Chapter 8 – Redox Reactions

Redox Reactions

NCERT Exercises

Question 1.
Assign oxidation number to the underlined elements in each of the following species :
(a) NaH₂PO₄
(b) NaHSO₄
(c) H₄P₂O₇
(d) K₂MnO₄
(e) CaO₂
(f) NaBH₄
(g) H₂S₂O₇
(h) KAI(SO₄)₂.12H₂O

Solution.
Let the oxidation no. of underlined element in all the given compounds = x

Question 2.
What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results?
(a) KI₃
(b) H₂S₄O₆
(c) Fe₃O₄
(d) CH₃CH₂OH
(e) CH₃COOH

Solution.
(a) In KI₃, since the oxidation number of K is +1, therefore, the average oxidation number of iodine = -1/3. In the structure, K⁺[I – I <— I]^–, a coordinate bond is formed between I2 molecule and I ion. The oxidation number of two iodine atoms forming the I2 molecule is zero while that of iodine ion forming the coordinate bond is -1. Thus, the O.N. of three iodine atoms in KI₃ are 0, 0 and -1 respectively.

Question 3.
Justify that the following reactions are redox reactions:
(a) CuO(s) + H_2(g) ➝ Cu₍₅₎ + H₂O(g)
(b) Fe₂O_3(s) + 3CO_(g) ➝ 2Fe_(s) + 3CO_2(g)
(c) 4BCI_3(g) + 3LiAIH_4(s) ➝ 2B₂H_6(g) + 3LiCI_(s) + 3AICI_3(s)
(d) 2K_(s) + F_2(g) ➝ 2K⁺F^–_(s)
(e) 4NH₃(g) + 5O_2(g) ➝ 4NO_(g) + 6H₂O_(g)

Solution.

Question 4.
Fluorine reacts with ice and results in the change:
H₂O_(s) + F_2(g) ➝ HF_(g) + HOF_(g)
Justify that this reaction is a redox reaction.

Solution.

Question 5.
Calculate the oxidation number of sulphur, chromium and nitrogen in H₂SO₅, Cr₂O₇^2- and NO₃^–. Suggest structure of these compounds. Count for the fallacy.

Solution.

Question 6.
Write formulas for the following compounds:
(a) Mercury(II) chloride
(b) Nickel(II) sulphate
(c) Tin(IV) oxide
(d) Thallium(I) sulphate
(e) Iron(III) sulphate
(f) Chromium(III) oxide

Solution.
(a) HgCl₂
(b) NiSO₄
(c) SnO₂
(d) Tl₂SO₄
(e) Fe₂(SO₄)₃
(f) Cr₂O₃

Question 7.
Suggest a list of the substances where carbon can exhibit oxidation states from -4 to +4 and nitrogen from -3 to +5.

Solution.

Question 8.
While Sulphur dioxide and hydrogen peroxide can act as oxidizing as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants. Why?

Solution.
The oxidation state of Sulphur in Sulphur dioxide is +4. It can be oxidized to +6 oxidation state or reduced to +2. Therefore, Sulphur dioxide acts as a reducing agent as well as oxidizing agent. Similarly, the oxidation state of oxygen in hydrogen peroxide is -1. It can be oxidized to O₂ (zero oxidation state) or reduced to H₂O or OH^– (-2 oxidation state) and therefore, acts as reducing as well as oxidizing agents.
However, both ozone and nitric acid can only decrease their oxidation number and therefore, act only as oxidizing agents.

Question 9.
Consider the reactions :
(a) 6CO₂(g) + 6H₂O_(l) ➝ C₆H₁₂O_6(ag) + 6O_2(g)
(b) O_3(g) + H₂O_2(l) ➝ H₂O_(l) + 2O_2(g)
Also suggest a technique to investigate the path of the above (a) and (b) redox reactions.

Solution.

Question 10.
The compound AgF₂ is unstable compound. However, if formed, the compound acts as a very strong oxidizing agent. Why ?

Solution.
In AgF₂ oxidation state of Ag is +2 which is very unstable. Therefore, it quickly, accepts an electron to form the more stable +1 oxidation state.
Ag²⁺ + e^– ➝ Ag⁺
Therefore, AgF₂, if formed, will act as a strong oxidizing agent.

