Electrochemistry: Notes, Solved Examples & Exam Questions | Grade 12 Chemistry Unit 2

Grade 12 Chemistry Unit 2: Electrochemistry

Welcome back, dear student! In Unit 1 we studied acid-base equilibria. Now in Unit 2, we will explore a very exciting topic — Electrochemistry. This is the branch of chemistry that deals with the relationship between electrical energy and chemical changes. Think about batteries that power your phone, or the process of electroplating jewelry — all of these depend on electrochemistry! Let’s learn it step by step.

2.1 Oxidation-Reduction Reactions

Before we talk about electrochemical cells, we need to understand the type of reactions that happen inside them. These are called oxidation-reduction (redox) reactions.

2.1.1 Oxidation

What comes to your mind when you hear “oxidation”? Many students think of iron rusting or a piece of paper burning. You are on the right track! But in chemistry, oxidation has a very precise meaning.

Oxidation is a process in which a substance loses electrons or has its oxidation number increased.

Remember: “LEO” — Loss of Electrons is Oxidation.

Let’s look at a simple example. When sodium metal reacts with chlorine gas:

2Na(s) + Cl₂(g) → 2NaCl(s)

Let’s examine what happens to sodium. A neutral Na atom has 1 valence electron. In NaCl, sodium becomes Na⁺ — it has lost its electron!

Na → Na⁺ + e⁻

This loss of electrons is oxidation. The oxidation number of Na increases from 0 to +1.

2.1.2 Reduction

Now, what happens to chlorine in the same reaction?

Reduction is a process in which a substance gains electrons or has its oxidation number decreased.

Remember: “GER” — Gain of Electrons is Reduction.

Together: “LEO the lion says GER” — Loss of Electrons is Oxidation, Gain of Electrons is Reduction.

Chlorine gains the electron that sodium lost:

Cl₂ + 2e⁻ → 2Cl⁻

The oxidation number of Cl decreases from 0 to −1. This is reduction.

Important: Oxidation and reduction always happen TOGETHER. You cannot have one without the other. If something loses electrons, something else must gain them!

Key Exam Notes — Oxidation and Reduction:

Oxidation = loss of electrons = increase in oxidation number.
Reduction = gain of electrons = decrease in oxidation number.
The substance that is oxidized is called the reducing agent (it causes reduction of another substance).
The substance that is reduced is called the oxidizing agent (it causes oxidation of another substance).
In our example: Na is oxidized (reducing agent), Cl₂ is reduced (oxidizing agent).

Practice Question 1: In the reaction Zn + CuSO₄ → ZnSO₄ + Cu, identify what is oxidized, what is reduced, the oxidizing agent, and the reducing agent.

Answer:
Zn → Zn²⁺ + 2e⁻ (oxidation number increases from 0 to +2) → Zn is oxidized (reducing agent).
Cu²⁺ + 2e⁻ → Cu (oxidation number decreases from +2 to 0) → Cu²⁺ is reduced (oxidizing agent).

Practice Question 2: Determine the oxidation number of Mn in KMnO₄.

Answer: KMnO₄ is neutral, so the sum of oxidation numbers = 0.
K = +1, O = −2 (each), let Mn = x
(+1) + x + 4(−2) = 0
1 + x − 8 = 0
x = +7
The oxidation number of Mn in KMnO₄ is +7.

2.1.3 Balancing Oxidation-Reduction (Redox) Reactions

Balancing redox reactions is a very important skill for your exam. There are two main methods: the oxidation number method and the half-reaction (ion-electron) method. Let’s learn both carefully.

Method 1: Oxidation Number Method

Follow these steps:

Assign oxidation numbers to all elements.
Identify which elements are oxidized and which are reduced.
Calculate the increase and decrease in oxidation numbers.
Use coefficients to make the increase equal to the decrease.
Balance the remaining atoms by inspection.
Balance O by adding H₂O and balance H by adding H⁺ (for acidic) or OH⁻ (for basic).

Worked Example 1

Question: Balance the following redox reaction in acidic medium: MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺

Solution (step-by-step):

Step 1: Assign oxidation numbers:

Mn in MnO₄⁻: x + 4(−2) = −1 → x = +7

Mn in Mn²⁺: +2

Fe in Fe²⁺: +2; Fe in Fe³⁺: +3

Step 2: Identify changes:

Mn: +7 → +2 (decrease of 5) → Reduction

Fe: +2 → +3 (increase of 1) → Oxidation

Step 3: Make increase = decrease. LCM of 5 and 1 is 5.

So: 1 MnO₄⁻ and 5 Fe²⁺

Step 4: Write the unbalanced equation:

MnO₄⁻ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺

Step 5: Balance O by adding H₂O (left side has 4 O):

MnO₄⁻ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O

Step 6: Balance H by adding H⁺ (right side has 8 H):

MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O

Step 7: Check charge balance: Left = −1 + 10 + 8 = +17; Right = +2 + 15 + 0 = +17 ✓

Key Exam Notes — Oxidation Number Method:

First find the total increase and total decrease in oxidation numbers.
Use LCM to make them equal — this gives you the coefficients.
Balance O with H₂O, then H with H⁺ (acidic) or OH⁻ (basic).
Always verify charge balance at the end!

Method 2: Half-Reaction (Ion-Electron) Method

This method is very systematic. Follow these steps:

Write the two half-reactions (oxidation and reduction separately).
Balance all atoms EXCEPT O and H.
Balance O by adding H₂O.
Balance H by adding H⁺ (acidic) or OH⁻ (basic).
Balance charge by adding electrons (e⁻).
Multiply the half-reactions so that electrons cancel.
Add the half-reactions and simplify.

