What amount of electric charge is required for the reduction of 1 mole of MnO₄⁻ into Mn²⁺ ? (A) 1 F (B) 5 F (C

Question

MnO_4^{-}

CBSECBSE Class XII Board 2025MCQ· 1mImportance★★★★★

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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 …

Start your 14-day free trial to unlock the full solution →

The reduction of $\text{MnO}_4^-$ to $\text{Mn}^{2+}$ involves a 5‑electron change per ion, so 1 mole requires 5 faradays of charge. The correct option is (B).

The key to this question lies in Faraday’s laws of electrolysis — specifically, the idea that the amount of charge needed to reduce or oxidise a substance is directly proportional to the number of electrons transferred per mole. But before we plug numbers, let’s understand why the electron count is what it is.

Why the electron count matters

Faraday’s first law says: the mass of a substance liberated at an electrode is proportional to the quantity of electricity passed. But for a mole of ions, the charge required is simply:

Q = n \cdot F

where $n$ is the number of electrons transferred per ion (or molecule) and $F$ is the Faraday constant (≈ 96485 C/mol). So the problem reduces to finding $n$ for the half‑reaction:

\text{MnO}_4^- \rightarrow \text{Mn}^{2+}

Step‑by‑step reasoning

Identify the oxidation states In $\text{MnO}_4^-$ , oxygen is always –2 (except in peroxides, but not here). Let the oxidation state of Mn be $x$ .

x + 4(-2) = -1 \quad \Rightarrow \quad x - 8 = -1 \quad \Rightarrow \quad x = +7

So Mn is in the +7 state.

In $\text{Mn}^{2+}$ , the oxidation state is clearly +2.

Find the change in oxidation state The change is from +7 to +2:

\Delta = (+7) - (+2) = +5

A decrease of 5 in oxidation number means the Mn atom has gained 5 electrons.

Watch out

A common mistake is to think the change is 7 – 2 = 5 but then forget the sign. The number of electrons gained is the magnitude of the change — here 5 — regardless of sign. The sign tells you reduction (gain of electrons) vs oxidation (loss).

Balance the half‑reaction (to confirm) …

Start your 14-day free trial to unlock the full solution →

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

CBSE 2026Set 56/1/11 markMCQ
$CuCl_2$
›Reveal solutionSolution
In the electrolysis of aqueous $CuCl_2$ with inert platinum electrodes, $Cu^{2+}$ ions are reduced to Cu metal at the cathode, and $Cl^-$ ions are oxidised to $Cl_2$ gas at the anode. The correct option is (D).

Why this approach works — the core idea

Electrolysis is about forcing a non-spontaneous redox reaction using electrical energy. At the cathode (negative electrode), reduction happens — species gain electrons. At the anode (positive electrode), oxidation happens — species lose electrons.

The tricky part with aqueous solutions is that water itself can compete with the dissolved ions. You have to compare the standard reduction potentials of all possible reactions to decide which one actually occurs at each electrode. The rule is simple: at the cathode, the species with the higher (more positive) reduction potential gets reduced first. At the anode, the species with the lower (more negative) reduction potential gets oxidised first (or equivalently, the one that is easiest to oxidise).

For aqueous $CuCl_2$ , the ions present are $Cu^{2+}$ and $Cl^-$ , plus $H^+$ and $OH^-$ from water's autoionisation.
$Cu^{2+} + 2e^- \rightarrow Cu(s) \quad E^\circ = +0.34\ \text{V}$
Step-by-step reasoning

1. What happens at the cathode?

The cathode is negative, so it attracts positive ions ( $Cu^{2+}$ and $H^+$ from water). Two reduction reactions are possible:

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

$2H_2O + 2e^- \rightarrow H_2(g) + 2OH^-$ with $E^\circ = -0.83\ \text{V}$

The copper reduction has a much higher (more positive) reduction potential. That means $Cu^{2+}$ is far more willing to accept electrons than water is. So copper metal deposits on the cathode, and no hydrogen gas is produced.

Watch out

A common mistake is to assume that because water can be reduced to $H_2$ , it always happens. But $Cu^{2+}$ has a significantly higher reduction potential, so it gets reduced first. Hydrogen gas evolution at the cathode would only occur if $Cu^{2+}$ were absent or present in very low concentration.

2. What happens at the anode?

The anode is positive, so it attracts negative ions ( $Cl^-$ and $OH^-$ from water). Two oxidation reactions are possible:

$2Cl^- \rightarrow Cl_2(g) + 2e^-$ — this is the reverse of the $Cl_2/Cl^-$ reduction, so its oxidation potential is $-1.36\ \text{V}$ (the negative of the reduction potential).

