the higher reduction potential is reduced. For aqueous NaCl, $H_{2} O$ ( $E^{\circ} = - 0.83$ V) wins over $N a^{+}$ ( $- 2.71$ V), so hydrogen is released: $2 H_{2} O + 2 e^{-} \to H_{2} + 2 O H^{-}$ .
At the anode, by $E^{\circ}$ values water should be oxidised to $O_{2}$ (1.23 V) before $C l^{-}$ (1.36 V). But oxygen needs an extra voltage, the overpotential, so chlorine is released from brine. The overall change is $2 N a C l + 2 H_{2} O \to 2 N a O H + H_{2} + C l_{2}$ .
For dilute sulphuric acid, oxygen is released at the anode; for concentrated acid, $S O_{4}^{2 -}$ is oxidised to peroxodisulphate, $S_{2} O_{8}^{2 -}$ .

With an active anode such as copper in $C u S O_{4}$ solution, the anode itself dissolves ( $C u \to C u^{2 +} + 2 e^{-}$ ) and copper deposits on the cathode; this is the basis of electrolytic refining and electroplating.

Where the topic shows up: industrial production of aluminium, sodium, chlorine and sodium hydroxide, refining of copper, and electroplating.

Electrolysis and Faraday's Laws: formulas and key facts

Charge: $Q = I \times t$; $I$ in ampere, $t$ in second, $Q$ in coulomb.
Faraday constant: $1 F = N_{A} \times e = 96487 C mol^{- 1} \approx 96500 C mol^{- 1}$; Charge on one mole of electrons.
Faraday's first law: $m = Z I t$; $Z$ = electrochemical equivalent (mass deposited by 1 C).
Mass deposited: $m = \frac{I t}{n F} \times M$; $n$ = electrons per ion in the electrode reaction; $M$ = molar mass.
Faraday's second law: $\frac{m _{1}}{m _{2}} = \frac{E _{1}}{E _{2}}, E = \frac{M}{n}$; Same charge through cells in series.
Electrolysis of brine: $2 N a C l (a q) + 2 H_{2} O (l) \to 2 N a O H (a q) + H_{2} (g) + C l_{2} (g)$; $H_{2}$ at the cathode, $C l_{2}$ at the anode.

How to solve Electrolysis and Faraday's Laws questions

Every Faraday numerical can be done with one chain: current and time → charge → moles of electrons → moles of substance → mass or volume. Write the electrode reaction first to get $n$ .

1Write the electrode half-reaction for the substance asked, e.g. $C u^{2 +} + 2 e^{-} \to C u$ , and read $n$ .
2Convert time to seconds and find $Q = I t$ .
3Moles of electrons $= Q / F$ (take $F = 96500$ C mol⁻¹ unless told otherwise).
4Moles of substance $= \frac{moles of electrons}{n}$ ; multiply by molar mass for mass.
5For time or current, run the chain backwards: mass → moles → moles of electrons ( $\times n$ ) → charge ( $\times F$ ) → $t = Q / I$ .
6For cells in series, the same charge passes through each: find moles of electrons from one cell, then use each cell's own $n$ .
7For product questions, compare reduction potentials at the cathode and remember the oxygen overpotential at the anode.

Electrolysis and Faraday's Laws in real life

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.

Charge needed to make a metal by electrolysis

Suppose an aluminium plant wants to produce 1.0 kg of aluminium by electrolysing alumina dissolved in molten cryolite. A trainee asks how much electric charge that needs.

Calculate the charge required to deposit 1.0 kg of Al (27 g mol⁻¹), and the time if the current is 10 000 A.

Try it first, then show the solutionSolution

Cathode: $A l^{3 +} + 3 e^{-} \to A l$ , so $n = 3$ .
Moles of Al $= 1000/27 = 37.04$ mol; moles of electrons $= 3 \times 37.04 = 111.1$ mol.
$Q = 111.1 \times 96500 = 1.07 \times 1 0^{7}$ C.
$t = Q / I = 1.07 \times 1 0^{7} /10000 = 1072$ s, about 18 minutes.

Answer: About $1.07 \times 1 0^{7}$ C; about 18 minutes at 10 000 A.

Memory hook: Three electrons for every aluminium atom: that is why aluminium smelters sit next to power stations.