Worked Example 2

Question: Balance using the half-reaction method in acidic medium: Cr₂O₇²⁻ + Fe²⁺ → Cr³⁺ + Fe³⁺

Solution:

Step 1: Write half-reactions:

Reduction: Cr₂O₇²⁻ → Cr³⁺

Oxidation: Fe²⁺ → Fe³⁺

Step 2: Balance atoms (except O and H):

Cr₂O₇²⁻ → 2Cr³⁺ (balance Cr)

Fe²⁺ → Fe³⁺ (already balanced)

Step 3: Balance O with H₂O:

Cr₂O₇²⁻ → 2Cr³⁺ + 7H₂O

Step 4: Balance H with H⁺:

Cr₂O₇²⁻ + 14H⁺ → 2Cr³⁺ + 7H₂O

Step 5: Balance charge with electrons:

Left charge: −2 + 14 = +12; Right charge: 2(+3) = +6

Add 6e⁻ to the left: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O

For oxidation: Fe²⁺ → Fe³⁺ + e⁻

Step 6: Multiply oxidation half-reaction by 6:

6Fe²⁺ → 6Fe³⁺ + 6e⁻

Step 7: Add and cancel electrons:

Cr₂O₇²⁻ + 14H⁺ + 6e⁻ + 6Fe²⁺ → 2Cr³⁺ + 7H₂O + 6Fe³⁺ + 6e⁻

Balanced equation:

Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O

Worked Example 3

Question: Balance in basic medium: MnO₄⁻ + I⁻ → MnO₂ + I₂

Solution:

Half-reactions:

Reduction: MnO₄⁻ → MnO₂

Oxidation: I⁻ → I₂

Balance atoms: 2I⁻ → I₂

Balance O with H₂O: MnO₄⁻ → MnO₂ + 2H₂O

Balance H with H⁺ (first balance as acidic): MnO₄⁻ + 4H⁺ → MnO₂ + 2H₂O

Balance charge: Left = −1 + 4 = +3; Right = 0. Add 3e⁻ to left:

MnO₄⁻ + 4H⁺ + 3e⁻ → MnO₂ + 2H₂O

Oxidation: 2I⁻ → I₂ + 2e⁻

Multiply to cancel electrons: LCM of 3 and 2 is 6.

2 × (MnO₄⁻ + 4H⁺ + 3e⁻ → MnO₂ + 2H₂O)

3 × (2I⁻ → I₂ + 2e⁻)

Add: 2MnO₄⁻ + 8H⁺ + 6I⁻ → 2MnO₂ + 4H₂O + 3I₂

Now convert to basic medium: Add 8OH⁻ to both sides to neutralize H⁺:

2MnO₄⁻ + 8H⁺ + 6I⁻ + 8OH⁻ → 2MnO₂ + 4H₂O + 3I₂ + 8OH⁻

8H⁺ + 8OH⁻ → 8H₂O. Simplify: 8H₂O on left, cancel 4 with right → 4H₂O on left:

2MnO₄⁻ + 6I⁻ + 4H₂O → 2MnO₂ + 3I₂ + 8OH⁻

Practice Question 3: Balance in acidic medium: MnO₄⁻ + Sn²⁺ → Mn²⁺ + Sn⁴⁺

Answer:
Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
Oxidation: Sn²⁺ → Sn⁴⁺ + 2e⁻
LCM of 5 and 2 = 10. Multiply reduction by 2, oxidation by 5:
2MnO₄⁻ + 16H⁺ + 10e⁻ → 2Mn²⁺ + 8H₂O
5Sn²⁺ → 5Sn⁴⁺ + 10e⁻
Balanced: 2MnO₄⁻ + 5Sn²⁺ + 16H⁺ → 2Mn²⁺ + 5Sn⁴⁺ + 8H₂O

Practice Question 4: Balance in basic medium: Cl₂ → ClO⁻ + Cl⁻ (this is a disproportionation reaction where Cl₂ is both oxidized and reduced!)

Answer:
This is a disproportionation — Cl goes from 0 to +1 (oxidation) and from 0 to −1 (reduction).
Reduction: Cl₂ + 2e⁻ → 2Cl⁻
Oxidation: Cl₂ + 4OH⁻ → 2ClO⁻ + 2H₂O + 2e⁻
(Balanced for basic: O with H₂O, H with OH⁻, then charge with e⁻)
Add: 2Cl₂ + 4OH⁻ → 2Cl⁻ + 2ClO⁻ + 2H₂O
Simplify: Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O

2.2 Electrolysis of Aqueous Solutions

2.2.1 Electrolytic Cells

Now let’s learn about using electrical energy to drive non-spontaneous chemical reactions. This process is called electrolysis, and the device used is called an electrolytic cell.

Electrolysis is the process of passing electric current through an electrolyte (molten or in solution) to cause a chemical reaction that would NOT occur on its own.


Key terms:
Anode: The positive electrode where oxidation occurs.
Cathode: The negative electrode where reduction occurs.
Electrolyte: The substance (molten salt or solution) that conducts electricity through the movement of ions.

Can you see the difference from what we learned before? In electrolytic cells, the anode is positive and the cathode is negative. This is because an external power source (battery) forces electrons to flow in a particular direction.

External Battery (+) ──────────────── (−) | | Anode Cathode (Positive) (Negative) OXIDATION REDUCTION | | └──── Electrolyte ───┘

Remember for electrolytic cells: Anode = Positive = Oxidation; Cathode = Negative = Reduction.

2.2.2 Preferential Discharge

Here is a very important question: when you have an aqueous solution, there are multiple ions present — the ions from the salt AND the ions from water (H⁺ and OH⁻). Which ion gets discharged (loses its charge) at each electrode?

The answer depends on the electrode potential (a measure of how easily an ion gains or loses electrons). There is a specific order called preferential discharge:

Order of Discharge at the Anode (where oxidation occurs):
If halide ions (Cl⁻, Br⁻, I⁻) are present → halide is discharged preferentially over OH⁻.
If NO halide ions are present (only SO₄²⁻, NO₃⁻, etc.) → OH⁻ is discharged: 4OH⁻ → O₂ + 2H₂O + 4e⁻

Order of Discharge at the Cathode (where reduction occurs):
If the metal ion is below hydrogen in the reactivity series (Cu²⁺, Ag⁺, Au⁺, etc.) → metal ion is discharged.
If the metal ion is above hydrogen (Na⁺, K⁺, Ca²⁺, Mg²⁺, Zn²⁺, etc.) → H⁺ from water is discharged: 2H⁺ + 2e⁻ → H₂

Position	Cations (cathode)	Anions (anode)
Discharged FIRST	Ag⁺, Cu²⁺, Au⁺ (below H₂)	I⁻, Br⁻, Cl⁻ (halides)
Discharged SECOND	H⁺ (from water)	OH⁻ (from water) → O₂
NOT discharged	Na⁺, K⁺, Ca²⁺, Mg²⁺, Al³⁺	SO₄²⁻, NO₃⁻, CO₃²⁻

Key Exam Notes — Preferential Discharge:

At cathode: metals below H₂ in reactivity series are discharged; above H₂, hydrogen gas is produced.
At anode: halides (Cl⁻, Br⁻, I⁻) are discharged over OH⁻; otherwise O₂ is produced from OH⁻.
Sulfate (SO₄²⁻) and nitrate (NO₃⁻) are NEVER discharged in aqueous solution.
The concentration of ions and the nature of the electrode can sometimes affect which ion is discharged.

