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.

COMEDK 2026Set 2026-A1 markMCQ
$\mathrm{CaCl}_2$
›Reveal solutionSolution
The key idea is that equal quantities of electricity produce equal moles of electrons, so the mass of metal produced is proportional to its equivalent weight. Since both Mg and Ca are divalent, the mass of Ca produced is (40/24) × 4.8 g = 8.0 g.

Concept and Intuition

This problem is about Faraday’s laws of electrolysis. The same amount of charge (coulombs) will liberate the same number of moles of electrons. For a metal ion $M^{n+}$ , the number of moles of metal produced is (moles of electrons) / n. So the mass of metal = (charge / (nF)) × atomic mass. Since the charge is the same for both, the mass ratio of two metals is simply (atomic mass₁ / n₁) : (atomic mass₂ / n₂). Here both Mg and Ca form +2 ions, so n is the same (2), and the mass ratio equals the atomic mass ratio.

Step-by-step reasoning

Identify the relevant half-reactions In molten $\mathrm{MgCl}_2$ , magnesium ions are reduced:

$\mathrm{Mg^{2+} + 2e^- \rightarrow Mg}$
In molten $\mathrm{CaCl}_2$ , calcium ions are reduced:
$\mathrm{Ca^{2+} + 2e^- \rightarrow Ca}$
Both require 2 moles of electrons per mole of metal.

Find moles of Mg produced Given mass of Mg = 4.8 g, atomic mass = 24 g/mol.

$\text{Moles of Mg} = \frac{4.8}{24} = 0.20 \text{ mol}$

Find moles of electrons used Each mole of Mg requires 2 moles of electrons.

$\text{Moles of electrons} = 0.20 \times 2 = 0.40 \text{ mol}$

Same charge means same moles of electrons for Ca The same quantity of electricity provides 0.40 mol of electrons. Each mole of Ca also requires 2 moles of electrons.
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COMEDK 2026Set 2026-M1 markMCQ
$3.482 \times 10^{-1} \mathrm{~g}$
›Reveal solutionSolution
This is a Faraday’s law problem where iron is deposited from Fe³⁺ ions. The key is that each Fe³⁺ requires 3 moles of electrons to become Fe(s). Using the mass deposited, time, and atomic mass, we solve for current. The answer is 1.5 A.

Concept & Intuition

Electrolysis is about forcing a non-spontaneous redox reaction with electricity. The amount of substance deposited at an electrode is directly proportional to the charge passed. Faraday’s law says:
$m = \frac{Q \cdot M}{n \cdot F}$
where $m$ = mass deposited, $Q$ = total charge (current × time), $M$ = molar mass, $n$ = number of electrons per ion, and $F$ = Faraday constant (96485 C/mol).

Here, ferric sulphate gives Fe³⁺ ions. Each Fe³⁺ gains 3 electrons to become Fe metal:
$\text{Fe}^{3+} + 3e^- \rightarrow \text{Fe(s)}$
So $n = 3$ . The pitfall? Many students mistakenly use $n=2$ (thinking of Fe²⁺) or forget to convert time to seconds.

Step-by-step solution

Identify the half-reaction and $n$ Ferric ion is Fe³⁺. Reduction to metal:

$\text{Fe}^{3+} + 3e^- \rightarrow \text{Fe}$
Hence, $n = 3$ moles of electrons per mole of Fe.

Write Faraday’s law

$m = \frac{I \cdot t \cdot M}{n \cdot F}$
where $I$ = current (A), $t$ = time (s), $M$ = atomic mass (g/mol), $F = 96485 \, \text{C/mol}$ .

Convert given data to consistent units

Mass: $m = 3.482 \times 10^{-1} \, \text{g} = 0.3482 \, \text{g}$

Time: $t = 20 \, \text{min} = 20 \times 60 = 1200 \, \text{s}$

Atomic mass: $M = 56 \, \text{g/mol}$

$n = 3$

$F = 96485 \, \text{C/mol}$

Solve for current $I$

Rearranging:

$I = \frac{m \cdot n \cdot F}{t \cdot M}$
Substitute values:
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COMEDK 2025Set 2025-A1 markMCQ
$\mathrm{PdX} \mathrm{n}_{\mathrm{n}}$
›Reveal solutionSolution
The charge on the Pd ion is found by equating the total charge passed (from current and time) to the charge required to deposit the given mass of Pd, using Faraday’s laws. The result shows the ion carries a +4 charge, so the correct option is (A).

Concept & Intuition

Electrolysis is governed by Faraday’s laws: the mass of a substance deposited at an electrode is proportional to the total charge passed. For a metal ion $\mathrm{Pd}^{n+}$ , the reduction reaction is
$\mathrm{Pd}^{n+} + n e^- \rightarrow \mathrm{Pd}$
So each mole of Pd deposited consumes $n$ moles of electrons. By calculating the total charge passed and the moles of Pd deposited, we can solve for $n$ , the charge number (oxidation state) of the palladium ion.

Calculate total charge passed

Current $I = 1.5 \, \text{A}$ , time $t = 2 \, \text{hours} = 2 \times 3600 = 7200 \, \text{s}$ .

