Unit notes and practice questions

Grade 12 Chemistry unit 2: Electrochemistry

Everything here is for unit 2 of the Grade 12 Chemistry textbook: what the chapter covers, the short notes, practice questions with their answers, and the lesson video where one exists.

5
Cards: 10
Practice questions on unit 2: 57–135
The textbook pages for unit 2

What unit 2 covers

1.Oxidation-Reduction Reactions
2.Oxidation
3.Reduction
4.Balancing Oxidation-Reduction (Redox) Reactions
5.Electrolysis of Aqueous Solutions
6.Electrolytic Cells
7.Preferential Discharge
8.Electrolysis of Some Selected Aqueous Solutions
9.Quantitative Aspects of Electrolysis
10.Faraday's First Law of Electrolysis
11.Faraday's Second Law of Electrolysis
12.Industrial Application of Electrolysis
13.Volatic Cells

Short notes for unit 2

The cards filed under this chapter on the Grade 12 Chemistry page. Open one at a time.

Balancing Redox Reactions in Basic MediumMethod

When you use it

Use when balancing oxidation-reduction equations occurring in basic or alkaline solution using the half-reaction method.

Watch out

First balance fully as if in acidic medium using H+ and H2O, then add equal numbers of OH- to BOTH sides for every H+, and combine H+ and OH- into H2O.

Drafted from Grade 12 Chemistry, pages 57-135, then checked twice before it went up

Electrolysis Products of Aqueous SolutionsReference tableDilute NaCl (inert) · Concentrated NaCl / Brine (inert) · Dilute H_2SO_4 (inert) · +2

Electrolyte and Electrodes	Cathode Product	Anode Product	Net Result
Dilute $\text{NaCl}$ (inert)	$\text{H}_2\text{ (g)}$	$\text{O}_2\text{ (g)}$	Decomposition of water
Concentrated $\text{NaCl}$ / Brine (inert)	$\text{H}_2\text{ (g)}$	$\text{Cl}_2\text{ (g)}$	$\text{NaOH}$ remains in solution
Dilute $\text{H}_2\text{SO}_4$ (inert)	$\text{H}_2\text{ (g)}$	$\text{O}_2\text{ (g)}$	Decomposition of water
$\text{CuSO}_4$ with inert electrodes (Pt)	$\text{Cu (s)}$	$\text{O}_2\text{ (g)}$	Solution becomes acidic ( $\text{H}_2\text{SO}_4$ )
$\text{CuSO}_4$ with active $\text{Cu}$ electrodes	$\text{Cu (s)}$ deposited	$\text{Cu}^{2+}\text{ (aq)}$ dissolved	$\text{CuSO}_4$ concentration unchanged

When you use it

Use to determine the products liberated at the cathode and anode based on ion concentrations and electrode activity.

Watch out

Active copper anodes dissolve into Cu2+ ions instead of oxidizing water or hydroxide ions to oxygen gas.

Drafted from Grade 12 Chemistry, pages 57-135, then checked twice before it went up

Faraday First Law of ElectrolysisFormula

m = \frac{M I t}{n F}

m = \frac{M I t}{n F}

$m$: mass of substance deposited or liberated $\text{g}$
$M$: molar mass of the substance $\text{g\,mol}^{-1}$
$I$: electric current $\text{A}$
$t$: time of electrolysis $\text{s}$
$n$: valence (number of electrons transferred per ion) $1$
$F$: Faraday constant (96,500 C/mol) $\text{C\,mol}^{-1}$

When you use it

Use to calculate the mass of a substance deposited or liberated at an electrode during electrolysis given current and time.

Watch out

Time t must be converted to seconds (s) before calculating, not left in minutes or hours.

Drafted from Grade 12 Chemistry, pages 57-135, then checked twice before it went up

Faraday Second Law of ElectrolysisFormula

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

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

$m_1$: mass of first substance $\text{g}$
$E_1$: equivalent weight of first substance (M/n) $\text{g\,eq}^{-1}$
$m_2$: mass of second substance $\text{g}$
$E_2$: equivalent weight of second substance (M/n) $\text{g\,eq}^{-1}$

When you use it

Use when the same quantity of electric current passes through two or more electrolytic cells connected in series.

Watch out

Equivalent weight E equals molar mass divided by valence (n). Do not confuse valence with the formula subscript of the salt.

Drafted from Grade 12 Chemistry, pages 57-135, then checked twice before it went up

Metallic versus Electrolytic ConductionReference tableCharge carriers · Matter transfer · Chemical change · +1

Property	Metallic Conduction	Electrolytic Conduction
Charge carriers	Delocalized mobile electrons	Mobile cations and anions
Matter transfer	No transfer of matter	Actual transfer of matter
Chemical change	No chemical decomposition	Accompanied by chemical decomposition
Effect of temperature	Conductivity decreases as temperature rises	Conductivity increases as temperature rises

When you use it

Use to distinguish how electrical charge moves through metals compared to molten or aqueous electrolytes.

Watch out

Remember that raising temperature increases electrolytic conductivity due to higher ion mobility, but decreases metallic conductivity due to lattice vibrations.

Drafted from Grade 12 Chemistry, pages 57-135, then checked twice before it went up

Practice questions on unit 2

10 multiple-choice questions written against this chapter, each with its answer and the reason. Try one before you open the answer.

Question 1

In electrochemistry, what happens to an atom or ion during an oxidation process?