Practice Question 5: Predict the products at each electrode during the electrolysis of aqueous CuSO₄ using inert electrodes.

Answer:
Ions present: Cu²⁺, SO₄²⁻, H⁺, OH⁻
Cathode (−): Cu²⁺ is below H₂ in reactivity series → Cu²⁺ + 2e⁻ → Cu(s) (copper deposits)
Anode (+): No halide present → OH⁻ is discharged: 4OH⁻ → O₂(g) + 2H₂O(l) + 4e⁻
Products: Copper at cathode, oxygen gas at anode.

Practice Question 6: Predict the products at each electrode during the electrolysis of aqueous NaCl using inert electrodes.

Answer:
Ions present: Na⁺, Cl⁻, H⁺, OH⁻
Cathode (−): Na⁺ is above H₂ in reactivity series → H⁺ is discharged: 2H⁺ + 2e⁻ → H₂(g)
Anode (+): Cl⁻ is a halide → Cl⁻ is discharged: 2Cl⁻ → Cl₂(g) + 2e⁻
Products: Hydrogen gas at cathode, chlorine gas at anode. Na⁺ and OH⁻ remain in solution (NaOH is produced).

2.2.3 Electrolysis of Some Selected Aqueous Solutions

Let’s now study specific examples in detail. This is heavily tested in exams!

(a) Electrolysis of Dilute H₂SO₄ (Acidified Water)

Ions present: H⁺, SO₄²⁻, OH⁻

Cathode: 2H⁺ + 2e⁻ → H₂(g)

Anode: 4OH⁻ → O₂(g) + 2H₂O(l) + 4e⁻

Overall: 2H₂O(l) → 2H₂(g) + O₂(g)

This is really the electrolysis of water! The ratio of H₂ : O₂ produced is 2 : 1 by volume.

(b) Electrolysis of Concentrated NaCl Solution (Brine)

Ions present: Na⁺, Cl⁻, H⁺, OH⁻

Cathode: 2H⁺ + 2e⁻ → H₂(g)

Anode: 2Cl⁻ → Cl₂(g) + 2e⁻

Na⁺ and OH⁻ remain → NaOH solution is left behind.

Products: H₂ (cathode), Cl₂ (anode), NaOH (solution) — this is the chlor-alkali process!

(c) Electrolysis of CuSO₄ Solution with Inert Electrodes

Ions: Cu²⁺, SO₄²⁻, H⁺, OH⁻

Cathode: Cu²⁺ + 2e⁻ → Cu(s) (copper metal deposits)

Anode: 4OH⁻ → O₂(g) + 2H₂O(l) + 4e⁻

As Cu²⁺ is removed and H⁺ builds up, the solution becomes increasingly acidic.

(d) Electrolysis of CuSO₄ Solution with Copper Electrodes

This is a special and important case! The copper anode is not inert — it participates in the reaction!

Cathode: Cu²⁺ + 2e⁻ → Cu(s) (copper deposits on cathode)

Anode: Cu(s) → Cu²⁺ + 2e⁻ (copper anode dissolves!)

The concentration of CuSO₄ stays constant! This process is called electroplating or electrorefining of copper. The impure copper anode loses copper, and pure copper deposits on the cathode.

Worked Example 4

Question: During the electrolysis of aqueous CuSO₄ using copper electrodes, a student noticed that the blue color of the solution did NOT change. Explain why.

Answer: With copper electrodes, Cu dissolves from the anode at the same rate that Cu²⁺ deposits at the cathode. The concentration of Cu²⁺ in solution remains constant, so the blue color (due to Cu²⁺) does not change.

Key Exam Notes — Electrolysis Examples:

Dilute H₂SO₄ → electrolysis of water → H₂ and O₂ in 2:1 ratio.
Concentrated NaCl → H₂, Cl₂, and NaOH (chlor-alkali process).
CuSO₄ with inert electrodes → Cu deposits (cathode), O₂ evolves (anode), solution becomes acidic.
CuSO₄ with Cu electrodes → Cu transfers from anode to cathode; concentration stays constant.
When the anode is made of the same metal as the cation in solution, the anode dissolves instead of OH⁻ being discharged.

Practice Question 7: What happens to the pH of the solution during the electrolysis of aqueous CuSO₄ using inert electrodes? Explain.

Answer: The pH decreases (solution becomes more acidic). At the anode, OH⁻ is discharged (4OH⁻ → O₂ + 2H₂O + 4e⁻), removing OH⁻ from the solution. At the cathode, Cu²⁺ is discharged instead of H⁺, so H⁺ remains in the solution. The removal of OH⁻ and retention of H⁺ makes the solution increasingly acidic.

2.3 Quantitative Aspects of Electrolysis

How much product will form during electrolysis? The answer was given by Michael Faraday through his two famous laws.

2.3.1 Faraday’s First Law of Electrolysis

Faraday’s First Law: The mass of a substance deposited or liberated at an electrode during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte.

$$m = Z \times Q = Z \times I \times t$$

Where:
m = mass of substance deposited (grams)
Z = electrochemical equivalent (g/C)
Q = quantity of electricity (coulombs, C)
I = current (amperes, A)
t = time (seconds, s)

We can also express this using Faraday’s constant (F). One faraday (F) = 96,500 C/mol, which is the charge carried by one mole of electrons.

$$m = \frac{M \times I \times t}{n \times F} = \frac{M \times Q}{n \times 96500}$$

Where M = molar mass (g/mol), n = number of electrons transferred per ion.

Worked Example 5

Question: Calculate the mass of copper deposited when a current of 5.0 A is passed through CuSO₄ solution for 30 minutes. (M_Cu = 63.5 g/mol)

Solution:

Step 1: Identify n: Cu²⁺ + 2e⁻ → Cu, so n = 2

Step 2: Convert time: t = 30 min × 60 = 1800 s

Step 3: Calculate Q = I × t = 5.0 × 1800 = 9000 C

Step 4: Calculate mass:

$$m = \frac{63.5 \times 9000}{2 \times 96500} = \frac{571500}{193000} = 2.96 \text{ g}$$

Worked Example 6

Question: What current is needed to deposit 2.0 g of silver from AgNO₃ solution in 20 minutes? (M_Ag = 108 g/mol)

Solution:

n = 1 (Ag⁺ + e⁻ → Ag), t = 20 × 60 = 1200 s, m = 2.0 g

$$I = \frac{m \times n \times F}{M \times t} = \frac{2.0 \times 1 \times 96500}{108 \times 1200} = \frac{193000}{129600} = 1.49 \text{ A}$$

Key Exam Notes — Faraday’s First Law:

Always convert time to seconds!
Identify n correctly: Cu²⁺ → n = 2; Ag⁺ → n = 1; Al³⁺ → n = 3.
1 Faraday (F) = 96,500 C/mol = charge of 1 mole of electrons.
Mass is directly proportional to current and time.