$2H_2O \rightarrow O_2(g) + 4H^+ + 4e^-$ — this is the reverse of the $O_2/H_2O$ reduction, so its oxidation potential is $-1.23\ \text{V}$ .

When comparing oxidation reactions, the one with the less negative (higher) oxidation potential occurs more readily. Here, $-1.23\ \text{V}$ is greater than $-1.36\ \text{V}$ , so you might think water oxidation to $O_2$ should happen first.

But there's a catch: the $O_2$ evolution reaction involves 4 electrons and has a significant overpotential on platinum electrodes. Overpotential is an extra voltage needed to overcome the kinetic barrier for a reaction to occur at a practical rate. On platinum, the overpotential for oxygen evolution is substantial (around 0.4–0.6 V), while for chlorine evolution it is very small. …
Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. The amount of electricity required to obtain 1 mol of Cu from CuSO4 is$
(a) 3 F
(b) 2 F
(c) 1 F
(d) 4 F
›Reveal solutionSolution

Faraday's law: the charge needed to deposit one mole of a metal equals its ionic charge (in Faradays), since that many moles of electrons are needed for the reduction half-reaction.

In CuSO4 solution, copper exists as the Cu2+ ion. Electrodeposition at the cathode proceeds by the half-reaction:

Cu2+ + 2e- -> Cu

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set 56/4/11 markMCQ
$MnO_4^{-}$
›Reveal solutionSolution
The reduction of $\text{MnO}_4^-$ to $\text{Mn}^{2+}$ involves a 5‑electron change per ion, so 1 mole requires 5 faradays of charge. The correct option is (B).

The key to this question lies in Faraday’s laws of electrolysis — specifically, the idea that the amount of charge needed to reduce or oxidise a substance is directly proportional to the number of electrons transferred per mole. But before we plug numbers, let’s understand why the electron count is what it is.

Why the electron count matters

Faraday’s first law says: the mass of a substance liberated at an electrode is proportional to the quantity of electricity passed. But for a mole of ions, the charge required is simply:
$Q = n \cdot F$
where $n$ is the number of electrons transferred per ion (or molecule) and $F$ is the Faraday constant (≈ 96485 C/mol). So the problem reduces to finding $n$ for the half‑reaction:
$\text{MnO}_4^- \rightarrow \text{Mn}^{2+}$
Step‑by‑step reasoning

Identify the oxidation states In $\text{MnO}_4^-$ , oxygen is always –2 (except in peroxides, but not here). Let the oxidation state of Mn be $x$ .

$x + 4(-2) = -1 \quad \Rightarrow \quad x - 8 = -1 \quad \Rightarrow \quad x = +7$
So Mn is in the +7 state.

In $\text{Mn}^{2+}$ , the oxidation state is clearly +2.

Find the change in oxidation state The change is from +7 to +2:

$\Delta = (+7) - (+2) = +5$
A decrease of 5 in oxidation number means the Mn atom has gained 5 electrons.

Watch out

A common mistake is to think the change is 7 – 2 = 5 but then forget the sign. The number of electrons gained is the magnitude of the change — here 5 — regardless of sign. The sign tells you reduction (gain of electrons) vs oxidation (loss).

Balance the half‑reaction (to confirm) …
Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$Q. The amount of electricity required to obtain one mole of Zn from ZnSO4 solution will be$
(a) 3F
(b) 4F
(c) 1F
(d) 2F
›Reveal solutionSolution

Zinc in ZnSO4 exists as Zn2+, so reducing one mole of Zn2+ to Zn metal needs 2 moles of electrons, i.e. 2 Faradays of charge.

At the cathode during electrolysis of ZnSO4 solution:

$\text{Zn}^{2+} + 2e^- \rightarrow \text{Zn(s)}$

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$Q. How many Coulombs of Electricity are required for the oxidation of FeO to Fe2O3:$
(a) 289500 C
(b) 482500 C
(c) 96500 C
(d) None of these.
›Reveal solutionSolution

In FeO $\to$ Fe $_2$ O $_3$ , iron's oxidation state changes from $+2$ to $+3$ — a loss of exactly one electron per Fe atom, so oxidizing 1 mole of FeO needs 1 Faraday $= 96500\ \text{C}$ .

