Yaadannoo boqonnaa fi gaaffii shaakalaa

Boqonnaa 2 Keemistirii kutaa 12: Elektirookeemistirii

Wanti fuula kana irra jiru hundi boqonnaa 2 kitaaba Keemistirii kutaa 12 tiif: wanta boqonnaan haguugu, yaadannoo gabaabaa, gaaffii shaakalaa deebii isaanii waliin, fi viidiyoo barnootaa bakka jirutti.

5
Kaardii: 10
Gaaffii shaakalaa boqonnaa 2: 57–135
Fuula kitaabaa boqonnaa 2

Boqonnaan 2 maal haguuga

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

Yaadannoo gabaabaa boqonnaa 2

Kaardiiwwan fuula Keemistirii kutaa 12 irratti boqonnaa kana jala galmaa'an. Tokko tokkoon bani.

Balancing Redox Reactions in Basic MediumMala

Yoom akka itti fayyadamtu

Walqixxoo redox kan furmaata basic yookiin alkaline keessatti gaggeeffamu half-reaction method fayyadamuun madaaluuf itti fayyadamaa.

Of eeggadhu

Jalqaba akka waan acidic medium keessa jiruutti H+ fi H2O fayyadamuun madaalaa, sana booda lakkoofsa H+ jiruuf OH- wal-qixa gara lamaanitti ida'uun gara H2O tti jijjiiraa.

Grade 12 Chemistry, fuula 57-135 irraa qophaa'e; osoo hin ba'in dura yeroo lama mirkanaa'e

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

Electrolyte fi Electrodes	Cathode Product	Anode Product	Bu'aa
Dilute $\text{NaCl}$ (inert)	$\text{H}_2\text{ (g)}$	$\text{O}_2\text{ (g)}$	Decomposition bishaanii
Concentrated $\text{NaCl}$ / Brine (inert)	$\text{H}_2\text{ (g)}$	$\text{Cl}_2\text{ (g)}$	$\text{NaOH}$ solution keessatti hafa
Dilute $\text{H}_2\text{SO}_4$ (inert)	$\text{H}_2\text{ (g)}$	$\text{O}_2\text{ (g)}$	Decomposition bishaanii
$\text{CuSO}_4$ inert electrodes (Pt) wajjin	$\text{Cu (s)}$	$\text{O}_2\text{ (g)}$	Solution acidic ta'a ( $\text{H}_2\text{SO}_4$ )
$\text{CuSO}_4$ active $\text{Cu}$ electrodes wajjin	$\text{Cu (s)}$ deposited ta'a	$\text{Cu}^{2+}\text{ (aq)}$ dissolved ta'a	Concentration $\text{CuSO}_4$ hin jijjiiramu

Yoom akka itti fayyadamtu

Yeroo aqueous electrolysis bu'aawwan cathode fi anode irratti uumaman concentration fi amala electrode irratti hundaa'uun beekuuf itti fayyadamaa.

Of eeggadhu

Active copper anode yoo jiraate anodichi ofumaa gara Cu2+ tti baqa malee oxygen gas hin uumu.

Grade 12 Chemistry, fuula 57-135 irraa qophaa'e; osoo hin ba'in dura yeroo lama mirkanaa'e

Faraday First Law of ElectrolysisFoormulaa

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

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

$m$: mass meeshaa kuufame yookiin bahe $\text{g}$
$M$: molar mass meeshichaa $\text{g\,mol}^{-1}$
$I$: yaa'a elektiriikii $\text{A}$
$t$: yeroo electrolysis $\text{s}$
$n$: valence (baay'ina electron ion tokkoon darbu) $1$
$F$: Dhaabbataa Faraday (96,500 C/mol) $\text{C\,mol}^{-1}$

Yoom akka itti fayyadamtu

Yeroo electrolysis mass meeshaa electrode irratti kuufame yookiin bahe electric current fi time fayyadamuun shallaguuf itti fayyadamaa.

Of eeggadhu

Shallaggii dura time t gara seconds (s) tti jijjiiramuu qaba malee daqiiqaa yookiin sa'aatiidhaan dhiifamuu hin qabu.