Practice Question 8: How long will it take to deposit 0.50 g of iron from FeCl₂ solution using a current of 3.0 A? (M_Fe = 56 g/mol)

Answer:
Fe²⁺ + 2e⁻ → Fe, so n = 2
m = (M × I × t) / (n × F)
0.50 = (56 × 3.0 × t) / (2 × 96500)
0.50 = 168t / 193000
t = (0.50 × 193000) / 168 = 96500 / 168 = 574 s = 9.57 minutes

2.3.2 Faraday’s Second Law of Electrolysis

Faraday’s Second Law: When the same quantity of electricity is passed through different electrolytes in series, the masses of different substances deposited are proportional to their chemical equivalent weights (molar mass divided by valency).

$$\frac{m_1}{m_2} = \frac{E_1}{E_2} = \frac{M_1/n_1}{M_2/n_2}$$

Where E = equivalent weight = M/n (molar mass divided by number of electrons transferred).

Worked Example 7

Question: A current is passed through two cells in series: one containing AgNO₃ and the other containing CuSO₄. If 1.08 g of silver is deposited in the first cell, what mass of copper is deposited in the second cell? (M_Ag = 108, M_Cu = 63.5)

Solution:

Equivalent weight of Ag = 108/1 = 108 g/eq

Equivalent weight of Cu = 63.5/2 = 31.75 g/eq

$$\frac{m_{Cu}}{m_{Ag}} = \frac{E_{Cu}}{E_{Ag}} = \frac{31.75}{108}$$

$$m_{Cu} = 1.08 \times \frac{31.75}{108} = 0.3175 \text{ g}$$

Worked Example 8

Question: The same quantity of electricity deposited 0.32 g of copper from CuSO₄ solution. What mass of zinc would be deposited from ZnSO₄ solution by the same quantity? (M_Cu = 63.5, M_Zn = 65)

Solution:

E_Cu = 63.5/2 = 31.75; E_Zn = 65/2 = 32.5

$$\frac{m_{Zn}}{0.32} = \frac{32.5}{31.75}$$

$$m_{Zn} = 0.32 \times \frac{32.5}{31.75} = 0.328 \text{ g}$$

Key Exam Notes — Faraday’s Second Law:

Applies when the SAME quantity of electricity passes through different cells connected in series.
Equivalent weight E = M/n (NOT just molar mass!).
For Ag⁺: E = 108/1 = 108; for Cu²⁺: E = 63.5/2 = 31.75; for Al³⁺: E = 27/3 = 9.
The ratio of masses equals the ratio of equivalent weights.

Practice Question 9: When the same current was passed through solutions of AgNO₃ and Al(NO₃)₃ in series for the same time, 2.16 g of silver was deposited. Calculate the mass of aluminum deposited. (M_Ag = 108, M_Al = 27)

Answer:
E_Ag = 108/1 = 108; E_Al = 27/3 = 9
m_Al / m_Ag = E_Al / E_Ag
m_Al / 2.16 = 9 / 108 = 1/12
m_Al = 2.16 / 12 = 0.18 g

2.4 Industrial Application of Electrolysis

Electrolysis is not just a laboratory technique — it has many important industrial applications. Let’s look at the key ones.

1. Extraction of Metals

Highly reactive metals (Na, K, Ca, Mg, Al) cannot be extracted by reduction with carbon. They are extracted by electrolysis of their molten salts.

Extraction of Aluminum (Hall-Héroult Process):

Aluminum is extracted by electrolysis of molten alumina (Al₂O₃) dissolved in molten cryolite (Na₃AlF₆) at about 950°C.

Cathode: Al³⁺ + 3e⁻ → Al(l)

Anode: 2O²⁻ → O₂ + 4e⁻

The O₂ reacts with the carbon anode: C + O₂ → CO₂. So the carbon anodes must be replaced periodically!

Extraction of Sodium (Down’s Cell):

Molten NaCl is electrolyzed at about 600°C.

Cathode: Na⁺ + e⁻ → Na(l)

Anode: 2Cl⁻ → Cl₂(g) + 2e⁻

A steel gauze diaphragm prevents Na and Cl₂ from recombining.

2. Electroplating

Electroplating is the process of coating a metal object with a thin layer of another metal by electrolysis. For example, coating iron with chromium to make it shiny and rust-resistant.

Rules for Electroplating:
The object to be plated is the cathode (negative electrode).
The plating metal is the anode (positive electrode).
The electrolyte contains ions of the plating metal.

Example — Chromium plating of iron:

Cathode: Iron object
Anode: Chromium metal
Electrolyte: Cr₂(SO₄)₃ solution
Cathode reaction: Cr³⁺ + 3e⁻ → Cr
Anode reaction: Cr → Cr³⁺ + 3e⁻

3. Electrorefining of Metals

Impure metals can be purified by electrolysis. For example, copper is refined using CuSO₄ solution.

Anode: Impure copper
Cathode: Pure copper (thin sheet)
Electrolyte: CuSO₄ solution with H₂SO₄
As current flows, Cu dissolves from the impure anode and deposits on the pure cathode.
Impurities either remain in solution or fall to the bottom as “anode mud” (which contains valuable metals like Ag and Au!).

4. Chlor-Alkali Industry

Electrolysis of concentrated NaCl solution (brine) produces three important chemicals:

Hydrogen gas (cathode) — used in making ammonia, margarine
Chlorine gas (anode) — used in water treatment, PVC production
Sodium hydroxide (remains in solution) — used in soap, paper, aluminum production

Key Exam Notes — Industrial Applications:

Al is extracted from molten Al₂O₃ in cryolite (Hall-Héroult process); carbon anodes are consumed.
Na is extracted from molten NaCl (Down’s cell); uses a diaphragm.
In electroplating: object = cathode, plating metal = anode.
Electrorefining: impure metal = anode, pure metal = cathode.
Chlor-alkali process: brine → H₂ + Cl₂ + NaOH.

Practice Question 10: Why is cryolite used in the extraction of aluminum?

Answer: Pure Al₂O₃ has a very high melting point (about 2050°C), which would require too much energy. Adding cryolite (Na₃AlF₆) lowers the melting point to about 950°C, making the process much more energy-efficient and cost-effective. Cryolite also improves the conductivity of the molten mixture.

Practice Question 11: In the electrorefining of copper, why do silver and gold collect as “anode mud” instead of dissolving?