Charge $Q = I \times t = 1.5 \times 7200 = 10800 \, \text{C}$ .

Calculate moles of Pd deposited

Mass deposited $m = 2.977 \, \text{g}$ , atomic mass $M = 106.4 \, \text{g/mol}$ .

Moles of Pd:

$\text{moles} = \frac{2.977}{106.4} \approx 0.02798 \, \text{mol}$

Relate charge to moles of electrons Faraday’s constant $F = 96485 \, \text{C/mol e}^-$ . Total moles of electrons used:
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COMEDK 2025Set 2025-M1 markMCQ
$\mathrm{CuSO}_4$
›Reveal solutionSolution

Effective charge $=7200\ \text{C}$ deposits $0.0373\ \text{mol Cu}^{2+}$ ; remaining $\text{Cu}^{2+}$ in the unchanged 0.8 L gives molarity $\approx 0.433\ \text{M}$ — option (C).

Effective charge passed (80% current efficiency)
$Q = I\,t\,\eta = 2.5\ \text{A} \times 3600\ \text{s} \times 0.80 = 7200\ \text{C}.$
Moles of $\text{Cu}^{2+}$ deposited — $\text{Cu}^{2+} + 2e^{-} \rightarrow \text{Cu}$ :
$n_{e^-} = \frac{7200}{96500} = 0.0746\ \text{mol}, \qquad n_{\text{Cu}} = \frac{0.0746}{2} = 0.0373\ \text{mol}.$
Remaining $\text{Cu}^{2+}$ …
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COMEDK 2024Set 2024-A1 markMCQ
$\mathrm{MnO}_4{ }^{2-}$
›Reveal solutionSolution

$\mathrm{MnO_4^{2-}\to MnO_4^{-}}$ is a one-electron oxidation (Mn goes +6 → +7), so 0.1 mol requires $0.1\times F = 0.1\times96500 = 9650$ C.

Manganese changes oxidation state from +6 in $\mathrm{MnO_4^{2-}}$ to +7 in $\mathrm{MnO_4^{-}}$ , i.e. loss of 1 electron per ion: …
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COMEDK 2024Set 2024-E1 markMCQ
$\mathrm{Cr}_2 \mathrm{O} 7^{2-}$
›Reveal solutionSolution
The key is to find the total change in oxidation number per mole of dichromate, then multiply by moles and convert to Faradays. For 3.5 moles of $\mathrm{Cr}_2\mathrm{O}_7^{2-}$ reduced to $\mathrm{Cr}^{3+}$ in acid, the charge required is 21.0 Faradays, so the correct option is (C).

Concept & Intuition

In electrochemistry, one Faraday (F) is the charge carried by one mole of electrons — about 96,485 coulombs. When we talk about "reduction" of an ion, we mean it gains electrons. The number of Faradays needed equals the total moles of electrons transferred. So the problem reduces to: How many electrons does one dichromate ion take up when it is reduced in acid? That number comes from the change in oxidation state of chromium.

Step-by-step solution

Determine the oxidation state of Cr in dichromate

In $\mathrm{Cr}_2\mathrm{O}_7^{2-}$ , oxygen is always -2 (except in peroxides). Let the oxidation state of Cr be $x$ .

For the ion: $2x + 7(-2) = -2$

$\Rightarrow 2x - 14 = -2 \Rightarrow 2x = +12 \Rightarrow x = +6$ .

So each Cr is in the +6 state.

Identify the reduction product in acid medium

In acidic solution, dichromate is reduced to chromium(III) ions, $\mathrm{Cr}^{3+}$ .

So each Cr goes from +6 to +3, a gain of 3 electrons per Cr atom.

Electrons per dichromate ion

Since one $\mathrm{Cr}_2\mathrm{O}_7^{2-}$ contains two Cr atoms, the total electrons gained per ion is $2 \times 3 = 6$ electrons.

Total electrons for 3.5 moles

For 3.5 moles of dichromate:

$\text{moles of electrons} = 3.5 \times 6 = 21.0$ moles of electrons. …
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COMEDK 2024Set 2024-E1 markMCQ
$_4$
›Reveal solutionSolution

Effective charge 19440 C deposits 0.1007 mol Cu; remaining $\mathrm{Cu^{2+}}=0.3375-0.1007=0.2368$ mol in 0.75 L → 0.316 M.

Effective charge passed (90% efficiency):
$Q=I\,t\,\eta=3.0\times(2\times3600)\times0.90=19440\ \mathrm{C}$
Moles of electrons:
$n_e=\frac{19440}{96500}=0.2015\ \mathrm{mol}$
Copper deposited ( $\mathrm{Cu^{2+}}+2e^-\to\mathrm{Cu}$ ):

$n_{Cu}=\frac{0.2015}{2}=0.1007\ \mathrm{mol}$ …
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COMEDK 2024Set 2024-M1 markMCQ
$\mathrm{O}_2$
›Reveal solutionSolution

Burning $44\ \text{g}$ ( $1\ \text{mol}$ ) of propane needs $5\ \text{mol}$ of $\text{O}_2$ ; producing that much $\text{O}_2$ by electrolysis takes $20\ \text{mol}$ of electrons, i.e. $t = \dfrac{20 \times 96500}{200} = 9650\ \text{s} = \mathbf{2.68\ hours}$ — option (A).

