Question:

Depict the galvanic cell in which the reaction Zn(s)+2Ag+(aq)→Zn2+(aq)+2Ag(s) takes place. Further show the carriers of the current in the inner and outer circuit of the cell.

Answer:

1. The galvanic cell for the reaction Zn(s)+2Ag+(aq)→Zn2+(aq)+2Ag(s) is shown below:

2. The anode of the cell is Zn metal and the cathode is 2Ag+ ions in aqueous solution.

3. The carriers of the current in the inner circuit of the cell are electrons which flow from the anode (Zn) to the cathode (2Ag+).

4. The carriers of the current in the outer circuit of the cell are the ions (Zn2+ and 2Ag+) which flow from the cathode (2Ag+) to the anode (Zn).

Question:

Depict the galvanic cell in which the reaction Zn(s)+2Ag+(aq)→Zn2+(aq)+2Ag(s) takes place. Further show which of the electrode is negatively charged.

Answer:

1. The galvanic cell for the reaction Zn(s)+2Ag+(aq)→Zn2+(aq)+2Ag(s) will be represented as:

Zn (s) | Zn2+ (aq) || Ag+ (aq) | Ag (s)

2. The electrode on the left side of the cell is negatively charged, which is the Zn (s) electrode, and the electrode on the right side of the cell is positively charged, which is the Ag (s) electrode.

Question:

XeO64−(aq)+2F−(aq)+6H+(aq)→XeO3(g)+3H2O(l) What conclusion about the compound Na4XeO6(of which XeO^(4−)6 is a part) can be drawn from the reaction?

Answer:

1. XeO64−(aq) is a part of the compound Na4XeO6.

2. The reaction of XeO64−(aq) with 2F−(aq) and 6H+(aq) produces XeO3(g) and 3H2O(l).

3. Therefore, the conclusion that can be drawn from the reaction is that Na4XeO6 undergoes a redox reaction when exposed to 2F−(aq) and 6H+(aq).

Question:

From the following reactions, determine if Ag+ is a stronger oxidizing agent than Cu2+. (a) H3PO2(aq)+4AgNO3(aq)+2H2O(l)→H3PO(aq)+4Ag(s)+4HNO3(aq) (b) H3PO2(aq)+2CuSO4(aq)+2H2O(l)→H3PO4(aq)+2Cu(s)+H2SO4(aq) (c) C6H5CHO(l)+2[Ag(NH3)2]+(aq)+3OH−(aq)→C6H5COO−(aq) +2Ag(s)+4NH3(aq)+2H2O(l) (d) C6H5CHO(l)+2Cu2+(aq)+5OH−(aq)→ No Change observed.

Answer:

Answer: Yes, Ag+ is a stronger oxidizing agent than Cu2+. This can be seen from the reaction products in (a), (c), and (d). In reaction (a), Ag+ is oxidized to Ag(s). In reaction (c), Ag+ is oxidized to Ag(s) and C6H5CHO(l) is oxidized to C6H5COO−(aq). In reaction (d), no change is observed, indicating that Cu2+ is not a strong enough oxidizing agent to oxidize C6H5CHO(l).

Question:

Write formulas for the following compounds. Thallium (I) sulphate

Answer:

Formula: TlSO4

Question:

Identify the substance oxidised, reduced, oxidising agent and reducing agent for the following reaction- 2AgBr(s)+C6H6O2(aq)→2Ag(s)+2HBr(aq)+C6H4O2(aq)

Answer:

Substance oxidised: AgBr Substance reduced: C6H6O2 Oxidising agent: C6H6O2 Reducing agent: AgBr

Question:

The Mn3+ ion is unstable in solution and undergoes disproportionation to give Mn2+, MnO2 and H+ ion. Write a balanced ionic equation for the reaction.

Answer:

Mn3+ + H2O → Mn2+ + MnO2 + H+

Question:

Balance the following redox reactions by ion electron method. Cr2O72−+SO2(g)→Cr3+(aq)+SO42−(aq) (in acidic solution)

Answer:

Answer:

Step 1: Balance the atoms other than oxygen and hydrogen. Cr2O72−+SO2(g)→2Cr3+(aq)+SO42−(aq)

Step 2: Balance the oxygen atoms. Add 4H2O molecules on the right side of the equation. Cr2O72−+SO2(g)→2Cr3+(aq)+SO42−(aq)+4H2O

Step 3: Balance the hydrogen atoms. Add 8H+ ions on the left side of the equation. 8H+(aq)+Cr2O72−+SO2(g)→2Cr3+(aq)+SO42−(aq)+4H2O

Question:

Assign oxidation number to the underlined elements in each of the following species. CaO2

Answer:

Oxidation number of Calcium (Ca): +2 Oxidation number of Oxygen (O): -2

Question:

Assign oxidation number to the underlined element in NaBH4

Answer:

1. NaBH4 is an ionic compound, so the oxidation number of the elements will be determined by the ionic charge.

2. Sodium is a group 1A element, and its oxidation number is +1.

3. Boron is a group 3A element, and its oxidation number is -3.

4. Hydrogen is a group 1A element, and its oxidation number is +1.

Therefore, the oxidation number of the underlined element (Na) in NaBH4 is +1.

