What will happen during the electrolysis of aqueous solution of $CuSO_4$ in the presence of $Cu$ electrodes? (Two or more than two options may be correct.) (i) Copper will deposit at cathode. (ii) Copper will dissolve at anode. (iii) Oxygen will be released at anode. (iv) Copper will deposit at anode.

> [!TLDR] > In the electrolysis of aqueous $CuSO_4$ with copper electrodes, the anode itself is oxidised (copper dissolves) and $Cu^{2+}$ ions from the solution are reduced at the cathode (copper deposits). No gas evolution occurs. The correct options are **(i)** and **(ii)**. This is a classic case of **electrolysis with active electrodes**. The key idea is that when the electrodes are made of the same metal as the cation in the electrolyte, the electrode reactions change completely — the anode is no longer inert, so it participates in the reaction. Let’s break it down. --- ### 1. What are the species present in the solution? Aqueous $CuSO_4$ dissociates into $Cu^{2+}$ and $SO_4^{2-}$ ions. Water itself is also present, providing $H^+$ and $OH^-$ ions (though in very low concentration). So at the electrodes, we have competing possibilities: - **At cathode (reduction):** $Cu^{2+}$ ions can be reduced to $Cu$ metal, or $H^+$ from water can be reduced to $H_2$ gas. - **At anode (oxidation):** $SO_4^{2-}$ ions can be oxidised (very difficult), $OH^-$ from water can be oxidised to $O_2$ gas, **or** the copper metal of the anode itself can be oxidised to $Cu^{2+}$. The standard reduction potentials tell us which reaction is favoured. --- ### 2. Why copper deposits at the cathode — not hydrogen The reduction potentials (at 298 K, 1 M, vs SHE) are: $$ Cu^{2+} + 2e^- \rightarrow Cu(s) \quad E^\circ = +0.34\ \text{V} $$ $$ 2H^+ + 2e^- \rightarrow H_2(g) \quad E^\circ = 0.00\ \text{V} $$ A more positive $E^\circ$ means a greater tendency to be reduced. $Cu^{2+}$ reduction is much more favourable than $H^+$ reduction. So at the cathode, **copper ions are reduced to copper metal**, which deposits on the cathode. > [!WARNING] > A common mistake is to think that because water is present, hydrogen gas must evolve. But unless the $Cu^{2+}$ concentration is extremely low, copper deposition is strongly favoured. In standard conditions, copper plates out first. So optio

What will happen during the electrolysis of aqueous solution of CuSO₄ in the presence of Cu electrodes? (Two o

NCERT Exemplar · Q49

CuSO_4

(i) Copper will deposit at cathode.

(ii) Copper will dissolve at anode.

(iii) Oxygen will be released at anode.

(iv) Copper will deposit at anode.

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Concept understanding — Faradays Laws Electrolysis

Faraday's Laws of Electrolysis

Imagine you're trying to plate a copper spoon with silver. You drop the spoon into a solution containing silver ions, connect it to a battery, and wait. How much silver actually deposits? Does it depend on how long you wait? On how strong the battery is? On what metal you're using?

Faraday's laws answer exactly these questions. They connect the invisible world of electrons flowing through a wire to the visible world of atoms depositing on a surface.

The Intuition First

Think of electrolysis as a counting problem. Each silver ion ( $\text{Ag}^+$ ) needs exactly one electron to become a neutral silver atom ( $\text{Ag}$ ). So if you push a certain number of electrons through the circuit, you should get exactly that many silver atoms deposited.

The first law says: more charge → more mass deposited. Double the charge, double the mass. It's a direct proportionality.

The second law says: different elements need different amounts of charge per atom. A copper ion ( $\text{Cu}^{2+}$ ) needs two electrons to become neutral copper, so for the same amount of charge, you get half as many copper atoms as silver atoms.

The Precise Statements

m \propto Q \quad \text{or} \quad m = ZQ

\frac{m_1}{E_1} = \frac{m_2}{E_2}

Here $E$ is the equivalent weight: atomic mass divided by the number of electrons transferred per ion ( $n$ ). For silver ( $\text{Ag}^+$ , $n=1$ ), $E = 107.87$ g. For copper ( $\text{Cu}^{2+}$ , $n=2$ ), $E = 63.55/2 = 31.77$ g.