In FeO, iron is present as $Fe^{2+}$ . In Fe $_2$ O $_3$ , iron is present as $Fe^{3+}$ . The oxidation half-reaction per Fe atom is:

$Fe^{2+} \rightarrow Fe^{3+} + e^-$

This shows exactly one electron is lost per Fe atom oxidized. By Faraday's first law, the quantity of charge needed to bring about a 1-electron change in 1 mole of a substance is exactly 1 Faraday: …

Start your 14-day free trial to see the full solution →
CBSE 2024Set 56/2/11 markMCQ
$H_2$
(g) is liberated at cathode (B) Na is formed at cathode (C) $O_2$
(g) is liberated at anode (D) $Cl_2$
(g) is liberated at cathode
›Reveal solutionSolution
During the electrolysis of an aqueous sodium chloride solution, water is preferentially reduced at the cathode to produce hydrogen gas, and chloride ions are preferentially oxidized at the anode to produce chlorine gas, primarily due to the overpotential for oxygen evolution. Therefore, $H_2$ (g) is liberated at the cathode.

Electrolysis involves using electrical energy to drive non-spontaneous chemical reactions. When an aqueous solution is electrolyzed, there's often a competition between water molecules and the dissolved ions to be oxidized or reduced at the electrodes. The outcome depends on their standard electrode potentials and kinetic factors like overpotential.

At the cathode (where reduction occurs), the species with a more positive (or less negative) standard reduction potential will be preferentially reduced.

At the anode (where oxidation occurs), the species with a more negative (or less positive) standard reduction potential (or more positive standard oxidation potential) will be preferentially oxidized.

Let's break down the process for aqueous NaCl.

Identify Species Present in Aqueous NaCl Solution:

When NaCl is dissolved in water, it dissociates into $Na^+$ and $Cl^-$ ions. Water itself is also present. So, the species available for reaction are:

Cations: $Na^+(aq)$ , $H^+(aq)$ (from water autoionization)

Anions: $Cl^-(aq)$ , $OH^-(aq)$ (from water autoionization)

Solvent: $H_2O(l)$

Reactions at the Cathode (Reduction):

The cathode is the negative electrode where reduction takes place. Possible species to be reduced are $Na^+$ ions and $H_2O$ molecules.

Reduction of $Na^+$ : $Na^+(aq) + e^- \rightarrow Na(s)$ $E^\circ = -2.71 \text{ V}$

Reduction of $H_2O$ : $2H_2O(l) + 2e^- \rightarrow H_2(g) + 2OH^-(aq)$ $E^\circ = -0.83 \text{ V}$ (at standard conditions, $1 \text{ M } OH^-$ )

Comparing the standard reduction potentials, $-0.83 \text{ V}$ is significantly less negative (more positive) than $-2.71 \text{ V}$ . This means that water is much easier to reduce than $Na^+$ ions. Therefore, $H_2$ gas will be liberated at the cathode.

Watch out

A common misconception is that $Na^+$ will be reduced because it's a cation. However, in aqueous solutions, the reduction potential of water often dictates the cathode product if the metal ion is very reactive (has a very negative reduction potential).

Reactions at the Anode (Oxidation):

The anode is the positive electrode where oxidation takes place. Possible species to be oxidized are $Cl^-$ ions and $H_2O$ molecules.

Oxidation of $Cl^-$ : $2Cl^-(aq) \rightarrow Cl_2(g) + 2e^-$ $E^\circ_{ox} = -1.36 \text{ V}$ (or $E^\circ_{red} = +1.36 \text{ V}$ )

Oxidation of $H_2O$ : $2H_2O(l) \rightarrow O_2(g) + 4H^+(aq) + 4e^-$ $E^\circ_{ox} = -1.23 \text{ V}$ (or $E^\circ_{red} = +1.23 \text{ V}$ )

Based purely on standard oxidation potentials, $H_2O$ (with $E^\circ_{ox} = -1.23 \text{ V}$ ) appears to be easier to oxidize than $Cl^-$ (with $E^\circ_{ox} = -1.36 \text{ V}$ ) because $-1.23 \text{ V}$ is less negative (more positive). This would suggest $O_2$ should be liberated.

However, there's a crucial kinetic factor called overpotential.

Important

Overpotential: The extra voltage required beyond the theoretical standard electrode potential to initiate a reaction at a reasonable rate. This is particularly significant for the evolution of gases like $O_2$ and $H_2$ on certain electrode surfaces. …
Start your 14-day free trial to see the full solution →
CBSE 2023Set ANNUAL1 markMCQ
$Q. The charge required for the reduction of 1 mol of Al3+ to Al is$
(a) 96500 C
(b) 193000 C
(c) 289500 C
(d) 386000 C
›Reveal solutionSolution

The charge needed equals the number of moles of electrons transferred (n) times Faraday's constant F (96500 C/mol).