Answer: Silver and gold are less reactive (more noble) than copper. They have higher reduction potentials, meaning they are NOT oxidized at the anode under the conditions used. While copper dissolves (Cu → Cu²⁺ + 2e⁻), Ag and Au simply fall off the anode and settle at the bottom as “anode mud.” This anode mud is actually a valuable by-product since it contains precious metals!

2.5 Voltaic Cells

So far we have studied electrolytic cells, where electrical energy drives a non-spontaneous reaction. Now let’s look at the opposite: voltaic (galvanic) cells, where a spontaneous chemical reaction produces electrical energy. This is how batteries work!

Voltaic Cell: A device that converts chemical energy from a spontaneous redox reaction into electrical energy.


Key difference from electrolytic cell:
In a voltaic cell, the anode is NEGATIVE and the cathode is POSITIVE.
This is the OPPOSITE of an electrolytic cell!

Structure of a Voltaic Cell

A simple voltaic cell consists of:

Two half-cells (each containing an electrode in an electrolyte solution)
A salt bridge (or porous barrier) connecting the two half-cells
An external wire connecting the two electrodes (where electrons flow)

The Daniell Cell (Zn-Cu Cell)

This is the classic example of a voltaic cell:

Zn electrode Salt bridge Cu electrode (Anode) (KNO₃ solution) (Cathode) ┌────────┐ ┌──────┐ ┌────────┐ │ Zn │ │ K⁺ │ │ Cu │ │ ↓ │ │ ↕ │ │ ↑ │ │ ZnSO₄ │───── wire ──│ NO₃⁻ │───────────│ CuSO₄ │ │(aq) │ e⁻ → │ │ → e⁻ │(aq) │ └────────┘ └──────┘ └────────┘ Zn → Zn²⁺ + 2e⁻ Cu²⁺ + 2e⁻ → Cu OXIDATION REDUCTION Anode (−) Cathode (+)

Half-reactions:

Anode (oxidation): Zn(s) → Zn²⁺(aq) + 2e⁻

Cathode (reduction): Cu²⁺(aq) + 2e⁻ → Cu(s)

Overall: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

What does the salt bridge do?

It completes the circuit by allowing ions to flow between the two half-cells.
Without it, the solutions would build up charge and the reaction would stop.
Anions (like NO₃⁻) flow toward the anode (to balance the extra Zn²⁺ produced).
Cations (like K⁺) flow toward the cathode (to replace the Cu²⁺ being consumed).

Electrode Potential and Cell Potential

Each half-cell has an electrode potential (E°), measured in volts. It tells us the tendency of a half-reaction to occur. Standard electrode potentials are measured against the Standard Hydrogen Electrode (SHE), which is assigned E° = 0.00 V.

Half-Reaction	E° (V)
Li⁺ + e⁻ → Li	−3.05
K⁺ + e⁻ → K	−2.93
Na⁺ + e⁻ → Na	−2.71
Zn²⁺ + 2e⁻ → Zn	−0.76
Fe²⁺ + 2e⁻ → Fe	−0.44
2H⁺ + 2e⁻ → H₂	0.00 (SHE)
Cu²⁺ + 2e⁻ → Cu	+0.34
Ag⁺ + e⁻ → Ag	+0.80
Au³⁺ + 3e⁻ → Au	+1.50

The cell potential (E°_cell) is calculated as:

$$E°_{cell} = E°_{cathode} – E°_{anode}$$

Remember: Always subtract the anode potential from the cathode potential!

If E°_cell > 0, the reaction is spontaneous (the cell will produce electricity). If E°_cell < 0, the reaction is non-spontaneous.

Worked Example 9

Question: Calculate the standard cell potential for the Daniell cell (Zn-Cu cell).

Solution:

Anode (oxidation): Zn → Zn²⁺ + 2e⁻ → E° = −0.76 V

Cathode (reduction): Cu²⁺ + 2e⁻ → Cu → E° = +0.34 V

$$E°_{cell} = E°_{cathode} – E°_{anode} = 0.34 – (-0.76) = 0.34 + 0.76 = 1.10 \text{ V}$$

Since E°_cell = +1.10 V > 0, the reaction is spontaneous. This cell can produce electricity!

Worked Example 10

Question: Will a voltaic cell made with Mg and Ag electrodes work? Calculate E°_cell.

Solution:

Mg is more easily oxidized (more negative E°), so Mg is the anode and Ag is the cathode.

Anode: Mg → Mg²⁺ + 2e⁻; E° = −2.37 V

Cathode: Ag⁺ + e⁻ → Ag; E° = +0.80 V

$$E°_{cell} = 0.80 – (-2.37) = 3.17 \text{ V}$$

Yes! E°_cell = 3.17 V > 0 — this cell definitely works and produces a large voltage.

Key Exam Notes — Voltaic Cells:

Voltaic cell: chemical energy → electrical energy (spontaneous reaction).
Anode = negative (oxidation); Cathode = positive (reduction).
Electrons flow from anode to cathode through the external wire.
Salt bridge: completes the circuit; anions → anode, cations → cathode.
E°_cell = E°_cathode − E°_anode; must be positive for spontaneous reaction.
The more negative E° value identifies the anode (more easily oxidized).
Electrolytic cell: anode = positive, cathode = negative (OPPOSITE of voltaic!).

Practice Question 12: For a voltaic cell using Fe and Cu electrodes, calculate E°_cell and write the overall cell reaction.

Answer:
Fe is more easily oxidized (E° = −0.44 V) than Cu (E° = +0.34 V).
Anode: Fe → Fe²⁺ + 2e⁻ (E° = −0.44 V)
Cathode: Cu²⁺ + 2e⁻ → Cu (E° = +0.34 V)
E°_cell = 0.34 − (−0.44) = 0.78 V
Overall: Fe(s) + Cu²⁺(aq) → Fe²⁺(aq) + Cu(s)

Practice Question 13: A student sets up a cell with Zn|Zn²⁺ and Ag|Ag⁺ half-cells. Calculate E°_cell. Which electrode is the anode?

Answer:
Zn has E° = −0.76 V and Ag has E° = +0.80 V.
Zn is more easily oxidized (more negative E°), so Zn is the anode and Ag is the cathode.
E°_cell = 0.80 − (−0.76) = 1.56 V

Practice Question 14: Can a cell made with Ag|Ag⁺ and Au|Au³⁺ half-cells produce electricity? Explain.