The Combined Law

These two laws merge into one powerful equation:

m = \frac{Q \times E}{F}

where $F$ is Faraday's constant — the charge carried by one mole of electrons: $F = 96485$ coulombs per mole.

Since $Q = I \times t$ (current × time), you can write:

m = \frac{I \times t \times E}{F}

This is the working formula for every electrolysis calculation in your exams.

Tip

To avoid confusion: equivalent weight $E$ is always atomic mass divided by $n$ (the number of electrons gained or lost per ion). For $\text{Al}^{3+}$ , $n=3$ ; for $\text{O}_2$ gas (from water), each oxygen atom loses 2 electrons, but the molecule has 2 atoms, so $n=4$ per $\text{O}_2$ molecule.

A Worked Example

Problem: How much copper deposits when a current of 2.0 A flows through a copper sulfate solution for 30 minutes? (Atomic mass of Cu = 63.5 g/mol, $n=2$ )

Step 1: Find the equivalent weight.

E = \frac{63.5}{2} = 31.75 \text{ g/mol}

Step 2: Find total charge.

Q = I \times t = 2.0 \times (30 \times 60) = 3600 \text{ C}

Step 3: Apply the combined law.

m = \frac{Q \times E}{F} = \frac{3600 \times 31.75}{96485} = 1.185 \text{ g}

So about 1.2 grams of copper deposits. …

Why this formula?

Faraday's Laws of Electrolysis: Why the Formulas Hold

Faraday's Laws of Electrolysis describe the quantitative relationship between the amount of electricity passed through an electrolyte and the mass of substance liberated at the electrodes. Let's build the reasoning step-by-step.

1. The Core Idea: Charge Carries Matter

Electrolysis works because ions (charged particles) move toward electrodes and undergo redox reactions.

At the cathode (negative electrode), cations gain electrons (reduction).
At the anode (positive electrode), anions lose electrons (oxidation).

The key insight: Each ion that reacts carries a fixed amount of charge.

For a monovalent ion (e.g., $\text{Na}^+$ ), charge = $1.602 \times 10^{-19} \, \text{C}$ (the elementary charge $e$ ).
For a divalent ion (e.g., $\text{Cu}^{2+}$ ), charge = $2e$ .

Thus, the total charge passed ( $Q$ ) is directly proportional to the number of ions that have reacted ( $N$ ):

Q = N \cdot z e

where:

$z$ = valency (number of electrons transferred per ion)
$e$ = elementary charge ( $1.602 \times 10^{-19} \, \text{C}$ )

2. From Number of Ions to Mass

The number of ions $N$ is related to the mass ( $m$ ) of substance liberated via Avogadro's number ( $N_A$ ) and molar mass ( $M$ ):

N = \frac{m}{M} \cdot N_A

Substitute into $Q = N z e$ :

Q = \left( \frac{m}{M} \cdot N_A \right) \cdot z e

3. Introducing Faraday's Constant

The product $N_A e$ appears repeatedly — it's called Faraday's constant ( $F$ ):

F = N_A e \approx 96485 \, \text{C mol}^{-1}

So:

Q = \frac{m}{M} \cdot z F

Rearrange for mass:

\boxed{m = \frac{Q M}{z F}}

This is the unified formula for both of Faraday's laws.

4. Why Two "Laws"? — They Are the Same Idea

Faraday originally stated two laws, but they are logical consequences of the same charge–mass relationship:

First Law (Direct Proportionality)

Mass liberated is directly proportional to the charge passed.

From $m = \frac{Q M}{z F}$ , if $M$ , $z$ , and $F$ are constant, then:

m \propto Q

Why? Because each ion needs a fixed charge to react — more charge means more ions, hence more mass.

Second Law (Electrochemical Equivalent)

For the same charge, masses liberated are proportional to equivalent weights.

Equivalent weight $E = \frac{M}{z}$ (mass per mole of electrons transferred).