Question:

Justify that the following reactions are redox reactions: Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g)

Answer:

1. Identify the oxidation states of the elements in the reactants: Fe2O3(s): Fe= +3, O= -2 3CO(g): C= +2, O= -2

2. Identify the oxidation states of the elements in the products: 2Fe(s): Fe= 0 3CO2(g): C= +4, O= -2

3. Compare the oxidation states of the reactants and products: Fe went from +3 to 0, C went from +2 to +4, O went from -2 to -2.

4. Conclusion: Since the oxidation states of the reactants and products are different, this is a redox reaction.

Question:

Suggest a list of the substances where carbon can exhibit oxidation states from -4 to +4 and nitrogen from -3 to +5.

Answer:

1. Carbon: -4 (CO4 2-), -3 (CO3 2-), -2 (CO2), -1 (CO), 0 (elemental carbon), +1 (CH4), +2 (CH3OH), +3 (CH2O), +4 (CH3Cl)

2. Nitrogen: -3 (NO3-), -2 (NO2-), -1 (NO), 0 (elemental nitrogen), +1 (NH3), +2 (NH2OH), +3 (NH2O), +4 (NHCl2), +5 (NH4Cl)

Question:

While sulphur dioxide and hydrogen peroxide can act as oxidising as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants. Why ?

Answer:

Step 1: Understand what an oxidising agent and a reducing agent are.

Oxidising agents are substances that gain electrons during a chemical reaction and are reduced in the process. Reducing agents are substances that lose electrons during a chemical reaction and are oxidised in the process.

Step 2: Understand why sulphur dioxide and hydrogen peroxide can act as both oxidising and reducing agents.

Sulphur dioxide and hydrogen peroxide can act as both oxidising and reducing agents because they can either gain or lose electrons in different reactions.

Step 3: Understand why ozone and nitric acid can only act as oxidants.

Ozone and nitric acid can only act as oxidants because they can only gain electrons during a chemical reaction, meaning they can only be reduced in the process.

Question:

Depict the galvanic cell in which the reaction Zn(s)+2Ag+(aq)→Zn2+(aq)+2Ag(s) takes place. Further show individual reaction at each electrode.

Answer:

Step 1: The galvanic cell for the given reaction is shown below:

Step 2: The individual reactions at each electrode are: Anode (Oxidation): Zn(s) → Zn2+(aq) + 2e- Cathode (Reduction): 2Ag+(aq) + 2e- → 2Ag(s)

Question:

Using the standard electrode potentials given in the Table, predict if the reaction between the following is feasible. Ag(s) and Fe3+(aq)

Answer:

1. Find the standard electrode potentials of Ag(s) and Fe3+(aq) in the Table.

2. Calculate the difference between the standard electrode potentials of Ag(s) and Fe3+(aq).

3. Compare the difference to the standard reduction potential of 0.7 V.

4. If the difference is greater than 0.7 V, the reaction is feasible.

Question:

Assign oxidation number to the underlined element in H4P2O7

Answer:

Oxidation number of hydrogen (H): +1 Oxidation number of phosphorus (P): +5 Oxidation number of oxygen (O): -2

Question:

Assign oxidation number to the underlined element in KAl(SO4)2.12H2O

Answer:

Oxidation number of K = +1 Oxidation number of Al = +3 Oxidation number of S = +6 Oxidation number of O = -2

Question:

Assign oxidation number to the underlined element in H2S2O7

Answer:

Oxidation number of hydrogen (H): +1 Oxidation number of sulfur (S): +6 Oxidation number of oxygen (O): -2

Question:

Calculate the oxidation no. of sulphur in H2S4O6.

Answer:

Step 1: Calculate the total charge of the compound.

H2S4O6 = 2(1) + 4(-2) + 6(-2) = -12

Step 2: Calculate the total number of atoms of sulphur (S).

H2S4O6 = 2 + 4 = 6 atoms of sulphur

Step 3: Calculate the oxidation number of sulphur.

Oxidation number of sulphur = -12/6 = -2

Question:

Justify that the following reactions are redox reactions: CuO(s)+H2(g)→Cu(s)+H2O(g)

Answer:

1. Identify the oxidation numbers of each element in the reactants and products: CuO(s): Cu(+2), O(-2) H2(g): H(+1) Cu(s): Cu(+2) H2O(g): H(+1), O(-2)

2. Determine if there is a change in the oxidation numbers of any elements: Yes, the oxidation number of Cu changes from +2 to 0 in the products.