Answer:
Ag: E° = +0.80 V; Au: E° = +1.50 V
Ag is more easily oxidized (less positive E°), so Ag is the anode and Au is the cathode.
E°_cell = 1.50 − 0.80 = +0.70 V
Yes! Since E°_cell = +0.70 V > 0, the cell is spontaneous and will produce electricity.
Overall: 3Ag(s) + Au³⁺(aq) → 3Ag⁺(aq) + Au(s)

Revision Notes — Exam Focus

Oxidation and Reduction at a Glance

	Oxidation	Reduction
Electrons	Lost	Gained
Oxidation number	Increases	Decreases
Agent	Reducing agent	Oxidizing agent
Memory aid	LEO	GER

Key Formulas

$$\text{Faraday’s First Law: } m = \frac{M \times I \times t}{n \times 96500}$$
$$\text{Faraday’s Second Law: } \frac{m_1}{m_2} = \frac{E_1}{E_2} = \frac{M_1/n_1}{M_2/n_2}$$
$$\text{Cell Potential: } E°_{cell} = E°_{cathode} – E°_{anode}$$
$$\text{Quantity of electricity: } Q = I \times t \text{ (coulombs)}$$
$$\text{1 Faraday } = 96,500 \text{ C/mol}$$
$$\text{Equivalent weight: } E = \frac{M}{n}$$

Important Definitions

Oxidation: Loss of electrons or increase in oxidation number.
Reduction: Gain of electrons or decrease in oxidation number.
Reducing agent: The substance that is oxidized (causes another substance to be reduced).
Oxidizing agent: The substance that is reduced (causes another substance to be oxidized).
Electrolysis: Chemical change caused by passing electric current through an electrolyte.
Electrolytic cell: Device that converts electrical energy to chemical energy (non-spontaneous).
Voltaic cell: Device that converts chemical energy to electrical energy (spontaneous).
Anode: Electrode where oxidation occurs.
Cathode: Electrode where reduction occurs.
Salt bridge: Connects two half-cells; allows ion flow to maintain electrical neutrality.
Electroplating: Coating an object with a metal layer using electrolysis.
Electrorefining: Purifying a metal by using it as the anode in electrolysis.
Preferential discharge: Order in which ions are discharged at electrodes during electrolysis.

Electrolytic vs. Voltaic Cells — Quick Comparison

Feature	Electrolytic Cell	Voltaic Cell
Energy conversion	Electrical → Chemical	Chemical → Electrical
Reaction type	Non-spontaneous	Spontaneous
Anode charge	Positive (+)	Negative (−)
Cathode charge	Negative (−)	Positive (+)
External source	Required (battery)	Not required
Salt bridge	Not always needed	Required

Preferential Discharge — Quick Rules

AT THE CATHODE (reduction): Metal below H₂ (Cu²⁺, Ag⁺, Au⁺) → Metal discharged Metal above H₂ (Na⁺, K⁺, Ca²⁺, Mg²⁺, Zn²⁺) → H₂ from water AT THE ANODE (oxidation): Halide present (Cl⁻, Br⁻, I⁻) → Halogen discharged No halide (SO₄²⁻, NO₃⁻) → O₂ from OH⁻

Standard Electrode Potentials to Remember

Half-Reaction	E° (V)
K⁺ + e⁻ → K	−2.93
Na⁺ + e⁻ → Na	−2.71
Mg²⁺ + 2e⁻ → Mg	−2.37
Al³⁺ + 3e⁻ → Al	−1.66
Zn²⁺ + 2e⁻ → Zn	−0.76
Fe²⁺ + 2e⁻ → Fe	−0.44
2H⁺ + 2e⁻ → H₂	0.00
Cu²⁺ + 2e⁻ → Cu	+0.34
Ag⁺ + e⁻ → Ag	+0.80
Au³⁺ + 3e⁻ → Au	+1.50

Common Mistakes to Avoid

Mistake 1: Confusing anode/cathode signs in electrolytic vs. voltaic cells. Electrolytic: anode = +, cathode = −. Voltaic: anode = −, cathode = +.
Mistake 2: Forgetting to convert time to seconds in Faraday’s law calculations.
Mistake 3: Using the wrong value of n (number of electrons). For Cu2+, n = 2; for Al3+, n = 3; for Ag+, n = 1.
Mistake 4: In Faraday’s Second Law, using molar mass instead of equivalent weight (E = M/n).
Mistake 5: Getting the wrong order in E°cell formula. It is ALWAYS E°cathode − E°anode, not the reverse!
Mistake 6: Forgetting to balance charge when balancing half-reactions. Always check that total charge on left = total charge on right.
Mistake 7: Thinking SO42− can be discharged at the anode. It CANNOT — only halides or OH− are discharged.
Mistake 8: Forgetting to add OH− to BOTH sides when converting from acidic to basic medium in half-reaction balancing.
Mistake 9: Forgetting that in CuSO4 with Cu electrodes, the anode dissolves (not O2 production).
Mistake 10: Confusing reducing agent and oxidizing agent. The substance OXIDIZED is the REDUCING agent.

Challenge Exam Questions

These questions are designed to test your deep understanding. Try them all before checking the answers!

Section A: Multiple Choice Questions

Question 1: In the reaction 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂, which substance is the oxidizing agent?
(a) Fe³⁺ (b) I⁻ (c) Fe²⁺ (d) I₂

Answer: (a) Fe³⁺
Fe³⁺ is reduced to Fe²⁺ (gains electrons, oxidation number decreases from +3 to +2). The substance that is reduced is the oxidizing agent. I⁻ is oxidized to I₂ (it is the reducing agent).

Question 2: During the electrolysis of dilute Na₂SO₄ using inert electrodes, the products are:
(a) Na and O₂ (b) H₂ and O₂ (c) Na and SO₂ (d) H₂ and SO₂

Answer: (b) H₂ and O₂
Na⁺ is above H₂ → H⁺ from water is discharged at cathode: 2H⁺ + 2e⁻ → H₂. SO₄²⁻ is not a halide → OH⁻ from water is discharged at anode: 4OH⁻ → O₂ + 2H₂O + 4e⁻. This is essentially electrolysis of water.

Question 3: In a voltaic cell, electrons flow from:
(a) Cathode to anode through external circuit (b) Anode to cathode through external circuit
(c) Anode to cathode through the salt bridge (d) Salt bridge to the electrodes

Answer: (b) Anode to cathode through external circuit
In a voltaic cell, oxidation at the anode produces electrons. These electrons flow through the external wire from the anode (negative) to the cathode (positive) where reduction occurs. Ions flow through the salt bridge, NOT electrons.