From $m = \frac{Q}{F} \cdot \frac{M}{z} = \frac{Q}{F} \cdot E$ , if $Q$ is fixed:

m \propto E

Why? For the same charge, the number of electrons transferred is fixed. A substance with a smaller $z$ (fewer electrons per ion) will liberate more moles of substance, hence more mass per mole.

5. Practical Formula for Exams …

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In the electrolysis of aqueous $CuSO_4$ with copper electrodes, the anode itself is oxidised (copper dissolves) and $Cu^{2+}$ ions from the solution are reduced at the cathode (copper deposits). No gas evolution occurs. The correct options are (i) and (ii).

This is a classic case of electrolysis with active electrodes. The key idea is that when the electrodes are made of the same metal as the cation in the electrolyte, the electrode reactions change completely — the anode is no longer inert, so it participates in the reaction.

Let’s break it down.

1. What are the species present in the solution?

Aqueous $CuSO_4$ dissociates into $Cu^{2+}$ and $SO_4^{2-}$ ions. Water itself is also present, providing $H^+$ and $OH^-$ ions (though in very low concentration).

So at the electrodes, we have competing possibilities:

At cathode (reduction): $Cu^{2+}$ ions can be reduced to $Cu$ metal, or $H^+$ from water can be reduced to $H_2$ gas.
At anode (oxidation): $SO_4^{2-}$ ions can be oxidised (very difficult), $OH^-$ from water can be oxidised to $O_2$ gas, or the copper metal of the anode itself can be oxidised to $Cu^{2+}$ .

The standard reduction potentials tell us which reaction is favoured.

2. Why copper deposits at the cathode — not hydrogen

The reduction potentials (at 298 K, 1 M, vs SHE) are:

Cu^{2+} + 2e^- \rightarrow Cu(s) \quad E^\circ = +0.34\ \text{V}

2H^+ + 2e^- \rightarrow H_2(g) \quad E^\circ = 0.00\ \text{V}

A more positive $E^\circ$ means a greater tendency to be reduced. $Cu^{2+}$ reduction is much more favourable than $H^+$ reduction. So at the cathode, copper ions are reduced to copper metal, which deposits on the cathode.

Watch out

A common mistake is to think that because water is present, hydrogen gas must evolve. But unless the $Cu^{2+}$ concentration is extremely low, copper deposition is strongly favoured. In standard conditions, copper plates out first.

So option (i) Copper will deposit at cathode is correct.

3. What happens at the anode — the crucial difference

If the anode were inert (like platinum or graphite), the only possible oxidation would be of $OH^-$ to $O_2$ :

4OH^- \rightarrow O_2 + 2H_2O + 4e^- \quad E^\circ = +0.40\ \text{V}

But here the anode is copper metal. Copper can itself be oxidised:

Cu(s) \rightarrow Cu^{2+} + 2e^- \quad E^\circ = -0.34\ \text{V}

The oxidation potential (the reverse of the reduction potential) is $+0.34\ \text{V}$ for copper dissolution, compared to $+0.40\ \text{V}$ for oxygen evolution. A lower oxidation potential means the reaction is easier. So copper metal from the anode dissolves into the solution as $Cu^{2+}$ ions, rather than oxygen being produced. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 22 on this concept.

AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
$Al_2O_3$
›Reveal solutionSolution
This tests Faraday's laws of electrolysis applied to $Al_2O_3$ electrolysis; the answer is 90 g.

Concept and Intuition

In electrolytic reduction of molten $Al_2O_3$ , $Al^{3+}$ ions gain 3 electrons at the cathode to form Al metal. The amount of substance deposited depends on the total charge passed and the number of electrons needed per mole of product (Faraday's second law).

Step-by-Step Solution

Charge passed: $Q = I \times t = 965\,A \times 1000\,s = 9.65\times10^{5}\,C$ .

Moles of electrons transferred: $n_e = Q/F = \dfrac{9.65\times10^{5}}{96500} = 10\,mol$ .

Cathode reaction: $Al^{3+} + 3e^- \to Al$ , so 3 mol electrons deposit 1 mol Al.

