# Remidi Kimia Kelas Xii.IPA. Redoks

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Quizzes Created: 1 | Total Attempts: 352
Questions: 20 | Attempts: 352

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• 1.

### Jika ke dalam 162 gram air (Mr= 18) dilarutkan 60 gram asam cuka ( Mr = 60) maka fraksi mol zat terlarut dan pelarutnya berturut-turut adalah...

• A.

0,05 dan 0,95

• B.

0,5 dan 0,5

• C.

0,1 dan 0,9

• D.

0,2 dan 0,8

• E.

0,3 dan 0,7

C. 0,1 dan 0,9
Explanation
The answer is 0.1 and 0.9 because the question asks for the mole fraction of the solute and the solvent. To calculate the mole fraction, we need to find the moles of each component. The moles of water can be calculated by dividing the mass of water by its molar mass (162 g / 18 g/mol = 9 mol). The moles of acetic acid can be calculated by dividing the mass of acetic acid by its molar mass (60 g / 60 g/mol = 1 mol). The mole fraction of the solute (acetic acid) is calculated by dividing the moles of acetic acid by the total moles (1 mol / 10 mol = 0.1). The mole fraction of the solvent (water) is calculated by dividing the moles of water by the total moles (9 mol / 10 mol = 0.9).

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• 2.

### Molaritas larutan 98 % H2SO4 (Mr=98, MJ= 1,8 g/mL) adalah....

• A.

6,4 M

• B.

8,3 M

• C.

16,0 M

• D.

17,5 M

• E.

18,0 M

E. 18,0 M
Explanation
The molarity of a solution is calculated by dividing the number of moles of solute by the volume of the solution in liters. In this case, the molar mass of H2SO4 is 98 g/mol. Since the solution is 98% H2SO4, it means that 98 g of H2SO4 is present in 100 g of solution. To calculate the moles of H2SO4, we divide the mass by the molar mass: 98 g / 98 g/mol = 1 mol. The density of the solution is given as 1.8 g/mL, so the volume of the solution containing 1 mol of H2SO4 is 1 mol / 1.8 g/mL = 0.556 L. Finally, we divide the moles by the volume to get the molarity: 1 mol / 0.556 L ≈ 1.80 M.

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• 3.

### Di antara kelima larutan 0,1 m di bawah ini yang mempunyai titik didih paling rendah pada tekanan 1 atm adalah . . . .

• A.

CO(NH2)2

• B.

KCl

• C.

MgCl2

• D.

NaCl

• E.

FeCl3

A. CO(NH2)2
Explanation
CO(NH2)2 is a compound known as urea. Urea is a non-electrolyte, meaning it does not dissociate into ions when dissolved in water. The presence of ions in a solution can increase the boiling point of the solution. Since urea does not dissociate into ions, it will have a lower boiling point compared to the other compounds listed, which are all ionic compounds and will dissociate into ions when dissolved in water. Therefore, CO(NH2)2 will have the lowest boiling point among the given options.

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• 4.

### Tekanan osmotik larutan nonelektrolit 0,50 M pada temperatur 0°C adalah . . . Atm ( R=0,082)

• A.

0,224

• B.

2,24

• C.

4,48

• D.

11,19

• E.

22,4

D. 11,19
Explanation
The correct answer is 11.19. Osmotic pressure is directly proportional to the concentration of the solute. In this case, the concentration of the nonelectrolyte solution is given as 0.50 M. Using the formula for osmotic pressure (π = MRT), where M is the molarity, R is the ideal gas constant, and T is the temperature in Kelvin, we can calculate the osmotic pressure. Plugging in the given values (M = 0.50, R = 0.082, T = 273), we get π = 0.50 * 0.082 * 273 = 11.19 atm.

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• 5.

### Titik didih larutan NaCl 2 m memiliki titik didih sebesar …….°C ( Kb = 0,52)

• A.

102,08

• B.

101,04

• C.

100,52

• D.

100,104

• E.