Tricks to remember Electrolysis and Faraday's Laws

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

I-T-Q-F-n-M chain
Current × Time = Charge; ÷ F = moles of e⁻; ÷ n = moles of metal; × M = mass.
Use it when: Forward or backward, every Faraday numerical.
RED CAT, AN OX
REDuction at the CAThode, oxidation at the ANode, in both galvanic and electrolytic cells.
Use it when: Writing electrode reactions.
Brine gives H, Cl and NaOH
Water beats sodium at the cathode; chloride beats oxygen at the anode (overpotential).
Use it when: Product questions on aqueous NaCl.

Electrolysis and Faraday's Laws: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Charge for a reduction · 1 mark · MCQBoard level

How much charge is needed to reduce 1 mol of $C r_{2} O_{7}^{2 -}$ to $C r^{3 +}$ in acidic medium?
(a) 3 F (b) 6 F (c) 2 F (d) 12 F

Try it first, then show the solutionSolution

Cr changes from +6 to +3: 3 electrons per Cr.
Two Cr atoms per $C r_{2} O_{7}^{2 -}$ : $C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to 2 C r^{3 +} + 7 H_{2} O$ , so 6 mol of electrons.

Answer: (b) 6 F ( $5.79 \times 1 0^{5}$ C).

Example 2: Mass of copper deposited · 2 marks · Very short answerBoard level

A current of 2.0 A is passed through aqueous copper sulphate for 30 minutes using platinum electrodes. Calculate the mass of copper deposited at the cathode. (Cu = 63.5 g mol⁻¹, $F$ = 96500 C mol⁻¹)

Try it first, then show the solutionSolution

$Q = 2.0 \times 30 \times 60 = 3600$ C (½ mark).
$C u^{2 +} + 2 e^{-} \to C u$ : 2 × 96500 C deposit 63.5 g (½ mark).
$m = \frac{3600 \times 63.5}{2 \times 96500} = 1.18$ g (1 mark).

Answer: About 1.18 g of copper.

Example 3: Time to deposit aluminium · 3 marks · Short answerHarder board level

How long will a current of 4.0 A take to deposit 1.80 g of aluminium from molten alumina? (Al = 27 g mol⁻¹, $F$ = 96500 C mol⁻¹)

Try it first, then show the solutionSolution

$A l^{3 +} + 3 e^{-} \to A l$ ; moles of Al $= 1.80/27 = 0.0667$ mol (1 mark).
Moles of electrons $= 3 \times 0.0667 = 0.200$ mol; $Q = 0.200 \times 96500 = 19300$ C (1 mark).
$t = Q / I = 19300/4.0 = 4825$ s $\approx 80.4$ min (1 mark).

Answer: 4825 s (about 80 minutes).

Example 4: Cells in series: Faraday's second law · 3 marks · Short answerHarder board level

Two electrolytic cells containing $A g N O_{3}$ and $C u S O_{4}$ solutions are connected in series. A steady current deposits 1.08 g of silver in the first cell. Calculate (a) the charge passed and (b) the mass of copper deposited in the second cell. (Ag = 108, Cu = 63.5 g mol⁻¹)

Try it first, then show the solutionSolution

$A g^{+} + e^{-} \to A g$ : moles Ag $= 1.08/108 = 0.0100$ mol = moles of electrons (1 mark).
(a) $Q = 0.0100 \times 96500 = 965$ C (½ mark).
(b) Same charge through the copper cell: $C u^{2 +} + 2 e^{-} \to C u$ , moles Cu $= 0.0100/2 = 0.00500$ mol (1 mark).
Mass Cu $= 0.00500 \times 63.5 = 0.318$ g. Check with the second law: $\frac{m _{C u}}{m _{A g}} = \frac{63.5/2}{108/1}$ (½ mark).

Answer: (a) 965 C. (b) 0.318 g of copper.

Example 5: Electrolysis of brine: products and amounts · 5 marks · Long answerHOTS / competency

Concentrated aqueous NaCl is electrolysed with inert electrodes using a current of 10 A for 965 s. (a) Write the reactions at the cathode and the anode and explain why sodium and oxygen are not the products. (b) Calculate the moles of $H_{2}$ and $C l_{2}$ formed and the mass of NaOH produced. ( $E^{\circ}$ : $N a^{+} / N a = - 2.71$ V, $H_{2} O / H_{2} = - 0.83$ V, $O_{2} / H_{2} O = + 1.23$ V, $C l_{2} / C l^{-} = + 1.36$ V; NaOH = 40 g mol⁻¹)