Moles of Al deposited $= 10/3\,mol$ . …
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AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
$Al^{3+}$
›Reveal solutionSolution
Faradays needed = (moles of ion) × (charge/electrons needed per ion); working both cases gives x=0.75, y=1.1.

Concept and Intuition

One Faraday (1 F) supplies one mole of electrons. To deposit/reduce a metal ion $M^{n+}$ completely, you need $n$ Faradays per mole of that ion, since $M^{n+}+ne^-\to M$ .

Step-by-Step Solution

$Al^{3+}+3e^-\to Al$ : for 0.25 mol Al³⁺, Faradays required $x = 3\times0.25 = 0.75$ F.

Moles of Cu²⁺ $= \dfrac{1100}{1000}\text{ L}\times0.5\text{ mol/L} = 0.55$ mol. …
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AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
$H_2$
›Reveal solutionSolution
Faraday's law gives moles of electrons passed; since dilute aqueous NaCl electrolysis reduces water (not Na $^+$ ) at the cathode, 2 electrons give 1 mole of H $_2$ , so $n(H_2)=0.01$ mol.

Concept and Intuition

In dilute aqueous NaCl, Na $^+$ is not reduced at the cathode (it is a much harder ion to reduce than water); instead water itself is reduced, liberating $H_2$ gas and $OH^-$ . Faraday's laws connect the charge passed to the moles of product via the number of electrons in the half-reaction.

Step-by-Step Solution

Charge passed: $Q = I\times t = (10\times10^{-3}\,A)\times(19.3\times10^4\,s) = 1930$ C.

Moles of electrons: $n_e = Q/F = 1930/96500 = 0.02$ mol.

Cathode half-reaction in dilute aqueous NaCl: $2H_2O + 2e^- \rightarrow H_2 + 2OH^-$ — 2 mol electrons give 1 mol $H_2$ . …
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AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
$\mathrm{M^{n+}}$
›Reveal solutionSolution
Applying Faraday's first law of electrolysis to the given charge, mass deposited, and atomic weight gives $n=4$ .

Concept and Intuition

Faraday's law relates the mass of metal deposited at the cathode to the charge passed:
$m = \frac{Q\,M}{n\,F}$
where $Q=It$ is total charge, $M$ is the atomic weight of the metal, $n$ is the number of electrons needed per metal ion (i.e. its charge $\mathrm{M}^{n+}$ ), and $F$ is Faraday's constant. Rearranging for $n$ lets you deduce the ionic charge state from an experimental deposition measurement.

Step-by-Step Solution

Total charge passed: $Q = It = 3\,\text{A} \times 3600\,\text{s} = 10800\,\text{C}$ .

Rearrange Faraday's law: $n = \dfrac{Q\,M}{m\,F}$ . …
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AP EAPCET 2024Set ap-2024-05-16-FN1 markMCQ
$Q. At 27 °C, the pH of 1 L of aqueous copper sulphate is 5.5. This solution was electrolyzed using two Pt electrodes for some time. What is the pH of remaining copper sulphate solution? (A) 5.5 (B) More than 5.5 but less than 7.0 (C) 7.5 (D) Less than 5.5 but more than zero$
›Reveal solutionSolution
Electrolysis with inert Pt electrodes deposits Cu at the cathode and liberates $O_2$ + $H^+$ at the anode, so the solution becomes progressively more acidic — pH drops below 5.5 but stays above zero.

Concept and Intuition

With Pt (inert) electrodes, $Cu^{2+}$ is preferentially reduced at the cathode (its reduction potential is far more favourable than reducing water/H⁺ to $H_2$ ), while at the anode water is oxidized to oxygen (sulfate is essentially inert to oxidation under these conditions). This liberates $H^+$ ions at the anode without any corresponding consumption of $H^+$ at the cathode, so the net effect is generation of acid.

Step-by-Step Solution

Cathode: $Cu^{2+} + 2e^- \to Cu(s)$ — copper deposits, no $H^+$ consumed.

Anode: $2H_2O \to O_2 + 4H^+ + 4e^-$ — $H^+$ ions are released into solution.

Overall: $2CuSO_4 + 2H_2O \xrightarrow{\text{electrolysis}} 2Cu + O_2 + 2H_2SO_4$ .