3. Conclude that the reaction is a redox reaction: Yes, the reaction is a redox reaction since there is a change in the oxidation number of Cu.

Question:

Using the standard electrode potentials given in the table, predict if the reaction between the following is possible. Ag(aq)+ and Cu(s)

Answer:

1. Find the standard electrode potentials for Ag(aq)+ and Cu(s) in the table.

2. Ag(aq)+: +0.80 V Cu(s): +0.34 V

3. Compare the two standard electrode potentials. The reaction is possible if the standard electrode potential of the oxidizing agent (Ag(aq)+ in this case) is greater than the standard electrode potential of the reducing agent (Cu(s) in this case).

4. Since +0.80 V > +0.34 V, the reaction is possible.

Question:

Predict the products of electrolysis in each of the following. A dilute solution of H2SO4 with platinum electrodes.

Answer:

1. At the anode (positive electrode): 2H2O –> 4H+ + 4e-

2. At the cathode (negative electrode): 2e- + 2H+ –> H2

3. Overall reaction: 2H2O –> H2 + 2H2SO4

Question:

Predict the products of electrolysis in each of the following. An aqueous solution AgNO3 with platinum electrodes.

Answer:

1. At the anode (positive electrode): Oxygen gas (O2) will be produced and Ag+ ions will be reduced to silver metal (Ag).

2. At the cathode (negative electrode): Hydrogen gas (H2) will be produced and NO3- ions will be oxidized to nitric oxide gas (NO).

Question:

Justify that the following reactions are redox reactions: CuO(s)+H2(g)→Cu(s)+H2O(g)

Answer:

1. Identify the oxidation states of the reactants and products: CuO (s): +2 H2 (g): 0 Cu (s): 0 H2O (g): 0

2. Determine if there is a change in oxidation states of any of the elements: Yes, there is a change in the oxidation state of copper from +2 to 0.

3. Conclusion: This is a redox reaction, as there is a change in the oxidation state of an element.

Question:

While sulphur dioxide and hydrogen peroxide can act as oxidising as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants. Why ?

Answer:

Answer:

1. Sulphur dioxide and hydrogen peroxide are molecules that contain oxygen atoms that can either gain or lose electrons when they react with other molecules. This means that they can act as both oxidising agents (when they gain electrons) and reducing agents (when they lose electrons).

2. Ozone and nitric acid, on the other hand, contain oxygen atoms that are more electronegative than the atoms of other molecules. This means that when they react with other molecules, they will always gain electrons, making them oxidising agents only.

Question:

Identify the substance oxidised, reduced, oxidising agent and reducing agent for each of the following reactions : Pb(s)+PbO2(s)+2H2SO4(aq)→2PbSO4(s)+2H2O(l)

Answer:

Substance oxidised: Pb Substance reduced: PbO2 Oxidising agent: H2SO4 Reducing agent: H2O

Question:

Consider the reactions: 2S2O32−(aq)+I2(s)⟶S4O62−(aq)+2I−(aq) S2O32−(aq)+2Br2(l)+5H2O(l)⟶2SO42−(aq)+4Br−(aq)+10H+(aq) Why does the same reductant thiosulphate react differently with iodine and bromine?

Answer:

Answer: The thiosulphate, S2O32−, acts as a reducing agent in both reactions. However, it reacts differently with iodine and bromine because iodine is a stronger oxidizing agent than bromine, and therefore requires more thiosulphate to be reduced.

Question:

Using the standard electrode potentials given in the table, predict if the reaction between the following is possible. Fe3+(aq) and Cu(s)

Answer:

1. Look up the standard electrode potentials for Fe3+(aq) and Cu(s) in the table.

2. Compare the standard electrode potentials for Fe3+(aq) and Cu(s).

3. If the standard electrode potential for Fe3+(aq) is greater than that for Cu(s), then the reaction is possible.

Question:

Given the standard electrode potentials of some metals. K+/K=−2.93V,
Ag+/Ag=0.80V, Hg2+/Hg=0.79V, Mg2+/Mg=−2.37V and Cr3+/Cr=−0.74V Arrange these metals in their increasing order of reducing power.

Answer:

1. K+/K=-2.93V
2. Mg2+/Mg=-2.37V
3. Cr3+/Cr=-0.74V
4. Hg2+/Hg=0.79V
5. Ag+/Ag=0.80V

Increasing order of reducing power: K+, Mg2+, Cr3+, Hg2+, Ag+

Question:

2S2O32−(aq)+I2(s)→S4O62−(aq)+2I−(aq) S2O32−(aq)+2Br2(l)+5H2O(l)→2SO42−(aq)+4Br−(aq)+10H+(aq) Why does the same reductant, thiosulphate react differently with iodine and bromine?

