Q.If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire?

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

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

In molten $\mathrm{CaCl}_2$, calcium ions are reduced:

Both require 2 moles of electrons per mole of metal.

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

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

4. 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.

5. Convert moles of Ca to mass Atomic mass of Ca = 40 g/mol.

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:

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:

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

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

Hence, $n = 3$ moles of electrons per mole of Fe.

2. Write Faraday’s law

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

3. 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: $M = 56 \, \text{g/mol}$
   • $n = 3$
   • $F = 96485 \, \text{C/mol}$

4. Solve for current $I$

   Rearranging:

Substitute values:

5. Calculate stepwise
   • Numerator: $0.3482 \times 3 = 1.0446$
   • Then $1.0446 \times 96485 \approx 100,800$ (precise: $1.0446 \times 96485 = 100,799.6$)
   • Denominator: $1200 \times 56 = 67,200$
   • So $I = \frac{100,800}{67,200} = 1.5 \, \text{A}$ exactly.

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

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.

1. Calculate total charge passed

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

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

2. Calculate moles of Pd deposited

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

   Moles of Pd:

3. Relate charge to moles of electrons Faraday’s constant $F = 96485 \, \text{C/mol e}^-$. Total moles of electrons used:

4. Find the charge number $n$ Each mole of Pd requires $n$ moles of electrons, so:

Thus $n = 4$.

Effective charge passed (80% current efficiency)

Moles of $\text{Cu}^{2+}$ deposited — $\text{Cu}^{2+} + 2e^{-} \rightarrow \text{Cu}$:

Remaining $\text{Cu}^{2+}$

Final molarity (volume unchanged at 0.800 L)

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:

Charge for 0.1 mol of ions:

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

1. 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.

2. 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.

3. 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.

4. 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.

5. Convert to Faradays

   1 mole of electrons = 1 Faraday.

   Therefore, charge required = 21.0 Faradays.

Effective charge passed (90% efficiency):

Moles of electrons:

Copper deposited ($\mathrm{Cu^{2+}}+2e^-\to\mathrm{Cu}$):

Initial $\mathrm{Cu^{2+}}=0.750\times0.45=0.3375\ \mathrm{mol}$. Remaining:

Final molarity (volume constant at 0.750 L):

Step 1 — Combustion stoichiometry

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:

So $4\ \text{mol}$ of electrons give $1\ \text{mol}\ \text{O}_2$; for $5\ \text{mol}\ \text{O}_2$ we need $5 \times 4 = 20\ \text{mol}$ of electrons.

Step 3 — Charge and time (Faraday's law)

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

Q = 1.786 x 10^-2 x 96500 = 1723 C

Current (t = 2 h = 7200 s):

I = Q/t = 1723 / 7200 = 0.2393 A

ANSWER: D

At the anode, electrolysis of the aqueous solution generates $\mathrm{H^+}$:

Target: $\mathrm{pH}=4 \Rightarrow [\mathrm{H^+}] = 10^{-4}\,\mathrm{M}$ in $100\,\mathrm{mL} = 0.1\,\mathrm{L}$:

Charge required (Faraday's law):

Time:

This matches option (A). (The same digits, $1.93\times10^{3}$, arise if one also fully neutralises the $10^{-2}\,\mathrm{mol}$ starting $\mathrm{OH^-}$; the intended numerical answer is $1.93$.)

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):

Cathode (reduction of oxygen):

Overall:

Without the Pt/Pd catalyst these reactions — particularly the sluggish oxygen reduction at the cathode — would be far too slow to draw useful current.

Step 4 — Rule out the distractors.

• (A) Zinc–Mercury: zinc and mercury(II) oxide are the electrode materials of the mercury cell, not fuel-cell catalysts.
• (B) Lead–Manganese: lead is the lead storage battery; $MnO_2$ is the dry (Leclanché) cell.
• (D) Nickel–Cadmium: the electrode materials of the rechargeable Ni–Cd cell.

All three are battery electrodes, not catalysts.

ANSWER: C