Try it first, then show the solutionSolution

Cathode: $2 H_{2} O (l) + 2 e^{-} \to H_{2} (g) + 2 O H^{-} (a q)$ . Water has a higher reduction potential (−0.83 V) than $N a^{+}$ (−2.71 V), so water is reduced, not sodium ions (1½ marks).
Anode: $2 C l^{-} (a q) \to C l_{2} (g) + 2 e^{-}$ . By $E^{\circ}$ alone, water (1.23 V) would be oxidised first, but oxygen evolution has a high overpotential, so chloride is oxidised (1½ marks).
(b) $Q = 10 \times 965 = 9650$ C; moles of electrons $= 9650/96500 = 0.100$ mol (½ mark).
$H_{2}$ : 2 electrons per molecule, 0.050 mol; $C l_{2}$ : 0.050 mol (½ mark).
$O H^{-}$ formed $= 0.100$ mol, so NaOH $= 0.100 \times 40 = 4.0$ g (1 mark).

Answer: (a) Cathode: $H_{2}$ from water; anode: $C l_{2}$ (oxygen overpotential). (b) 0.050 mol $H_{2}$ , 0.050 mol $C l_{2}$ , 4.0 g NaOH.

Where students lose marks in Electrolysis and Faraday's Laws

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

Mistake: Forgetting to convert minutes or hours to seconds
$Q = I t$ needs seconds.
Wrong working
$Q = 2.0 \times 30 = 60$ C
Right working
$Q = 2.0 \times 1800 = 3600$ C
Do this: Convert time in the first line.
Usually costs: The final answer mark.
Mistake: Wrong number of electrons
Students use 1 for every metal.
Wrong working
$m_{C u} = \frac{3600 \times 63.5}{96500}$
Right working
$m_{C u} = \frac{3600 \times 63.5}{2 \times 96500}$
Do this: Write the half-reaction; $n$ equals the charge on the ion being discharged.
Usually costs: 1 mark.
Mistake: Predicting sodium at the cathode for aqueous NaCl
Molten and aqueous electrolysis are mixed up.
Do this: Molten NaCl gives Na; aqueous NaCl gives $H_{2}$ because water is reduced more easily.
Usually costs: 1-2 marks.
Mistake: Explaining chlorine at the anode only with $E^{\circ}$
The $E^{\circ}$ values favour oxygen, so the answer contradicts itself.
Do this: Mention the overpotential of oxygen as the reason chlorine is released.
Usually costs: 1 mark in reasoning questions.

Exam tips

Begin every numerical with the electrode reaction; it carries a step mark and fixes $n$ .
Use the $F$ value given in the paper (96500 or 96487 C mol⁻¹).
For series cells, say explicitly 'the same quantity of electricity passes through both cells'.
In product questions, write both electrode reactions and the overall reaction.
Report time in the unit the question uses (seconds or minutes); show the conversion.