Step 1 — Combustion stoichiometry
$\text{C}_3\text{H}_8 + 5\,\text{O}_2 \rightarrow 3\,\text{CO}_2 + 4\,\text{H}_2\text{O}$
Moles of propane $= \dfrac{44}{44} = 1\ \text{mol}$ , so $\text{O}_2$ required $= 5\ \text{mol}$ .

Step 2 — Electrons needed to make $5\ \text{mol}\ \text{O}_2$

At the anode during water electrolysis:

$2\,\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^-$ …
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COMEDK 2023Set 2023-E1 markMCQ
$100 \mathrm{~ml}$
›Reveal solutionSolution

Current (t = 2 h = 7200 s): I = Q/t = 1723 / 7200 = 0.2393 A

Concept: Faraday's laws. At the anode in the electrolysis of water:

2 H2O -> O2 + 4 H+ + 4 e-

So 4 mol of electrons (4 F) liberate 1 mol O2.

Moles of O2 at STP:

n(O2) = 0.100 L / 22.4 L/mol = 4.464 x 10^-3 mol

Charge required:

n(e-) = 4 x 4.464 x 10^-3 = 1.786 x 10^-2 mol …
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COMEDK 2023Set 2023-M1 markMCQ
$\mathrm{pH}=4$
›Reveal solutionSolution

Reaching $\mathrm{pH}=4$ requires generating $\mathrm{H^+}$ ( $10^{-4}\,\mathrm{M}$ in $0.1\,\mathrm{L} = 10^{-5}\,\mathrm{mol}$ ). Since each $\mathrm{H^+}$ needs one electron ( $Q = nF$ ), $t = nF/I = (10^{-5}\times96500)/0.5 \approx 1.93$ , matching option (A).

At the anode, electrolysis of the aqueous solution generates $\mathrm{H^+}$ :
$2\mathrm{H_2O} \rightarrow \mathrm{O_2} + 4\mathrm{H^+} + 4e^-,\quad \text{so moles of } \mathrm{H^+} = \text{moles of } e^-.$
Target: $\mathrm{pH}=4 \Rightarrow [\mathrm{H^+}] = 10^{-4}\,\mathrm{M}$ in $100\,\mathrm{mL} = 0.1\,\mathrm{L}$ :
$n_{\mathrm{H^+}} = 10^{-4} \times 0.1 = 10^{-5}\,\mathrm{mol}.$
Charge required (Faraday's law): …
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KCET 2022Set B-31 markMCQ
$Q. In Fuel cells ______ are used as catalysts. (A) Zinc - Mercury (B) Lead - Manganese (C) Platinum - Palladium (D) Nickel - Cadmium$
›Reveal solutionSolution

The NCERT hydrogen–oxygen fuel cell uses finely divided platinum or palladium metal, incorporated into the porous carbon electrodes, as the catalyst.

Step 1 — What a fuel cell is.

A fuel cell is a galvanic cell that converts the energy of combustion of a fuel (here $H_2$ ) directly into electrical energy, instead of burning it to run a turbine. The classic example is the $H_2$ – $O_2$ fuel cell used in the Apollo space programme.

Step 2 — The construction.

$H_2$ and $O_2$ are bubbled through porous carbon electrodes into a concentrated aqueous $NaOH$ (or $KOH$ ) electrolyte. To make the electrode reactions fast enough at ordinary temperature, catalysts such as finely divided platinum or palladium metal are incorporated into the electrodes.

Step 3 — The electrode reactions being catalysed.

Anode (oxidation of the fuel):
$2H_2(g) + 4OH^-(aq) \longrightarrow 4H_2O(l) + 4e^-$
Cathode (reduction of oxygen):
$O_2(g) + 2H_2O(l) + 4e^- \longrightarrow 4OH^-(aq)$
Overall:

$2H_2(g) + O_2(g) \longrightarrow 2H_2O(l)$ …
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←Previous questionQ48 — What will happen during the electrolysis of aqueous solution of CuSO₄ …→Next questionQ50 — Conductivity kappa, is equal to ______. (Two or more than two options …

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.)

Concept understanding — Faradays Laws Electrolysis

Faraday's Laws of Electrolysis

The Intuition First

The Precise Statements

The Combined Law

A Worked Example

Why this formula?

Faraday's Laws of Electrolysis: Why the Formulas Hold

1. The Core Idea: Charge Carries Matter

2. From Number of Ions to Mass

3. Introducing Faraday's Constant

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

First Law (Direct Proportionality)

Second Law (Electrochemical Equivalent)

5. Practical Formula for Exams …

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

Method: Electrode Potential Comparison for Electrolysis (Active Electrodes)

Steps:

1. Assuming Oxygen Is Released at the Anode, As With Inert Electrodes

2. Forgetting to Compare the Electrode's Own Oxidation Potential

3. Thinking Copper Can Deposit at the Anode

4. Missing the Real-World Link to Electrorefining

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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