As electrolysis proceeds, $[H^+]$ increases, so pH decreases below the starting 5.5. …
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AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
$\mathrm{MF_2}$
›Reveal solutionSolution
This tests Faraday's laws of electrolysis applied to a molten metal fluoride salt to find the atomic mass of the deposited metal. The answer is (A) 63.47 u.

Concept and Intuition

In electrolysis, the amount of substance deposited at an electrode is directly related to the total charge passed via Faraday's laws. For a metal $M$ in the salt $MF_2$ , the metal exists as the $M^{2+}$ ion, which requires exactly 2 moles of electrons to be reduced to 1 mole of neutral metal atoms.

Step-by-Step Solution

Total charge passed: $Q = 8040\ \mathrm{C}$ .

Moles of electrons passed: $n_e = \dfrac{Q}{F} = \dfrac{8040}{96500} = 0.08332\ \mathrm{mol}$ .

Since $M^{2+} + 2e^- \rightarrow M$ , moles of metal deposited: $n_M = \dfrac{n_e}{2} = \dfrac{0.08332}{2} = 0.04166\ \mathrm{mol}$ . …
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AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
$Q. The anode and cathode used in electrolytic refining of copper respectively are (A) Pure copper, impure copper (B) Impure copper, pure copper (C) Pure copper, pure zinc (D) Impure copper, pure zinc$
›Reveal solutionSolution
Electrolytic refining of copper: the impure metal is oxidised (dissolved) at the anode and pure metal is deposited at the cathode.

Concept and Intuition

In electrorefining, the metal to be purified is made the anode so that it oxidises and goes into solution as $\mathrm{Cu^{2+}}$ , while a strip of the pure metal is made the cathode where $\mathrm{Cu^{2+}}$ is reduced and deposited as pure copper. The electrolyte is acidified $\mathrm{CuSO_4}$ solution, and the less noble impurities stay in solution or fall as "anode mud" while more noble impurities collect as anode mud too (e.g. Ag, Au).

Step-by-Step Solution

Anode reaction (oxidation): $\mathrm{Cu(s,\ impure)\rightarrow Cu^{2+}+2e^-}$ — this must be the impure block since it is consumed/dissolved. …
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AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
$\mathrm{CuSO_4}$
›Reveal solutionSolution
Faraday's law with $n=2$ electrons per Cu deposited gives $\approx 0.39\ \mathrm g$ of copper — answer (B).

Concept and Intuition

Faraday's first law: the mass deposited at an electrode is proportional to the total charge passed, via $m = \dfrac{Q}{F}\times \dfrac{M}{n}$ , where $n$ is the number of electrons needed to deposit one atom/ion (here $\mathrm{Cu^{2+}}+2e^-\to\mathrm{Cu}$ , so $n=2$ ).

Step-by-Step Solution

Charge passed: $Q = I\times t = 2\ \mathrm A \times (10\times60)\ \mathrm s = 2\times600 = 1200\ \mathrm C$ .

Moles of electrons: $n_e = Q/F = 1200/96500 = 0.012435\ \mathrm{mol}$ . …
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AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
$\mathrm{MgCl_2}$
›Reveal solutionSolution
Depositing all the $Mg^{2+}$ from 0.1 mol of $MgCl_2$ needs 2 electrons per ion, so $0.1 \times 2 = 0.2$ Faradays are required.

Concept and Intuition

Electrodeposition amount is governed by Faraday's laws of electrolysis: the moles of electrons (Faradays) needed equal the moles of the species times the number of electrons transferred in reducing (or oxidizing) that species. For a metal ion of charge $n+$ , depositing it as the neutral metal requires $n$ electrons per ion.

Step-by-Step Solution

Moles of $MgCl_2$ in 1 L of 0.1 M solution: $n = M \times V = 0.1\ \text{mol/L} \times 1\ \text{L} = 0.1$ mol, giving 0.1 mol of $Mg^{2+}$ ions (1:1 stoichiometry, $MgCl_2 \rightarrow Mg^{2+} + 2Cl^-$ ).

