Why Faraday's laws matter
Michael Faraday was the first scientist to describe the quantitative aspects of electrolysis. Between 1833 and 1834, after extensive study of solutions and molten electrolytes, he published two laws that remain the quantitative backbone of every electrolytic process — from the electrorefining of copper to the extraction of aluminium and sodium. In NCERT terms, these laws "flow" directly from the electrode reactions already discussed: electrolysis is simply a redox reaction driven by an external source, and Faraday's laws tell us exactly how much product that drive yields.
The conceptual leap is that an electrode reaction has a fixed stoichiometry in electrons. When a silver ion is reduced, $\ce{Ag+ + e- -> Ag}$, exactly one electron is consumed per atom of silver deposited. Count the electrons that flow and you have, in principle, counted the atoms produced. The current carries the electrons; time accumulates them; and a single conversion factor — the charge on one mole of electrons — turns that electron count into moles, and finally grams.
The first law: mass ∝ charge
Faraday's first law of electrolysis states that the amount of chemical reaction occurring at an electrode during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte. Expressed as the mass $m$ of substance liberated or deposited on passing a charge $Q$:
$$ m \propto Q \qquad\Longrightarrow\qquad m = Z\,Q $$
The proportionality constant $Z$ is the electrochemical equivalent of the substance. Because the charge supplied by a steady current $I$ flowing for a time $t$ is $Q = It$, the working form of the first law is the equation NEET expects you to deploy on sight:
$$ m = Z\,I\,t $$
Here $m$ is in grams, $I$ in amperes, $t$ in seconds, and $Z$ in grams per coulomb. The figure below traces the physical chain the law encodes: a measured current over a measured time fixes the charge, and the charge fixes the deposited mass.
Charge, current and time: Q = It
In Faraday's own era there were no constant-current sources. Chemists measured charge indirectly with a coulometer — a standard electrolytic cell in which the mass of silver or copper deposited reported the charge that had flowed. Coulometers are now obsolete; we have stable current sources and read the charge directly from $Q = It$, with $Q$ in coulombs when $I$ is in amperes and $t$ in seconds.
This single substitution is where most marks are won or lost. NEET problems quote current in amperes and time in minutes or hours; the formula demands seconds. Convert first, always.
Time must be in seconds before you touch Q = It
A current of 1.5 A passed "for 10 minutes" is 1.5 × 600 s = 900 C, not 1.5 × 10 = 15 C. A factor-of-60 (or 3600 for hours) error is the single most common slip on electrolysis sums, and the wrong value is almost always one of the four options.
Convert minutes → seconds (×60) and hours → seconds (×3600) before computing charge.
The Faraday constant and moles of electrons
The bridge between charge and chemistry is the charge carried by one mole of electrons. The charge on a single electron is $1.6021 \times 10^{-19}\ \text{C}$, so one mole of electrons carries:
$$ F = N_A \times e = 6.022 \times 10^{23}\ \text{mol}^{-1} \times 1.6021 \times 10^{-19}\ \text{C} = 96487\ \text{C mol}^{-1} $$
This quantity of electricity is called one faraday, symbol $F$. For NEET arithmetic the rounded value $F \approx 96500\ \text{C mol}^{-1}$ is used universally. The number of moles of electrons that have passed is therefore simply the charge divided by the Faraday constant:
$$ n_{e^-} = \dfrac{Q}{F} = \dfrac{It}{F} $$
The amount of charge needed for a given oxidation or reduction depends entirely on the stoichiometry of the electrode reaction. One mole of electrons reduces one mole of $\ce{Ag+}$, but two are required for magnesium and three for aluminium:
$\ce{Ag+ + e- -> Ag}$ (1 F) $\ce{Mg^2+ + 2e- -> Mg}$ (2 F) $\ce{Al^3+ + 3e- -> Al}$ (3 F)
The same logic governs the magnitude of industrial electrolysis. Commercial metal extraction runs currents as high as 50,000 A, which amounts to roughly 0.518 F passing every second. This is why the aluminium and chlor-alkali industries are among the largest single consumers of electricity in any economy: the throughput of metal is set, atom for atom, by the number of electrons the cell can be made to carry, and Faraday's first law fixes the exact conversion between kilowatt-hours of charge and tonnes of product.