Answer:

1. 2S2O32-(aq) + I2(s) → S4O62-(aq) + 2I-(aq)

2. S2O32-(aq) + 2Br2(l) + 5H2O(l) → 2SO42-(aq) + 4Br-(aq) + 10H+(aq)

3. The same reductant, thiosulphate, reacts differently with iodine and bromine because iodine is an element, while bromine is a compound. Iodine is more easily oxidized by thiosulphate than bromine, so when it reacts with thiosulphate, it produces sulfur dioxide and two iodide ions. On the other hand, when thiosulphate reacts with bromine, it produces sulfur dioxide, four bromide ions, and ten hydrogen ions.

Question:

Balance the following redox reactions by ion-electron method. H2O2(aq)+Fe2+(aq)→Fe3+(aq)+H2O(l) (in acidic solution)

Answer:

Step 1: Write the half-reactions for the oxidation and reduction processes.

Oxidation: Fe2+ → Fe3+

Reduction: H2O2 → H2O

Step 2: Balance the number of electrons in each half-reaction.

Oxidation: Fe2+ → Fe3+ + e-

Reduction: 2H2O2 + 2e- → 2H2O

Step 3: Balance the number of atoms in each half-reaction.

Oxidation: Fe2+ → Fe3+ + e-

Reduction: 2H2O2 + 2e- → 2H2O

Step 4: Balance the charges in each half-reaction.

Oxidation: Fe2+ → Fe3+ + e-

Reduction: 2H2O2 + 2e- → 2H2O

Step 5: Add the two half-reactions together.

Fe2+ + 2H2O2 → Fe3+ + 2H2O

Question:

When concentrated sulphuric acid is added to an inorganic mixture containing chloride, we get colorless pungent smell gas HCl, but if the mixture contains bromide then we get red vapor of bromine. Why?

Answer:

1. Sulphuric acid is a strong oxidizing agent and when it is added to an inorganic mixture containing chloride, it oxidizes the chloride to form hydrogen chloride gas (HCl), which has a colorless and pungent smell.

2. However, when the same mixture contains bromide, the sulphuric acid oxidizes the bromide to form bromine, which is a red-colored vapor.

Question:

Assign oxidation number to the underlined elements in each of the following species. (i) NaH2PO4 (ii) NaHSO4 (iii) H4P2O7 (iv) K2MnO4 (v) CaO2 (vi) NaBH4 (vii) H2S2O7 (viii) KAl(SO4)2.12H2O (ix) Cr2O72− (x) S2O32−

Answer:

(i) Na: +1, H: +1, P: +5, O: -2 (ii) Na: +1, H: +1, S: +4, O: -2 (iii) H: +1, P: +5, O: -2 (iv) K: +1, Mn: +7, O: -2 (v) Ca: +2, O: -2 (vi) Na: +1, B: +3, H: +1 (vii) H: +1, S: +4, O: -2 (viii) K: +1, Al: +3, S: +4, O: -2, H: +1 (ix) Cr: +6, O: -2 (x) S: +4, O: -2

Question:

Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. Justify this statement giving three illustrations.

Answer:

1. Oxidising agents are molecules or ions that can accept electrons from other molecules or ions, thus oxidising them. Reducing agents are molecules or ions that can donate electrons to other molecules or ions, thus reducing them.

2. Whenever a reaction between an oxidising agent and a reducing agent is carried out, the oxidising agent will accept electrons from the reducing agent, causing the oxidation state of the reducing agent to decrease. Conversely, the reducing agent will donate electrons to the oxidising agent, causing the oxidation state of the oxidising agent to increase.

3. If the reducing agent is in excess, then more electrons will be donated than accepted, and the oxidation state of the reducing agent will decrease more than the oxidation state of the oxidising agent will increase. Therefore, a compound of lower oxidation state will be formed.

4. Conversely, if the oxidising agent is in excess, then more electrons will be accepted than donated, and the oxidation state of the oxidising agent will increase more than the oxidation state of the reducing agent will decrease. Therefore, a compound of higher oxidation state will be formed.

5. Three illustrations of this statement are as follows:

i. In the reaction between hydrogen peroxide (H2O2) and iron (Fe), the oxidation state of iron decreases from +2 to 0, forming iron(II) oxide (FeO).

ii. In the reaction between copper(II) chloride (CuCl2) and sodium (Na), the oxidation state of sodium increases from 0 to +1, forming sodium chloride (NaCl).

iii. In the reaction between potassium permanganate (KMnO4) and iron (Fe), the oxidation state of iron increases from 0 to +3, forming iron(III) oxide (Fe2O3).