100,052

A. 102,08
Explanation
The boiling point of a solution is higher than that of the pure solvent due to the presence of solute particles. This phenomenon is known as boiling point elevation. The boiling point elevation is directly proportional to the molality of the solution. In this case, the molality of the NaCl solution is 2 m, and the boiling point elevation constant (Kb) is 0.52. By using the formula ΔTb = Kb * m, where ΔTb is the boiling point elevation and m is the molality, we can calculate the boiling point elevation to be 1.04°C. Adding this to the boiling point of pure water (100°C) gives us a boiling point of 101.04°C. Therefore, the correct answer is 101.04°C.

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• 6.

### Larutan isotonis adalah larutan yang memiliki ….

• A.

Titik didih sama

• B.

Titik beku sama

• C.

Tekanan osmotik sama

• D.

Konsentrasi sama

• E.

Tekanan uap sama

C. Tekanan osmotik sama
Explanation
Larutan isotonis adalah larutan yang memiliki tekanan osmotik sama. Tekanan osmotik adalah tekanan yang diberikan oleh partikel-partikel terlarut dalam larutan untuk menarik air masuk ke dalam larutan melalui membran semipermeabel. Dalam larutan isotonis, konsentrasi partikel-partikel terlarut di dalam larutan sama dengan konsentrasi partikel-partikel terlarut di dalam sel. Hal ini menyebabkan tekanan osmotik di dalam larutan isotonis sama dengan tekanan osmotik di dalam sel, sehingga tidak ada pergerakan air masuk atau keluar sel.

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• 7.

### Larutan urea 1 m akan membeku pada suhu …. ( kf = 1,86)

• A.

-5,58

• B.

-2,79

• C.

-2,5

• D.

-1,86

• E.

-1,68

D. -1,86
Explanation
The correct answer is -1,86. Urea is a solute that lowers the freezing point of a solvent. The freezing point depression is proportional to the molality of the urea solution. The equation for freezing point depression is ΔT = kf * m, where ΔT is the change in freezing point, kf is the cryoscopic constant, and m is the molality of the solution. In this case, the molality is 1 m, and kf is given as 1.86. Therefore, the freezing point depression is -1.86 degrees Celsius, which means that the solution will freeze at a temperature that is 1.86 degrees Celsius lower than the freezing point of the pure solvent.

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• 8.

### Larutan memiliki … dibandingkan pelarutnya.

• A.

Tekanan uap yang lebih tinggi

• B.

Titik didih lebih rendah

• C.

Titik beku yang lebih rendah

• D.

Tekanan osmotik yang sama

• E.

Konsentrasi yang lebih rendah

C. Titik beku yang lebih rendah
Explanation
Larutan memiliki titik beku yang lebih rendah dibandingkan pelarutnya karena adanya zat terlarut dalam larutan yang mengganggu ikatan antar molekul pelarut. Hal ini menyebabkan pelarut membutuhkan energi yang lebih tinggi untuk membentuk ikatan antar molekul dan mengubah fase menjadi padat. Sehingga, titik beku pelarut menjadi lebih rendah ketika terdapat zat terlarut dalam larutan.

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• 9.

### Diketahui reaksi redoks: aCu + bH+ + cNO3– → dCu2+ + H2O + eNO2. Setelah reaksi disetarakan maka harga a, b, c, d, dan e berturut-turut . . . .

• A.

1 – 4 – 2 – 1 – 2

• B.

2 – 1 – 4 – 1 – 2

• C.

1 – 2 – 4 – 2 – 1

• D.

1 – 4 – 1 – 2 – 1

• E.

2 – 1 – 1 – 2 – 1

A. 1 – 4 – 2 – 1 – 2
Explanation
The correct answer is 1 - 4 - 2 - 1 - 2. In the balanced redox reaction, the coefficient of Cu is 1, the coefficient of H+ is 4, the coefficient of NO3- is 2, the coefficient of Cu2+ is 1, the coefficient of H2O is 2, and the coefficient of NO2 is 2.

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• 10.

• A.

Pt dan C

• B.

Zn dan C

• C.

Pt dan PbO2

• D.

Zn dan Cu

• E.

Pb dan PbO2

B. Zn dan C
Explanation
The correct answer is Zn dan C. Zinc (Zn) is commonly used as the anode (negative electrode) in batteries, while carbon (C) is used as the cathode (positive electrode). This combination allows for the flow of electrons between the anode and cathode, creating a current and producing energy.