Question 4: What is the oxidation number of Cr in Cr₂O₇²⁻?
(a) +6 (b) +7 (c) +3 (d) +12

Answer: (a) +6
2x + 7(−2) = −2 → 2x − 14 = −2 → 2x = 12 → x = +6. Each Cr has oxidation number +6 in Cr₂O₇²⁻.

Question 5: The function of a salt bridge in a voltaic cell is to:
(a) Provide a path for electron flow (b) Maintain electrical neutrality
(c) Increase the cell voltage (d) Prevent the reaction from occurring

Answer: (b) Maintain electrical neutrality
As the cell operates, Zn²⁺ builds up in the anode half-cell and Cu²⁺ is depleted in the cathode half-cell. Without a salt bridge, charge would build up and stop the reaction. The salt bridge allows ions to flow and maintain electrical neutrality. Electrons flow through the external wire, NOT the salt bridge.

Section B: Fill in the Blanks

Question 6: The mass of a substance deposited during electrolysis is __________ proportional to the quantity of electricity passed (Faraday’s First Law).

Answer: directly

Question 7: In the electrolysis of aqueous CuSO₄ using copper electrodes, the __________ electrode decreases in mass and the __________ electrode increases in mass.

Answer: anode (decreases in mass as Cu dissolves); cathode (increases in mass as Cu deposits).

Question 8: One faraday equals __________ coulombs, which corresponds to __________ mole(s) of electrons.

Answer: 96,500 coulombs; 1 mole of electrons.

Question 9: In a voltaic cell, the __________ has the more negative standard electrode potential and serves as the site of __________.

Answer: anode; oxidation. The electrode with more negative E° is more easily oxidized, so it becomes the anode where oxidation occurs.

Question 10: During the electrolysis of concentrated NaCl solution, the three products are H₂, Cl₂, and __________.

Answer: NaOH (sodium hydroxide). H⁺ from water is reduced at the cathode (→ H₂), Cl⁻ is oxidized at the anode (→ Cl₂), and Na⁺ and OH⁻ remain in solution as NaOH.

Section C: Short Answer Questions

Question 11: Explain the difference between an electrolytic cell and a voltaic cell in terms of: (a) energy conversion, (b) spontaneity, (c) anode charge.

Answer:
(a) Energy conversion: Electrolytic cell converts electrical energy to chemical energy. Voltaic cell converts chemical energy to electrical energy.
(b) Spontaneity: Electrolytic cell drives a non-spontaneous reaction using external energy. Voltaic cell uses a spontaneous reaction to produce energy.
(c) Anode charge: In an electrolytic cell, the anode is positive (connected to + terminal of battery). In a voltaic cell, the anode is negative (source of electrons).

Question 12: Why is it important to use a diaphragm in the Down’s cell for sodium extraction?

Answer: In the Down’s cell, molten NaCl is electrolyzed to produce Na at the cathode and Cl₂ at the anode. Sodium metal is highly reactive and would react violently with chlorine gas if they mixed: 2Na + Cl₂ → 2NaCl. The steel gauze diaphragm separates the two products, preventing them from coming into contact and recombining.

Question 13: During the electrolysis of CuSO₄ with inert electrodes, the solution gradually becomes acidic. Explain this observation using half-reactions.

Answer:
At the cathode: Cu²⁺ + 2e⁻ → Cu (Cu²⁺ is removed from solution)
At the anode: 4OH⁻ → O₂ + 2H₂O + 4e⁻ (OH⁻ is removed from solution)
Since OH⁻ is being consumed but H⁺ is NOT being consumed (because Cu²⁺ is discharged instead of H⁺), the concentration of H⁺ increases relative to OH⁻. The solution becomes increasingly acidic (pH decreases).

Question 14: Define “disproportionation reaction” and give one example from this unit.

Answer: A disproportionation reaction is a redox reaction in which the same element is both oxidized AND reduced. Example: Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O. Here, Cl goes from 0 to −1 (reduction) and from 0 to +1 (oxidation) simultaneously.

Section D: Step-by-Step Calculation Questions

Question 15: Balance the following redox reaction in acidic medium using the half-reaction method: MnO₄⁻ + C₂O₄²⁻ → Mn²⁺ + CO₂

Answer:
Reduction half: MnO₄⁻ → Mn²⁺
Balance Mn: done
Balance O: MnO₄⁻ → Mn²⁺ + 4H₂O
Balance H: MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O
Balance charge: −1 + 8 = +7 (left); +2 (right). Add 5e⁻ to left:
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Oxidation half: C₂O₄²⁻ → 2CO₂
Balance C: done (2 on each side)
Balance O: done (4 on each side)
Balance charge: −2 (left); 0 (right). Add 2e⁻ to right:
C₂O₄²⁻ → 2CO₂ + 2e⁻

LCM of 5 and 2 = 10:
2 × (MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O)
5 × (C₂O₄²⁻ → 2CO₂ + 2e⁻)

Balanced: 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O

Question 16: Calculate the mass of aluminum deposited when a current of 10.0 A is passed through molten Al₂O₃ for 2 hours. (M_Al = 27 g/mol)

Answer:
Al³⁺ + 3e⁻ → Al, so n = 3
t = 2 hours × 3600 = 7200 s
Q = I × t = 10.0 × 7200 = 72,000 C

m = (M × Q) / (n × F) = (27 × 72000) / (3 × 96500) = 1,944,000 / 289,500 = 6.71 g

Question 17: A current of 2.5 A was passed through a solution of AgNO₃ for 45 minutes. Calculate: (a) the mass of silver deposited, (b) the volume of gas evolved at STP if the same current is passed through dilute H₂SO₄ for the same time. (M_Ag = 108; molar volume at STP = 22.4 L/mol)

Answer:
t = 45 × 60 = 2700 s; Q = 2.5 × 2700 = 6750 C

(a) Silver deposited: Ag⁺ + e⁻ → Ag, n = 1
m = (108 × 6750) / (1 × 96500) = 729000 / 96500 = 7.56 g

(b) Volume of H₂ from dilute H₂SO₄:
At cathode: 2H⁺ + 2e⁻ → H₂ (n = 2)
Moles of electrons = Q / F = 6750 / 96500 = 0.06995 mol
Moles of H₂ = 0.06995 / 2 = 0.03498 mol
Volume of H₂ = 0.03498 × 22.4 = 0.784 L

Question 18: The same quantity of electricity was passed through solutions of CuSO₄ and AgNO₃ in series. If 3.175 g of copper was deposited, calculate: (a) the mass of silver deposited, (b) the mass of zinc that would be deposited from ZnSO₄ by the same quantity. (M_Cu = 63.5, M_Ag = 108, M_Zn = 65)

Answer:
E_Cu = 63.5/2 = 31.75; E_Ag = 108/1 = 108; E_Zn = 65/2 = 32.5

(a) Mass of silver:
m_Ag / m_Cu = E_Ag / E_Cu
m_Ag / 3.175 = 108 / 31.75
m_Ag = 3.175 × (108/31.75) = 3.175 × 3.402 = 10.80 g

(b) Mass of zinc:
m_Zn / m_Cu = E_Zn / E_Cu
m_Zn / 3.175 = 32.5 / 31.75
m_Zn = 3.175 × (32.5/31.75) = 3.175 × 1.0236 = 3.25 g

Question 19: A voltaic cell is constructed using the following half-cells: Ni|Ni²⁺ (E° = −0.25 V) and Pb|Pb²⁺ (E° = −0.13 V). (a) Identify the anode and cathode. (b) Calculate E°_cell. (c) Write the overall cell reaction.