Question:

Balance the following redox reactions by the ion-electron method in basis medium MnO4−(aq)+I−(aq)→MnO2(s)+I2(s)

Answer:

Step 1: Balance the atoms other than oxygen and hydrogen.

MnO4−(aq)+I−(aq)→MnO2(s)+I2(s)

Mn: 1 atom on left side and 1 atom on right side I: 1 atom on left side and 2 atoms on right side

Step 2: Balance the oxygen atoms.

MnO4−(aq)+I−(aq)→MnO2(s)+I2(s)

O: 4 atoms on left side and 2 atoms on right side

Step 3: Balance the hydrogen atoms.

MnO4−(aq)+I−(aq)→MnO2(s)+I2(s)

H: 0 atoms on both sides

Step 4: Balance the charges.

MnO4−(aq)+I−(aq)→MnO2(s)+I2(s)

Charge: -1 on left side and 0 on right side

Step 5: The balanced equation is as follows:

MnO4−(aq) + 5I−(aq) → MnO2(s) + 2I2(s)

Question:

Predict the products of electrolysis in each of the following. An aqueous solution of CuCl2 with platinum electrodes.

Answer:

1. At the anode (positive electrode): Cu2+ ions will be oxidized to form copper metal (Cu)

2. At the cathode (negative electrode): Cl- ions will be reduced to form chlorine gas (Cl2)

Question:

Using the standard electrode potentials given in the table, predict if the reaction between the following is possible. Br2(aq) and Fe2+(aq)

Answer:

1. Look up the standard electrode potentials for Br2(aq) and Fe2+(aq) in the table.

2. Calculate the overall cell potential by subtracting the standard reduction potential of Fe2+(aq) from the standard oxidation potential of Br2(aq).

3. If the overall cell potential is positive, the reaction is possible. If the overall cell potential is negative, the reaction is not possible.

Question:

Using the standard electrode potentials given in the table, predict if the reaction between the following is possible. Fe3+(aq) and I−(aq)

Answer:

1. Look up the standard electrode potentials for Fe3+(aq) and I−(aq) in the table.

2. Calculate the overall standard electrode potential for the reaction.

3. Compare the overall standard electrode potential to the value of 0 V.

4. If the overall standard electrode potential is positive, the reaction is not possible. If the overall standard electrode potential is negative, the reaction is possible.

Question:

State whether True or False. Ozone is the oxidising agent in the following reaction: O3(g)+H2O2(l)→H2O(l)+2O2(g)

Answer:

False

Question:

From the periodic table select three possible non-metals which can show disproportionation reaction.

Answer:

Step 1: Open the periodic table and identify the elements which are non-metals. These are typically located on the right side of the table.

Step 2: Select three non-metals from the periodic table which can show disproportionation reaction. Examples of non-metals which can show disproportionation reaction include chlorine, sulfur, and oxygen.

Question:

Balance the following equation in basic medium by ion-electron method and oxidation number method and identify the oxidising agent and the reducing agent. N2H4(l)+ClO3−(aq)→NO(g)+Cl−(g)

Answer:

Ion-electron method:

Step 1: Balance the atoms other than oxygen and hydrogen.

N2H4(l) + ClO3−(aq) → NO(g) + Cl−(g)

Step 2: Balance the oxygen atoms.

N2H4(l) + 4ClO3−(aq) → NO(g) + 4Cl−(g)

Step 3: Balance the hydrogen atoms.

2N2H4(l) + 4ClO3−(aq) → 2NO(g) + 8Cl−(g)

Oxidation number method:

Step 1: Assign oxidation numbers to all atoms.

N2H4(l): N = +3; H = +1 ClO3−(aq): Cl = +7; O = -2 NO(g): N = +2; O = -2 Cl−(g): Cl = -1

Step 2: Balance the atoms other than oxygen and hydrogen.

N2H4(l) + ClO3−(aq) → NO(g) + Cl−(g)

Step 3: Balance the oxygen atoms.

N2H4(l) + 4ClO3−(aq) → NO(g) + 4Cl−(g)

Step 4: Balance the hydrogen atoms.

2N2H4(l) + 4ClO3−(aq) → 2NO(g) + 8Cl−(g)

The oxidising agent is ClO3−(aq) and the reducing agent is N2H4(l).

Question:

Consider the elements Cs, Ne, I and F. Identify the element which exhibits neither the negative nor does the positive oxidation state.

Answer:

Answer:

1. Cs: exhibits a positive oxidation state
2. Ne: exhibits a neutral oxidation state
3. I: exhibits both positive and negative oxidation states
4. F: exhibits a negative oxidation state

Therefore, the element which exhibits neither the negative nor does the positive oxidation state is Ne (Neon).