…

Start your 14-day free trial to see the full solution →
CBSE 2023Set ANNUAL1 markMCQ
$Q. Amount of electricity required to deposit one mole of Al from Al2O3 is$
(a) 1F
(b) 6F
(c) 3F
(d) 2F (F = faraday)
›Reveal solutionSolution

Al is deposited from Al2O3 as Al3+ + 3e- -> Al, so 1 mole of Al requires 3 Faraday of charge.

In the electrolytic reduction of alumina (Al2O3), aluminium exists as the Al3+ ion. The cathode half-reaction is:

$Al^{3+} + 3e^- \rightarrow Al$

…

Start your 14-day free trial to see the full solution →
CBSE 2022Set ANNUAL1 mark
$Q. How many coulomb is the charge of 1 mol electron? OR How many coulomb of electricity is carried by Al3+ ion?$
›Reveal solutionSolution

The charge on one mole of electrons is defined as 1 Faraday, equal to Avogadro's number times the charge on a single electron.

Calculation:

Charge on one electron, $e = 1.6\times10^{-19}\text{ C}$

Number of electrons in 1 mole $=N_A=6.022\times10^{23}$

$\text{Total charge}=N_A\times e=6.022\times10^{23}\times1.6\times10^{-19}\approx96352\text{ C}$ …

Start your 14-day free trial to see the full solution →
CBSE 2020Set 56/3/11 mark
$H_2O$
›Reveal solutionSolution

Water oxidation to oxygen gas releases electrons; counting them through the balanced half-reaction and applying Faraday's law gives the charge needed. 4 F = 386,000 C are required.

Why this approach works

Electrolysis problems hinge on a simple chain: chemical change → electrons transferred → charge required. Faraday's laws of electrolysis tell us that the amount of substance transformed at an electrode is directly proportional to the quantity of electricity passed through the electrolyte. The bridge between chemistry and electricity is the mole of electrons, quantified by the Faraday constant $F = 96{,}485 \, \text{C/mol}$ .

When we oxidize water to oxygen, we're stripping electrons away. The first step is always to write the balanced half-reaction so we know exactly how many electrons are involved per mole of product.

Step-by-step solution

1. Write the oxidation half-reaction for water.

Water molecules lose electrons to form oxygen gas. In acidic or neutral conditions, the half-reaction is:
$2 \, H_2O \longrightarrow O_2 + 4 \, H^+ + 4 \, e^-$
This tells us that producing 1 mole of $O_2$ requires the removal of 4 moles of electrons from 2 moles of water.

2. Identify what the question asks.

The question asks for the charge needed to oxidize 1 mol of $H_2O$ , not 1 mol of $O_2$ . This is a crucial distinction.

From the half-reaction above, 2 mol of $H_2O$ produce 1 mol of $O_2$ and release 4 mol of electrons. Therefore, 1 mol of $H_2O$ releases:
$\frac{4 \, \text{mol} \, e^-}{2 \, \text{mol} \, H_2O} = 2 \, \text{mol} \, e^- \, \text{per mol} \, H_2O$
Watch out

A common mistake is to assume 1 mol $H_2O$ → 1 mol $O_2$ , leading to 4 mol $e^-$ . Always check the stoichiometry of the balanced equation.

3. Convert moles of electrons to charge using Faraday's constant.

The charge $Q$ required is:

$Q = n \cdot F$ …
Start your 14-day free trial to see the full solution →
CBSE 2019Set ANNUAL1 markMCQ
$Q. Charge carried by 1 mole of PO4^3- ions is:$
(a) 96500 C
(b) 32167 C
(c) 289500 C
(d) 193000 C
›Reveal solutionSolution

Each mole of a triply-charged ion carries 3 Faradays of charge.

…

Start your 14-day free trial to see the full solution →

Q.What amount of electric charge is required for the reduction of 1 mole of MnO4−MnO_4^{-}MnO4−​ into Mn2+Mn^{2+}Mn2+ ? (A) 1 F (B) 5 F (C) 4 F (D) 6 F

Why the electron count matters

Step‑by‑step reasoning

🎓Unlock everything free for 14 days

Q.What amount of electric charge is required for the reduction of 1 mole of $MnO_4^{-}$ into $Mn^{2+}$ ? (A) 1 F (B) 5 F (C) 4 F (D) 6 F

Unlock everything free for 14 days