Practice questions on Electrolysis and Faraday's Laws

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

Q1 · 1 mark · MCQBoard level
One faraday of electricity is:
(a) the charge on one electron
(b) the charge on one mole of electrons
(c) 1 coulomb
(d) the charge needed to deposit 1 g of any metal
Hint
$N_{A} \times e$ .
Answer
(b), about 96487 C.
Q2 · 2 marks · Very short answerBoard level
The same quantity of electricity is passed through molten NaCl and molten $C a C l_{2}$ in series. State Faraday's second law and use it to find the ratio of the masses of sodium and calcium deposited. (Na = 23, Ca = 40 g mol⁻¹)
Hint
Equivalent mass = molar mass ÷ electrons per ion.
Answer
Second law: the masses of different substances liberated by the same quantity of electricity are proportional to their equivalent weights. $E_{N a} = 23/1 = 23$ and $E_{C a} = 40/2 = 20$ , so $m_{N a} : m_{C a} = 23 : 20$ (1.15 : 1).
Q3 · 2 marks · Very short answerHarder board level
Calculate the charge in coulombs required for (a) the reduction of 0.50 mol of $F e^{3 +}$ to $F e^{2 +}$ and (b) the deposition of 0.20 mol of Cu from $C u^{2 +}$ .
Hint
Electrons per ion: 1 and 2.
Answer
(a) $0.50 \times 1 \times 96500 = 48250$ C. (b) $0.20 \times 2 \times 96500 = 38600$ C.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): In the electrolysis of aqueous sodium chloride, hydrogen and not sodium is formed at the cathode.
Reason (R): The reduction potential of water (−0.83 V) is higher than that of $N a^{+}$ (−2.71 V).
(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
Higher reduction potential is reduced preferentially.
Answer
(a).
Q5 · 2 marks · Very short answerHarder board level
A solution of $N i S O_{4}$ is electrolysed with a current of 1.5 A for 1.0 hour. What mass of nickel is deposited at the cathode? (Ni = 58.7 g mol⁻¹)
Hint
$Q = 5400$ C, $n = 2$ .
Answer
$m = \frac{5400 \times 58.7}{2 \times 96500} = 1.64$ g.
Q6 · 2 marks · Very short answerHarder board level
A current of 0.50 A flows through a wire for 965 s. Calculate the number of electrons that pass through it. ( $N_{A} = 6.022 \times 1 0^{23}$ mol⁻¹)
Hint
Moles of electrons = $Q / F$ .
Answer
$Q = 482.5$ C; moles of electrons $= 0.00500$ ; number $= 0.00500 \times 6.022 \times 1 0^{23} = 3.01 \times 1 0^{21}$ .
Q7 · 3 marks · Short answerHOTS / competency
Predict the products at each electrode when aqueous $C u S O_{4}$ is electrolysed (a) with platinum electrodes and (b) with copper electrodes. Explain the difference and name one industrial use of (b).
Hint
An active anode can be oxidised itself.
Answer
(a) Cathode: $C u^{2 +} + 2 e^{-} \to C u$ ; anode: $2 H_{2} O \to O_{2} + 4 H^{+} + 4 e^{-}$ (solution becomes acidic, blue colour fades). (b) Cathode: Cu deposits; anode: $C u \to C u^{2 +} + 2 e^{-}$ , the copper anode dissolves, so the solution's concentration stays the same. Platinum is inert; copper is oxidised more easily than water. Use: electrolytic refining of copper (or copper plating).
Q8 · 4 marks · Case-basedHOTS / competency
Read and answer.
Suppose a small workshop silver-plates a copper medal. The medal has a total surface area of 10 cm², and the plating must be 0.010 mm thick. The plating bath contains a silver salt solution and a current of 0.20 A is used. (Density of silver = 10.5 g cm⁻³, Ag = 108 g mol⁻¹)
(a) Should the medal be the cathode or the anode? Write the electrode reaction. (1 mark)
(b) Calculate the mass of silver needed. (1 mark)
(c) Calculate the time required. (2 marks)
Hint
Volume = area × thickness; 0.010 mm = 0.0010 cm.
Answer
(a) Cathode: $A g^{+} + e^{-} \to A g$ . (b) Volume $= 10 \times 0.0010 = 0.010$ cm³; mass $= 0.010 \times 10.5 = 0.105$ g. (c) Moles Ag $= 0.105/108 = 9.72 \times 1 0^{- 4}$ mol = moles of electrons; $Q = 93.8$ C; $t = 93.8/0.20 = 469$ s, about 7.8 minutes.

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.Value of 1 faraday?Show answer
96487 C mol⁻¹ (about 96500 C).
2.Moles of electrons to deposit 1 mol Al?Show answer
3 mol (3 F).
3.Anode product in electrolysis of dilute $H_{2} S O_{4}$ ?Show answer
Oxygen.
4.Anode product with concentrated $H_{2} S O_{4}$ ?Show answer
Peroxodisulphate, $S_{2} O_{8}^{2 -}$ .
5.Formula for mass deposited?Show answer
$m = \frac{I tM}{n F}$ .
6.Why is Cl₂, not O₂, formed from brine?Show answer
Oxygen evolution needs a large overpotential.

Electrolysis and Faraday's Laws: frequently asked questions

What are Faraday's laws of electrolysis?

First law: mass of substance produced is proportional to the charge passed. Second law: for the same charge, masses of different substances produced are proportional to their equivalent weights.

What is 1 faraday in Class 12 Chemistry?

It is the charge carried by one mole of electrons, N_A \times e = 96487 C, usually rounded to 96500 C.

What are the products of electrolysis of aqueous NaCl?

Hydrogen at the cathode, chlorine at the anode, and sodium hydroxide in the solution.

What is overpotential?

The extra voltage, beyond the value predicted from E^\circ, needed to make some electrode reactions (such as oxygen evolution) go at a useful rate.

How do I find the time needed to deposit a given mass?

Find moles of substance, multiply by electrons per ion to get moles of electrons, multiply by F for charge, then divide by current.