Write the cathodic deposition reaction: $Mg^{2+} + 2e^- \rightarrow Mg$ — each $Mg^{2+}$ ion needs 2 electrons. …
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AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
$\mathrm{Fe^{3+}}$
›Reveal solutionSolution
Reducing $Fe^{3+}$ to $Fe^{2+}$ is a one-electron process; use $Q=nF$ and $t=Q/I$ , converting seconds to hours.

Concept and Intuition

Faraday's law connects moles of electrons transferred to charge passed: $Q = n_e \times F$ , where $n_e$ is moles of electrons and $F=96500$ C/mol. The reduction half-reaction $Fe^{3+}+e^-\to Fe^{2+}$ shows exactly 1 electron per Fe ion reduced.

Step-by-Step Solution

Moles of electrons needed for 3 mol $Fe^{3+}$ = 3 mol (1:1 ratio).

Charge required: $Q = 3\times96500 = 289500\ \text{C}$ .

Time: $t = Q/I = 289500/2.0 = 144750\ \text{s}$ . …
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AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
$\mathrm{CuSO_4}$
›Reveal solutionSolution
This tests Faraday's laws of electrolysis for a two-electrode cell (Cu deposited at cathode, O $_2$ evolved at the inert Pt anode). Answer: 1.27 g Cu, 0.224 L O $_2$ .

Concept and Intuition

When current flows through an electrolyte, the amount of substance deposited/liberated at each electrode is proportional to the charge passed, via $n_{e^-} = Q/F$ . With Pt (inert) electrodes in aqueous $\mathrm{CuSO_4}$ : at the cathode $\mathrm{Cu^{2+}}$ ions are reduced to Cu metal (2 electrons per Cu atom); since Pt itself does not dissolve at the anode, the anode reaction is oxidation of water to $\mathrm{O_2}$ (4 electrons per $\mathrm{O_2}$ molecule), because $\mathrm{SO_4^{2-}}$ is not easily oxidised compared to water.

Step-by-Step Solution

Charge passed: $Q = It = 38.6\ \mathrm{A}\times100\ \mathrm{s} = 3860\ \mathrm{C}$ .

Moles of electrons transferred: $n_{e^-} = Q/F = 3860/96500 = 0.04\ \mathrm{mol}$ .

Cathode: $\mathrm{Cu^{2+}+2e^-\to Cu}$ , so moles of Cu $= 0.04/2 = 0.02\ \mathrm{mol}$ .

Mass of Cu $= 0.02\ \mathrm{mol}\times63.54\ \mathrm{g\,mol^{-1}} = 1.2708\ \mathrm{g}\approx1.27\ \mathrm{g}$ . …
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AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
$\mathrm{AlCl_3}$
›Reveal solutionSolution
This is a straightforward Faraday's-law electrolysis calculation for aluminium deposition from molten $\mathrm{AlCl_3}$ ; the mass deposited is 0.90 g.

Concept and Intuition

Electrolysis obeys Faraday's laws: the amount of substance deposited at an electrode is proportional to the quantity of electricity (charge) passed, and the proportionality depends on the valency (charge) of the ion being discharged. One mole of electrons (1 Faraday $= 96500\ \text{C}$ ) discharges $1/n$ mole of an ion of charge $n^+$ . For $\mathrm{Al^{3+}}$ , $n=3$ , so three moles of electrons are needed to deposit one mole of Al metal.

Step-by-Step Solution

Charge passed: $Q = I \times t = 96.5\ \text{A} \times 100\ \text{s} = 9650\ \text{C}$ .

Moles of electrons passed: $n_e = Q/F = 9650/96500 = 0.1\ \text{mol}$ . …
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Q.What will happen during the electrolysis of aqueous solution of CuSO4CuSO_4CuSO4​ in the presence of CuCuCu electrodes? (Two or more than two options may be correct.)

1. What are the species present in the solution?

2. Why copper deposits at the cathode — not hydrogen

3. What happens at the anode — the crucial difference

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Q.What will happen during the electrolysis of aqueous solution of $CuSO_4$ in the presence of $Cu$ electrodes? (Two or more than two options may be correct.)

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