The stoichiometric coefficient $n$ is therefore the single most decision-critical number in any electrolysis sum. It is read directly off the balanced electrode half-reaction, never off the overall formula of the compound. For a simple metal cation the value of $n$ is just its charge; for a polyatomic redox couple it is the total change in oxidation number summed over every atom that is oxidised or reduced. Getting $n$ right is what separates a one-mark answer from a wrong option.
| Electrode reaction | Electrons (n) | Charge per mole product |
|---|---|---|
| $\ce{Ag+ + e- -> Ag}$ | 1 | 1 F = 96500 C |
| $\ce{Cu^2+ + 2e- -> Cu}$ | 2 | 2 F = 193000 C |
| $\ce{Mg^2+ + 2e- -> Mg}$ | 2 | 2 F = 193000 C |
| $\ce{Al^3+ + 3e- -> Al}$ | 3 | 3 F = 289500 C |
| $\ce{2H2O + 2e- -> H2 + 2OH-}$ | 2 (per H2) | 2 F = 193000 C |
| $\ce{2H2O -> O2 + 4H+ + 4e-}$ | 4 (per O2) | 4 F = 386000 C |
Electrochemical and chemical equivalents
Two "equivalents" appear in this topic and confusing them is costly. The electrochemical equivalent $Z$ is the mass of a substance liberated by one coulomb of charge (or by one ampere flowing for one second). The chemical equivalent — the equivalent weight — is the molar mass divided by the number of electrons exchanged at the electrode:
$$ \text{Equivalent weight} = \dfrac{\text{Molar mass}}{n} \qquad\text{and}\qquad Z = \dfrac{\text{Equivalent weight}}{F} = \dfrac{M}{nF} $$
The relationship $Z = M/nF$ is worth memorising because it lets you regenerate the first law from scratch: substituting into $m = ZIt$ gives the unified form below, which is the formula most students actually use.
$$ m = \dfrac{M}{nF}\,It = \dfrac{\text{Equivalent weight} \times Q}{96500} $$
Faraday's laws presuppose you know which species is reduced at the cathode and oxidised at the anode. Revisit electrochemical cells and electrolysis for the electrode-reaction rules these numericals rest on.
The second law: mass ∝ equivalent weight
Faraday's second law of electrolysis compares the masses of different substances liberated by the same quantity of electricity. It states that when the same charge is passed through different electrolytes — placed in separate cells connected in series — the masses of the substances liberated are proportional to their chemical equivalent weights. For two substances A and B:
$$ \dfrac{m_A}{m_B} = \dfrac{\text{Equivalent weight of A}}{\text{Equivalent weight of B}} $$
The physical reason is exactly the electron-counting argument again. The same charge means the same number of moles of electrons reach each cell. A monovalent ion captures one electron per atom, a divalent ion two, so for an identical electron supply the cell with the higher-valency ion deposits proportionally fewer moles — and the masses settle into the ratio of equivalent weights. The series schematic below shows why "same charge" is automatic in a series arrangement.
Worked numericals: the master chain
Every Faraday numerical reduces to four steps: find the charge $Q = It$; convert to moles of electrons $n_{e^-} = Q/F$; use the half-reaction to find moles of product; multiply by molar mass. The three examples below cover the standard NEET variants.
A solution of $\ce{CuSO4}$ is electrolysed for 10 minutes with a current of 1.5 A. What mass of copper is deposited at the cathode? (Cu = 63.5)
Charge: $t = 10 \times 60 = 600\ \text{s}$, so $Q = It = 1.5 \times 600 = 900\ \text{C}$.
Half-reaction: $\ce{Cu^2+ + 2e- -> Cu}$, so $n = 2$; depositing 1 mol (63.5 g) of Cu needs $2F = 2 \times 96500\ \text{C}$.
Mass: $m = \dfrac{63.5 \times 900}{2 \times 96500} \approx 0.296\ \text{g}$ of copper. (NCERT Example 2.10 reports 0.2938 g using Cu = 63.)
How many faradays are needed to produce 20 g of calcium from molten $\ce{CaCl2}$? (Ca = 40)
Half-reaction: $\ce{Ca^2+ + 2e- -> Ca}$, so $n = 2$ and the equivalent weight $= 40/2 = 20\ \text{g}$.