Question:

Why it is more appropriate to write reactions O3(g)+H2O2(l)→H2O(l)+2O2(g) as :O3(g)+H2O2(l)→H2O(l)+O2(g)+O2(g) ? Also suggest a technique to investigate the path of the above redox reaction.

Answer:

Answer:

It is more appropriate to write the reaction O3(g)+H2O2(l)→H2O(l)+2O2(g) as O3(g)+H2O2(l)→H2O(l)+O2(g)+O2(g) because it accurately reflects the fact that two molecules of O2 are produced in the reaction.

A technique to investigate the path of the above redox reaction is to use a technique called pulse radiolysis. Pulse radiolysis is a method of studying the reactions of short-lived chemical species (radicals) in solution. It involves the use of a high-energy radiation pulse to initiate the reaction, allowing for the study of the reaction in a controlled environment.

Question:

Calculate the oxidation number of sulphur, chromium and nitrogen in H2SO5,Cr2O72− and NO3−. Suggest structure of these compounds. Count for the fallacy.

Answer:

Oxidation number of Sulphur in H2SO5: The oxidation number of sulphur in H2SO5 is +6. This can be determined by assigning oxidation numbers to each atom in the compound. The oxidation number of hydrogen is +1, oxygen is -2 and sulphur is +6.

Oxidation number of Chromium in Cr2O72−: The oxidation number of chromium in Cr2O72− is +6. This can be determined by assigning oxidation numbers to each atom in the compound. The oxidation number of chromium is +6 and oxygen is -2.

Oxidation number of Nitrogen in NO3−: The oxidation number of nitrogen in NO3− is +5. This can be determined by assigning oxidation numbers to each atom in the compound. The oxidation number of nitrogen is +5 and oxygen is -2.

Structure of H2SO5: H2SO5 is a sulfuric acid. It is composed of two hydrogen atoms, one sulfur atom, and five oxygen atoms. The structure of H2SO5 can be represented as follows:

H2SO5

Structure of Cr2O72−: Cr2O72− is chromic acid. It is composed of two chromium atoms, seven oxygen atoms, and two negative charges. The structure of Cr2O72− can be represented as follows:

Cr2O72−

Structure of NO3−: NO3− is nitrate. It is composed of one nitrogen atom, three oxygen atoms, and one negative charge. The structure of NO3− can be represented as follows:

NO3−

Fallacy: There is no fallacy in the calculation of the oxidation numbers of sulphur, chromium and nitrogen in H2SO5, Cr2O72− and NO3−.

Question:

The compound AgF2 is unstable compound. However, if formed, the compound acts as a very strong oxidising agent. Why ?

Answer:

1. AgF2 is an unstable compound because it contains two different elements, silver and fluorine, which have different electronegativities. This causes the electrons in the compound to be unequally shared, resulting in an unstable compound.

2. When AgF2 is formed, it acts as a very strong oxidising agent because it is able to easily donate electrons to other molecules. This is due to the difference in electronegativity between silver and fluorine, which allows the electrons to be easily transferred to other molecules.

Question:

Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.

Answer:

Step 1: Explain the meaning of oxidant and reductant. Oxidant is any substance that can accept electrons or donate oxygen atoms, while reductant is any substance that can donate electrons or accept oxygen atoms.

Step 2: Explain why fluorine is the best oxidant among the halogens. Fluorine is the most electronegative of all the halogens, so it has the strongest tendency to accept electrons. This makes it the best oxidant among the halogens.

Step 3: Explain why hydroiodic acid is the best reductant among hydrohalic compounds. Hydroiodic acid is the most acidic of all the hydrohalic compounds, so it has the strongest tendency to donate electrons. This makes it the best reductant among the hydrohalic compounds.

Question:

What sorts of informations can you draw from the following reaction ? (CN)2(g)+2OH−(aq)→CN−(aq)+CNO−(aq)+H2O(l)

Answer:

1. The reactants in the reaction are (CN)2(g) and 2OH−(aq).

2. The products in the reaction are CN−(aq), CNO−(aq), and H2O(l).

3. The reaction involves the formation of two new molecules, CN−(aq) and CNO−(aq).

4. The reaction involves the breaking of one molecule, (CN)2(g).

5. The reaction involves the formation of water, H2O(l).

6. The reaction involves the transfer of electrons from (CN)2(g) to OH−(aq).

Question:

Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess of chlorine is removed by treating with sulphur dioxide. Present a balanced equation for the redox change taking place in water.

Answer:

Chlorine + Sulphur Dioxide → Chlorine Dioxide + Sulphur

Balanced equation: 2Cl2 + SO2 → ClO2 + 2SO3

Question:

Refer to the periodic table and write three metals that can show disproportionation reaction.