Answer:
(a) Ni has the more negative E° (−0.25 V), so Ni is more easily oxidized. Ni is the anode and Pb is the cathode.

(b) E°_cell = E°_cathode − E°_anode = (−0.13) − (−0.25) = −0.13 + 0.25 = +0.12 V

(c) Overall reaction:
Anode: Ni → Ni²⁺ + 2e⁻
Cathode: Pb²⁺ + 2e⁻ → Pb
Overall: Ni(s) + Pb²⁺(aq) → Ni²⁺(aq) + Pb(s)

Question 20: During the electroplating of a spoon with silver, a current of 0.5 A is passed through AgNO₃ solution for 1 hour. (a) Calculate the mass of silver deposited. (b) If the spoon has a surface area of 50 cm², what is the thickness of the silver coating? (Density of Ag = 10.5 g/cm³; M_Ag = 108)

Answer:
t = 1 × 3600 = 3600 s; Q = 0.5 × 3600 = 1800 C

(a) Mass of silver:
m = (108 × 1800) / (1 × 96500) = 194400 / 96500 = 2.015 g

(b) Thickness of coating:
Volume = mass / density = 2.015 / 10.5 = 0.1919 cm³
Thickness = Volume / surface area = 0.1919 / 50 = 0.00384 cm = 0.0384 mm
The silver coating is approximately 0.038 mm thick.

Question 21: Balance the following in basic medium: CrO₄²⁻ + Fe²⁺ → Cr(OH)₃ + Fe³⁺

Answer:
Reduction half: CrO₄²⁻ → Cr(OH)₃
Balance Cr: done
Balance O: CrO₄²⁻ → Cr(OH)₃ + H₂O (check: 4 O on left; 3+1=4 O on right ✓)
Balance H: CrO₄²⁻ + 5H⁺ → Cr(OH)₃ + H₂O
Balance charge: −2 + 5 = +3 (left); 0 (right). Add 3e⁻ to left:
CrO₄²⁻ + 5H⁺ + 3e⁻ → Cr(OH)₃ + H₂O

Oxidation half: Fe²⁺ → Fe³⁺ + e⁻

LCM of 3 and 1 = 3:
1 × (CrO₄²⁻ + 5H⁺ + 3e⁻ → Cr(OH)₃ + H₂O)
3 × (Fe²⁺ → Fe³⁺ + e⁻)
Add: CrO₄²⁻ + 5H⁺ + 3Fe²⁺ → Cr(OH)₃ + H₂O + 3Fe³⁺

Convert to basic: Add 5OH⁻ to both sides:
CrO₄²⁻ + 5H⁺ + 3Fe²⁺ + 5OH⁻ → Cr(OH)₃ + H₂O + 3Fe³⁺ + 5OH⁻
5H⁺ + 5OH⁻ → 5H₂O. Combine and simplify:
CrO₄²⁻ + 3Fe²⁺ + 5H₂O → Cr(OH)₃ + H₂O + 3Fe³⁺ + 5OH⁻
Cancel 1 H₂O:
CrO₄²⁻ + 3Fe²⁺ + 4H₂O → Cr(OH)₃ + 3Fe³⁺ + 5OH⁻

Question 22: How many minutes are required to produce 5.6 L of H₂ at STP by electrolysis of acidified water using a current of 5 A?

Answer:
Moles of H₂ = 5.6 / 22.4 = 0.25 mol
2H⁺ + 2e⁻ → H₂, so 1 mol H₂ requires 2 mol e⁻ = 2F charge
Total charge needed = 0.25 × 2 × 96500 = 48,250 C
Q = I × t → 48,250 = 5 × t
t = 48,250 / 5 = 9,650 s = 9,650 / 60 = 160.8 minutes ≈ 161 minutes

2.1 Oxidation-Reduction Reactions

2.1.1 Oxidation

2.1.2 Reduction

2.1.3 Balancing Oxidation-Reduction (Redox) Reactions

Method 1: Oxidation Number Method

Method 2: Half-Reaction (Ion-Electron) Method

2.2 Electrolysis of Aqueous Solutions

2.2.1 Electrolytic Cells

2.2.2 Preferential Discharge

2.2.3 Electrolysis of Some Selected Aqueous Solutions

(a) Electrolysis of Dilute H2SO4 (Acidified Water)

(b) Electrolysis of Concentrated NaCl Solution (Brine)

(c) Electrolysis of CuSO4 Solution with Inert Electrodes

(d) Electrolysis of CuSO4 Solution with Copper Electrodes

2.3 Quantitative Aspects of Electrolysis

2.3.1 Faraday’s First Law of Electrolysis

2.3.2 Faraday’s Second Law of Electrolysis

2.4 Industrial Application of Electrolysis

1. Extraction of Metals

2. Electroplating

3. Electrorefining of Metals

4. Chlor-Alkali Industry

2.5 Voltaic Cells

Structure of a Voltaic Cell

The Daniell Cell (Zn-Cu Cell)

Electrode Potential and Cell Potential

Revision Notes — Exam Focus

Oxidation and Reduction at a Glance

Key Formulas

Important Definitions

Electrolytic vs. Voltaic Cells — Quick Comparison

Preferential Discharge — Quick Rules

Standard Electrode Potentials to Remember

Common Mistakes to Avoid

Challenge Exam Questions

Section A: Multiple Choice Questions

Section B: Fill in the Blanks

Section C: Short Answer Questions

Section D: Step-by-Step Calculation Questions

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(a) Electrolysis of Dilute H₂SO₄ (Acidified Water)

(c) Electrolysis of CuSO₄ Solution with Inert Electrodes

(d) Electrolysis of CuSO₄ Solution with Copper Electrodes