Equivalents of product: $\dfrac{20\ \text{g}}{20\ \text{g/equiv}} = 1$ equivalent. By the first law, charge in faradays equals gram-equivalents of product, so $1 \times 1 = \mathbf{1\ F}$.
This is NEET 2020 Q.177 verbatim — the answer is 1 faraday, not 2. Multiplying 0.5 mol × 2 e− = 1 F; the valency cancels the half-mole.
What mass of silver is deposited when 500 C is passed through a solution of $\ce{AgNO3}$? (Ag = 108)
Half-reaction: $\ce{Ag+ + e- -> Ag}$, so 1 mol (108 g) needs $1F = 96500\ \text{C}$.
Mass: $m = \dfrac{108 \times 500}{96500} \approx 0.56\ \text{g}$ of silver. This is exactly how the historical silver coulometer reported charge.
Match the electron count to the species, not the formula
In the NEET 2024 match-list, "1 mol of $\ce{MnO4-}$ to $\ce{Mn^2+}$" needs 5 F because Mn goes from +7 to +2 — a five-electron change — while "1 mol of $\ce{H2O}$ to $\ce{O2}$" needs only 2 F (each O goes from −2 to 0, two electrons per water molecule). Always assign electrons from the change in oxidation state, never from the count of atoms.
Faradays per mole = change in oxidation number × number of redox atoms in the formula.
Series cells and gas volumes
When several cells share one circuit in series, the same current flows for the same time, so each cell receives an identical charge $Q = It$. This is precisely the condition Faraday's second law requires. The deposited masses then sit in the ratio of equivalent weights, and a single calculation in one cell fixes the products in all the others.
The same current that deposits 1.08 g of silver in a $\ce{AgNO3}$ cell passes through a $\ce{CuSO4}$ cell in series. What mass of copper is deposited? (Ag = 108, Cu = 63.5)
Second law: $\dfrac{m_{Cu}}{m_{Ag}} = \dfrac{E_{Cu}}{E_{Ag}}$ where $E_{Ag} = 108/1 = 108$ and $E_{Cu} = 63.5/2 = 31.75$.
Mass of Cu: $m_{Cu} = 1.08 \times \dfrac{31.75}{108} \approx 0.318\ \text{g}$ of copper — no need to know the actual charge.
Gas-evolution problems extend the same chain by one extra step: convert moles of gas to a volume using the molar volume of 22.4 L at STP. For oxygen at the anode, $\ce{2H2O -> O2 + 4H+ + 4e-}$, four moles of electrons release one mole of $\ce{O2}$; for hydrogen at the cathode, $\ce{2H2O + 2e- -> H2 + 2OH-}$, two moles of electrons release one mole of $\ce{H2}$.
A charge of 1 F is passed through acidified water with inert electrodes. What volume of $\ce{O2}$ is liberated at the anode at STP?
Moles of electrons: $n_{e^-} = Q/F = 96500/96500 = 1\ \text{mol}$.
Anode reaction: $\ce{2H2O -> O2 + 4H+ + 4e-}$ → 4 mol e− give 1 mol $\ce{O2}$, so 1 mol e− gives 0.25 mol $\ce{O2}$.
Volume: $0.25 \times 22.4 = 5.6\ \text{L}$ of oxygen at STP.
Faraday's laws in one screen
- First law: $m = ZIt$; mass deposited is proportional to charge $Q = It$.
- Faraday constant: $F = N_A e = 96487 \approx 96500\ \text{C mol}^{-1}$ — the charge on one mole of electrons.
- Moles of electrons: $n_{e^-} = Q/F$; one mole of $\ce{M^{n+}}$ needs $n$ faradays.
- Equivalents: equivalent weight $= M/n$; electrochemical equivalent $Z = M/nF$.
- Second law: same charge through series cells ⇒ masses in the ratio of equivalent weights.
- Master chain: $Q = It \;\to\; n_{e^-} = Q/F \;\to\; \text{mol product} \;\to\; \text{mass or volume}$.
- Constant slip: convert time to seconds; assign electrons from oxidation-state change.