Answer:

1. Copper (Cu)
2. Silver (Ag)
3. Mercury (Hg)

Question:

What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results? Fe3O4

Answer:

The oxidation number of iron (Fe) in Fe3O4 is +3. This can be rationalised by using the formula for the compound: Fe3O4. According to the formula, there are three atoms of iron and four atoms of oxygen. Since oxygen has an oxidation number of -2, the total oxidation number of oxygen atoms is -8. Since the total oxidation number of the compound is 0, the oxidation number of iron must be +8, divided by three atoms of iron, the oxidation number of iron is +3.

Question:

Fluorine reacts with ice and results in the change: H2O(s)+F2(g)⟶HF(g)+HOF(g) i) Justify that this reaction is a redox reaction. ii) Write the substances oxidized and reduced. iii) Write the name of the substances which acts as reducing agent and oxidizing agent.

Answer:

i) This reaction is a redox reaction because there is a transfer of electrons from one substance to another.

ii) The substance oxidized is Fluorine (F2) and the substance reduced is Water (H2O).

iii) The reducing agent is Water (H2O) and the oxidizing agent is Fluorine (F2).

Question:

Consider the elements Cs, Ne, I and F. Identify the element that exhibits only postive oxidation state.

Answer:

Answer:

1. Cs: Positive oxidation states: +1, +2

2. Ne: Positive oxidation states: +2

3. I: Positive oxidation states: +1, +3, +5

4. F: Positive oxidation states: +1

Therefore, the element that exhibits only positive oxidation state is F.

Question:

Identify the substance oxidised, reduced, oxidising agent and reducing agent for each of the following reactions: N2H4(l)+2H2O2(l)→N2(g)+4H2O(l)

Answer:

Substance oxidised: N2H4 (ammonia) Substance reduced: H2O2 (hydrogen peroxide) Oxidising agent: H2O2 (hydrogen peroxide) Reducing agent: N2H4 (ammonia)

Question:

Consider the reactions: Why it is more appropriate to write these reactions as: 6CO2(g)+12H2O(l)→C6H12O6(aq)+6O2(g)+6H2O(l) Also suggest a technique to investigate the path of the above (a) and (b) redox reactions. 6CO2(g)+6H2O(l)→C6H12O6(aq)+6O2(g)

Answer:

Answer: The reaction written in the question is a redox reaction, and it is more appropriate to write it as 6CO2(g)+12H2O(l)→C6H12O6(aq)+6O2(g)+6H2O(l) because it is a balanced equation. This means that the number of atoms of each element on the left side of the equation is equal to the number of atoms of each element on the right side of the equation.

A technique to investigate the path of the above (a) and (b) redox reactions is to use mass spectrometry. Mass spectrometry is a technique that can be used to measure the mass and composition of molecules, and thus can be used to determine the path of a reaction.

Question:

Though alkaline potassium permanganate and acidic potassium permanganate both are used as oxidants, yet in the manufacture of benzoic acid from toluene we use alcoholic potassium permanganate as an oxidant. Why? Write the balanced redox equation for the reaction.

Answer:

Answer:

1. The use of alcoholic potassium permanganate as an oxidant in the manufacture of benzoic acid from toluene is because it is a milder oxidizing agent than alkaline or acidic potassium permanganate and can be used to selectively oxidize the toluene molecule to benzoic acid without oxidizing other components in the reaction mixture.

2. The balanced redox equation for the reaction is:

C7H8 + 2KMnO4 + 3H2SO4 → C7H6O2 + 2KHSO4 + 2MnSO4 + 3H2O

Question:

Balance the following equation in basic medium by ion-electron method and oxidation number method and identify the oxidising agent and the reducing agent. P4(s)+OH−(aq)→PH3(g)+H2PO2−(aq)

Answer:

Ion-electron method:

Step 1: Balance the atoms on both sides of the equation:

P4(s) + 4OH−(aq) → 4PH3(g) + H2PO2−(aq)

Step 2: Balance the charges on both sides of the equation:

P4(s) + 4OH−(aq) → 4PH3(g) + H2PO2−(aq) + 3H+ (aq)

Oxidation number method:

Step 1: Assign oxidation numbers to each element in the equation:

P4(s) + 4OH−(aq) → 4PH3(g) + H2PO2−(aq) + 3H+ (aq) Oxidation number of P = +3 Oxidation number of O = -2 Oxidation number of H = +1

Step 2: Balance the charges on both sides of the equation:

P4(s) + 4OH−(aq) → 4PH3(g) + H2PO2−(aq) + 3H+ (aq) + 2e-

The oxidising agent is OH- (aq) and the reducing agent is P4 (s).

Question:

In Ostwald’s process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can be obtained starting only with 10.00 g. of ammonia and 20.00 g of oxygen ?