Grade 12 Chemistry, fuula 57-135 irraa qophaa'e; osoo hin ba'in dura yeroo lama mirkanaa'e

Faraday Second Law of ElectrolysisFoormulaa

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

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

$m_1$: mass meeshaa jalqabaa $\text{g}$
$E_1$: equivalent weight meeshaa jalqabaa (M/n) $\text{g\,eq}^{-1}$
$m_2$: mass meeshaa lammaffaa $\text{g}$
$E_2$: equivalent weight meeshaa lammaffaa (M/n) $\text{g\,eq}^{-1}$

Yoom akka itti fayyadamtu

Yeroo hamma electric charge wal-fakkaataan electrolytic cells walitti hidhamiinsa series qaban keessa darbu itti fayyadamaa.

Of eeggadhu

Equivalent weight E jechuun molar mass qoodu valence (n) dha. Valence lakkoofsa subscript soogiddaa wajjin walitti hin madaalinaa.

Grade 12 Chemistry, fuula 57-135 irraa qophaa'e; osoo hin ba'in dura yeroo lama mirkanaa'e

Metallic versus Electrolytic ConductionGabatee wabiiBaattuuwwan charge · Dabarsa matter · Chemical change · +1

Amala	Metallic Conduction	Electrolytic Conduction
Baattuuwwan charge	Electrons socho'oo delocalized ta'an	Cations fi anions socho'oo
Dabarsa matter	Dabarsi matter hin jiru	Dabarsa matter qabatamaatu jira
Chemical change	Chemical decomposition hin gaggeeffamu	Chemical decomposition waliin kan gaggeeffamu
Dhiibbaa temperature	Yeroo temperature dabalu conductivityn ni hir'ata	Yeroo temperature dabalu conductivityn ni dabala

Yoom akka itti fayyadamtu

Akkaataa electric charge metals fi electrolytes dhangala'aa keessa darbu gargar baasuuf itti fayyadamaa.

Of eeggadhu

Ho'i dabaluun ionic mobility waan dabaluuf electrolytic conductivity ni dabala, garuu metallic conductivity ni hir'isa.

Grade 12 Chemistry, fuula 57-135 irraa qophaa'e; osoo hin ba'in dura yeroo lama mirkanaa'e

Gaaffii shaakalaa boqonnaa 2

Gaaffii filannoo 10 boqonnaa kana irratti barreeffaman, tokkoon tokkoon isaanii deebii fi sababa qabu. Deebii banuu dura yaali. Gaaffiin kun Ingiliffaa tiin.

Gaaffii 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.

Deebii fi sababa agarsiisi

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

Sababa

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

Gaaffii 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.

Deebii fi sababa agarsiisi

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

Sababa

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.

Gaaffii 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

Deebii fi sababa agarsiisi

Deebii: B the quantity of electricity passed through the electrolyte

Sababa

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).

Gaaffii 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

Deebii fi sababa agarsiisi

Deebii: B A thin strip of pure copper

Sababa

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).

Gaaffii 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

Deebii fi sababa agarsiisi

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

Sababa

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.

Gaaffii 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.

Deebii fi sababa agarsiisi

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

Sababa

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.

Gaaffii 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

Deebii fi sababa agarsiisi

Deebii: C 3 electrons

Sababa

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.

Gaaffii 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.

Deebii fi sababa agarsiisi

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

Sababa

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).

Gaaffii 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

Deebii fi sababa agarsiisi

Deebii: A 0.507 g

Sababa

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.

Gaaffii 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)

A.0.032 g
B.0.064 g
C.0.108 g
D.0.367 g

Deebii fi sababa agarsiisi

Deebii: A 0.032 g

Sababa

According to Faraday's second law: m_Cu / E_Cu = m_Ag / E_Ag.
m_Cu = (E_Cu × m_Ag) / E_Ag = (31.75 × 0.108) / 108 = 0.032 g.

Appiin boqonnaa kana irratti gaaffii 12 qaba; utuu hojjettuu ni sirreessa. Kan asii jiran 10 jalqabaa ti.

Fuula kitaabaa boqonnaa 2

Boqonnaan 2 Maxxansa Ministeera Barnootaa keessatti fuula 57 hanga fuula 135 ti.

Maxxansa Ministeera Barnootaa, Ingiliffaa, Fuula 298

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Yaadannoo gabaabaa Keemistirii kutaa 12 hunda

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