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• 11.

### Logam yang dapat memberikan perlindungan katodik pada besi (Fe) adalah ….

• A.

Ni

• B.

Sn

• C.

Pb

• D.

Mg

• E.

Cu

D. Mg
Explanation
Mg (magnesium) dapat memberikan perlindungan katodik pada besi (Fe) karena memiliki potensial elektrode yang lebih negatif daripada besi. Ketika magnesium dan besi terhubung dalam larutan elektrolit, magnesium akan berkorosi lebih cepat daripada besi, sehingga magnesium berfungsi sebagai anode dan besi sebagai katode. Proses ini disebut perlindungan katodik, di mana magnesium akan mengorbankan dirinya sendiri untuk melindungi besi dari korosi.

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• 12.

### Pada reaksi redoks berikut, Mg + H2SO4 → MgSO4 + H2 , Oksidator reaksi tersebut adalah. . . .

• A.

Mg

• B.

H2SO4

• C.

MgSO4

• D.

H2

• E.

2H+

B. H2SO4
Explanation
In the given redox reaction, Mg is being oxidized from its elemental form to Mg2+ in MgSO4. The oxidizing agent is the substance that causes the oxidation, which in this case is H2SO4. H2SO4 donates protons (H+) to Mg, causing it to lose electrons and undergo oxidation. Therefore, H2SO4 is the oxidizing agent in this reaction.

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• 13.

### Pada reaksi berikut ini.NaOH + HCl → NaCl + H2O, Senyawa yang mengalami reaksi oksidasi adalah….

• A.

NaOH

• B.

HCl

• C.

NaCl

• D.

H2O

• E.

Tidak satu pun karena bukan reaksi redoks

E. Tidak satu pun karena bukan reaksi redoks
Explanation
In the given reaction, NaOH and HCl are both strong electrolytes that dissociate completely in water to form Na+ and OH- ions, and H+ and Cl- ions respectively. The reaction between NaOH and HCl is a neutralization reaction where the H+ ion from HCl combines with the OH- ion from NaOH to form water (H2O) and the Na+ ion from NaOH combines with the Cl- ion from HCl to form NaCl. There is no change in oxidation states of any elements in this reaction, therefore none of the compounds undergo oxidation. Hence, the correct answer is "tidak satu pun karena bukan reaksi redoks" which translates to "none of them because it is not a redox reaction".

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• 14.

### Zat yang bertindak sebagai pereduksi (reduktor) pada reaksi redoks berikut ini: CuSO4 + Zn → ZnSO4 + Cu adalah ….

• A.

Zn

• B.

CuSO4

• C.

Cu

• D.

ZnSO4

• E.

S

A. Zn
Explanation
In the given redox reaction, CuSO4 + Zn → ZnSO4 + Cu, Zn acts as the reducing agent. This is because Zn is being oxidized, as it loses electrons and forms Zn2+ ions, while Cu2+ ions in CuSO4 are being reduced to form Cu atoms. Therefore, Zn is causing the reduction of Cu2+ ions and is considered the reducing agent in this reaction.

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• 15.

### Diketahui : Zn2+ / Zn = –0,76 V; Pb2+ / Pb = –0,13 V; Cu2+ / Cu = +0,34 V; Ag+ / Ag = +0,80 V. Reaksi sel yang menghasilkan potensial terbesar adalah ….

• A.

Zn(s) / Zn2+(aq) // Cu2+(aq) / Cu(s)

• B.

Pb(s) / Pb2+(aq) // Ag2+(aq) / Ag(s)

• C.

Zn(s) / Zn2+(aq) // Ag2+(aq) / Ag(s)

• D.

Pb(s) / Pb2+(aq) // Cu2+(aq) / Cu(s)

• E.