Answer:

1. Calculate the molar mass of ammonia (NH3): Molar mass of ammonia = 17.03 g/mol

2. Calculate the molar mass of oxygen (O2): Molar mass of oxygen = 32.00 g/mol

3. Calculate the number of moles of ammonia (nNH3): nNH3 = 10.00 g/17.03 g/mol = 0.5883 mol

4. Calculate the number of moles of oxygen (nO2): nO2 = 20.00 g/32.00 g/mol = 0.625 mol

5. Calculate the number of moles of nitric oxide (NO) that can be obtained: NO = 2nNH3 + nO2 = (2*0.5883 mol) + 0.625 mol = 2.2083 mol

6. Calculate the maximum weight of nitric oxide that can be obtained: Maximum weight of nitric oxide = 2.2083 mol * 30.01 g/mol = 66.10 g

Question:

Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible: (i) Fe3+(aq) and I−(aq) (ii) Ag+(aq) and Cu(s) (iii) Fe3+(aq) and Br−(aq) (iv) Ag(s) and Fe3+(aq) (v) Br2(aq) and Fe2+(aq)

Answer:

(i) Fe3+(aq) and I−(aq): Fe3+(aq) + I−(aq) -> Fe2+(aq) + I2(aq) The reaction is feasible since the standard electrode potential for Fe3+(aq) + I−(aq) -> Fe2+(aq) + I2(aq) is 1.36V which is greater than 0V.

(ii) Ag+(aq) and Cu(s): Ag+(aq) + Cu(s) -> Ag(s) + Cu2+(aq) The reaction is feasible since the standard electrode potential for Ag+(aq) + Cu(s) -> Ag(s) + Cu2+(aq) is 0.80V which is greater than 0V.

(iii) Fe3+(aq) and Br−(aq): Fe3+(aq) + Br−(aq) -> Fe2+(aq) + Br2(aq) The reaction is feasible since the standard electrode potential for Fe3+(aq) + Br−(aq) -> Fe2+(aq) + Br2(aq) is 1.07V which is greater than 0V.

(iv) Ag(s) and Fe3+(aq): Ag(s) + Fe3+(aq) -> Ag+(aq) + Fe2+(aq) The reaction is feasible since the standard electrode potential for Ag(s) + Fe3+(aq) -> Ag+(aq) + Fe2+(aq) is 0.80V which is greater than 0V.

(v) Br2(aq) and Fe2+(aq): Br2(aq) + Fe2+(aq) -> Br−(aq) + Fe3+(aq) The reaction is feasible since the standard electrode potential for Br2(aq) + Fe2+(aq) -> Br−(aq) + Fe3+(aq) is 1.07V which is greater than 0V.

Question:

Predict the products of electrolysis in each of the following: (i) An aqueous solution of AgNO3 with silver electrodes. (ii) An aqueous solution of AgNO3 with platinum electrodes. (iii) A dilute solution of H2SO4 with platinum electrodes. (iv) An aqueous solution of CuCl2 with platinum electrodes

Answer:

(i) An aqueous solution of AgNO3 with silver electrodes: At the anode (positive electrode): Oxidation of silver ions (Ag+) to form silver atoms (Ag). At the cathode (negative electrode): Reduction of water molecules (H2O) to form hydrogen gas (H2).

(ii) An aqueous solution of AgNO3 with platinum electrodes: At the anode (positive electrode): Oxidation of silver ions (Ag+) to form silver atoms (Ag). At the cathode (negative electrode): Reduction of water molecules (H2O) to form oxygen gas (O2).

(iii) A dilute solution of H2SO4 with platinum electrodes: At the anode (positive electrode): Oxidation of sulfate ions (SO4 2-) to form sulfur dioxide gas (SO2). At the cathode (negative electrode): Reduction of water molecules (H2O) to form hydrogen gas (H2).

(iv) An aqueous solution of CuCl2 with platinum electrodes: At the anode (positive electrode): Oxidation of copper ions (Cu2+) to form copper atoms (Cu). At the cathode (negative electrode): Reduction of chloride ions (Cl-) to form chlorine gas (Cl2).

Question:

Arrange the following metals in the order in which they displace each other from the solution of their salts. Al,Cu,Fe,Mg and Zn.

Answer:

Mg, Fe, Al, Zn, Cu

Question:

Given the standard electrode potentials, K+/K=−2.93 V,Ag+/Ag=0.80 V, Hg2+/Hg=0.79 V, Mg2+/Mg=−2.37V,Cr3+/Cr=−0.74 V Arrange these metals in their increasing order of reducing power.

Answer:

1. Mg2+/Mg=−2.37 V
2. K+/K=−2.93 V
3. Cr3+/Cr=−0.74 V
4. Ag+/Ag=0.80 V
5. Hg2+/Hg=0.79 V

Increasing order of reducing power: Mg2+/Mg, K+/K, Cr3+/Cr, Ag+/Ag, Hg2+/Hg