A.It loses electrons and its oxidation number increases.
B.It gains electrons and its oxidation number increases.
C.It loses electrons and its oxidation number decreases.
D.It gains electrons and its oxidation number decreases.

Show the answer and why

Answer: A It loses electrons and its oxidation number increases.

Why

Oxidation is defined as a process in which an atom or element loses electrons, resulting in an increase in its oxidation number.

Question 2

Why does the electrical conductivity of a metallic conductor decrease as the temperature increases?

A.The metal atoms lose their valence electrons to the air.
B.Enhanced thermal vibration of metal atoms disrupts the movement of electrons.
C.The metal melts and stops conducting current.
D.The concentration of mobile positive ions decreases rapidly.

Show the answer and why

Answer: B Enhanced thermal vibration of metal atoms disrupts the movement of electrons.

Why

In metallic conduction, conductivity decreases with increasing temperature because the enhanced thermal vibrations of the metal atoms disrupt the smooth movement of free electrons passing through them.

Question 3

According to Faraday's first law of electrolysis, the mass of a substance deposited at an electrode is directly proportional to:

A.the volume of the electrolytic container
B.the quantity of electricity passed through the electrolyte
C.the atmospheric pressure above the solution
D.the total mass of the solvent used

Show the answer and why

Answer: B the quantity of electricity passed through the electrolyte

Why

Faraday's first law states that the mass (m) of a substance deposited or dissolved at an electrode is directly proportional to the quantity of electricity (Q) passed through the electrolyte (m ∝ Q).

Question 4

During the industrial electrolytic refining of copper, which material is used as the cathode?

A.A block of impure copper
B.A thin strip of pure copper
C.A carbon graphite plate
D.An iron nail

Show the answer and why

Answer: B A thin strip of pure copper

Why

In electrorefining, the impure copper is made the anode (which dissolves), and a thin strip of pure copper is made the cathode (where pure copper deposits).

Question 5

What products are collected at the cathode and anode during the electrolysis of dilute sulfuric acid (H₂SO₄)?

A.Cathode: H2 gas; Anode: O2 gas
B.Cathode: O2 gas; Anode: H2 gas
C.Cathode: SO2 gas; Anode: H2 gas
D.Cathode: H2 gas; Anode: SO3 gas

Show the answer and why

Answer: A Cathode: H2 gas; Anode: O2 gas

Why

In dilute H₂SO₄, H⁺ ions are reduced at the cathode to produce H₂ gas, while OH^- ions are preferentially oxidized at the anode to produce O₂ gas and water.

Question 6

During the electrolysis of concentrated sodium chloride solution (brine), chlorine gas is produced at the anode instead of oxygen gas. Why does this happen?

A.Sodium ions neutralize the hydroxide ions at the anode.
B.The high concentration of chloride ions favors their discharge over hydroxide ions.
C.Chloride ions have a much lower mobility than water molecules.
D.Hydroxide ions cannot migrate to a positive electrode.

Show the answer and why

Answer: B The high concentration of chloride ions favors their discharge over hydroxide ions.

Why

The concentration of an ion affects preferential discharge. In concentrated NaCl (brine), the concentration of Cl^- is much higher than OH^-, causing Cl^- to be discharged preferentially to form Cl₂ gas.

Question 7

When balancing the reduction half-reaction MnO₄^- → MnO₂ in an acidic medium, how many electrons (e^-) must be added to the reactant side?

A.1 electron
B.2 electrons
C.3 electrons
D.5 electrons

Show the answer and why

Answer: C 3 electrons

Why

Balancing atoms gives: MnO₄^- + 4H⁺ → MnO₂ + 2H₂O. The left side has a charge of +3 and the right side is neutral (0), so 3 electrons (3e^-) must be added to the reactant side.

Question 8

A student in Hawassa wants to electroplate an iron key with copper metal. How should the student connect the electrodes to a DC power source?

A.Connect the iron key to the negative terminal (cathode) and the copper plate to the positive terminal (anode).
B.Connect the iron key to the positive terminal (anode) and the copper plate to the negative terminal (cathode).
C.Connect both the iron key and the copper plate to the negative terminal.
D.Connect the iron key to the negative terminal and use a graphite rod as the cathode.

Show the answer and why

Answer: A Connect the iron key to the negative terminal (cathode) and the copper plate to the positive terminal (anode).

Why

In electroplating, the object to be plated (the iron key) must be the cathode (negative terminal), and the metal to be deposited (copper plate) must be the anode (positive terminal).

Question 9

In a workshop in Addis Ababa, a current of 0.452 A is passed through molten CaCl₂ for 1.5 hours (5,400 seconds). What mass of calcium metal is deposited at the cathode? (Use: 1 F = 96,500 C, molar mass of Ca = 40.08 g/mol)

A.0.507 g
B.1.014 g
C.0.253 g
D.2.440 g

Show the answer and why

Answer: A 0.507 g

Why

First, find the total charge: Q = I × t = 0.452 A × 5400 s = 2440.8 C.
Moles of electrons: 2440.8 C / 96,500 C/mol = 0.0253 mol e^-.
Since reduction of Ca²⁺ requires 2 moles of electrons per mole of Ca: mass = (0.0253 / 2) × 40.08 g = 0.507 g.

Question 10

Two electrolytic cells containing solutions of CuSO₄ and AgNO₃ are connected in series. When 0.108 g of Ag is deposited in the first cell, what mass of copper is deposited in the second cell? (Equivalent mass of Ag = 108 g/eq, Equivalent mass of Cu = 31.75 g/eq)