Zn(s) / Zn2+(aq) // Pb2+(aq) / Pb(s)

C. Zn(s) / Zn2+(aq) // Ag2+(aq) / Ag(s)
Explanation
The given answer is "Zn(s) / Zn2+(aq) // Ag2+(aq) / Ag(s)". This is because the reaction involving zinc (Zn) and silver (Ag) has the largest potential difference (0.80 V) compared to the other reactions. The potential difference is a measure of the tendency of a reaction to occur, with a higher potential difference indicating a greater tendency. Therefore, the reaction Zn(s) / Zn2+(aq) // Ag2+(aq) / Ag(s) will result in the largest potential.

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• 16.

### Berdasarkan deret Volta, reduktor yang lebih kuat dari Mn adalah ….

• A.

Zn

• B.

Mg

• C.

Fe

• D.

Ni

• E.

Pb

B. Mg
Explanation
Berdasarkan deret Volta, reduktor yang lebih kuat dari Mn adalah Mg. Deret Volta adalah urutan logam berdasarkan kemampuan mereka untuk mengurangi ion logam lain dalam larutan. Semakin tinggi suatu logam dalam deret Volta, semakin kuat reduktor itu. Dalam deret Volta, Mg berada di atas Mn, sehingga Mg lebih kuat sebagai reduktor daripada Mn.

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• 17.

### Jika leburan garam NaCl dielektrolisis dengan elektrode C, pada katode akan terbentuk ….

• A.

Gas clor

• B.

Gas oksigen

• C.

Logam Na

• D.

Uap air

• E.

Gas hidrogen

C. Logam Na
Explanation
When NaCl is electrolyzed with electrode C, sodium metal (Na) will be formed at the cathode. This is because during electrolysis, positively charged sodium ions (Na+) are attracted to the negatively charged cathode, where they gain electrons and are reduced to form sodium metal.

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• 18.

### Waktu yang diperlukan untuk melapisi suatu permukaan besi dengan 3,27 gram seng dalam larutan ion seng ( Zn2+) yang dialirkan arus listrik sebesar 5 ampere (Ar Zn = 65,4) adalah ….

• A.

3620 detik

• B.

1930 detik

• C.

1800 detik

• D.

1720 detik

• E.

1600 detik

B. 1930 detik
Explanation
The time required to coat an iron surface with 3.27 grams of zinc in a zinc ion solution with a current of 5 amperes can be calculated using Faraday's law of electrolysis. The formula for this is:

Time = (Mass of substance / (Current × Faraday's constant)) × 3600

Plugging in the values given in the question, we get:

Time = (3.27 / (5 × 96500)) × 3600 = 1930 seconds

Therefore, the correct answer is 1930 seconds.

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• 19.

### Muatan listrik yang diperlukan untuk mereduksi ion magnesiun (Mg2+) menjadi 12 gram logam magnesiumadalah . . . . (Ar Mg = 24)

• A.

1,0 F

• B.

1,5 F

• C.

2,0 F

• D.

3,0 F

• E.

4,0 F

A. 1,0 F
Explanation
The correct answer is 1,0 F. The question asks for the amount of electric charge required to reduce 12 grams of magnesium ions (Mg2+) to metallic magnesium. The atomic mass of magnesium is 24 g/mol, so 12 grams of magnesium corresponds to 0.5 moles. Since each magnesium ion gains two electrons during reduction, the total charge required is 2 * 0.5 = 1 F (Faraday).

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• 20.

### Listrik sebesar 9650 C digunakan untuk mereduksi lugam Ag+ menjadi Ag, maka perak yang dihasilkan … ( Ar Ag= 108)

• A.

108 gram

• B.

54 gram

• C.

22,4 gram

• D.

10,8 gram

• E.

5,4 gram

C. 22,4 gram
Explanation
The given question states that 9650 C of electricity is used to reduce Ag+ ions to Ag. The molar mass of Ag is given as 108 g/mol. To calculate the amount of silver produced, we can use Faraday's law of electrolysis which states that the amount of substance produced is directly proportional to the charge passed through the electrolyte. Since 1 mole of electrons carries a charge of 96500 C, we can calculate that 9650 C of charge corresponds to 0.1 mole of electrons. Since the reduction of 1 mole of Ag+ ions produces 1 mole of Ag, the amount of silver produced is 0.1 mole or 22.4 grams.

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• Mar 21, 2023
Quiz Edited by
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• Oct 31, 2018
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