Question 11.
Whenever a reaction between an oxidizing 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 oxidizing agent is in excess. Justify this statement giving three illustrations.

Solution.
(i) C is a reducing agent while O₂ is an oxidizing agent. If excess of carbon is burnt in a limited supply of O₂, CO is formed in which oxidation state of C is +2 but when O₂ is in excess CO formed gets oxidized to CO₂ in which oxidation state of C is + 4.

Question 12.
How do you count for the following observations ?
(a) 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 a balanced redox equation for the reaction.
(b) When concentrated sulphuric acid is added toaninorganicmixture containing chloride, we get colorless pungent smelling gas HCI, but if the mixture contains bromide then we get red vapor of bromine. Why?

Solution.
(a) In neutral medium, KMnO₄ acts as an oxidant as follows :
MnO₄^– + 2H₂O + 3e^– ➝ MnO₂ + 40H^–
In laboratory, alkaline KMnO₄ is used to oxidize toluene to benzoic acid.

Question 13.
Identify the substance oxidized, reduced, oxidizing agent and reducing agent for each of the following reactions:

Solution.

Question 14.
Consider the reactions:

Why does the same reductant, thiosulphate react differently with iodine and bromine?
Solution.

Question 15.
Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.

Solution.
The halogens (X₂) have strong electron accepting tendency and have positive standard oxidation potential values. They are therefore, powerful oxidizing agents. The decreasing order of oxidizing powers of halogens is :

Question 16.
Why does the following reaction occur ?

Solution.

Question 17.
Consider the reactions :

Solution.

Question 18.
Balance the following redox reactions by ion – electron method:

Solution.

Question 19.
Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidizing agent and the reducing agent.

Solution.

Question 20.
What sorts of informations can you draw from the following reaction?
(CN)_2(g) + 2OH^–_(aq) ➝ CN^–_(aq) + CNO^–_(aq) + H₂O_(l)

Solution.

Question 21.
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.

Solution.

Question 22.
Consider the elements: Cs, Ne, I and F
(a) Identify the element that exhibits only negative oxidation state.
(b) Identify the element that exhibits only positive oxidation state.
(c) Identify the element that exhibits both positive and negative oxidation states.
(d) Identify the element which exhibits neither the negative nor does the positive oxidation state.

Solution.
(a) F: Fluorine is the most electronegative element and shows only -1 oxidation state.
(b) Cs : Because of the presence of single electron in the valence shell, (alkali metals) Cs exhibits an oxidation state of +1 only.
(c) I: Because of the presence of seven electrons in the valence shell, I shows an oxidation state of-1 in compounds with more electropositive elements (such as H, Na, K, Ca, etc.) and oxidation states of +3, +5, +7 in compounds of I with more electronegative elements (such as, O, F, etc.)
(d) Ne: It is an inert gas (with high ionization enthalpy and highly positive electron gain enthalpy) and hence, it exhibits neither negative nor positive oxidation states.

Question 23.
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 this redox change taking place in water.

Solution.

Question 24.
Refer to the periodic table given in your book and now answer the following questions:
(a) Select the possible non metals that can show disproportionation reaction.
(b) Select three metals that can show disproportionation reaction.

Solution.

Question 25.
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?

Solution.
The reaction involved in the manufacturing process is :

Question 26.
Using the standard electrode potentials, predict if the reaction between the following is feasible:

Solution.

Question 27.
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.

Solution.

(1) An aqueous solution of AgNO₃ using platinum electrodes :     (P.I.S.A. Based)
Both AgNO₃ and water will ionise in aqueous solution

At cathode : Ag⁺ ions with less discharge potential are reduced in preference to H⁺ ions which will remain in solution. As a result, silver will be deposited at cathode.
Ag⁺ (aq) + e^– → Ag (deposited)
At anode : An equivalent amount of silver will be oxidized to Ag⁺ ions by releasing electrons.

As result of electrolysis, Ag from silver anode dissolves as Ag⁺(aq) ions while an equivalent amount of Ag⁺(aq) ions from the aqueous AgNO₃ solution get deposited on the cathode.

(2) An aqueous solution of AgNO₃ using platinum electrodes :

In the case, the platinum electrodes are the non-attackable electrodes. On passing current the following changes will occur at the electrodes.
At cathode : Ag⁺ ions will be reduced to Ag which will get deposited at the cathode.
At anode : Both NO^–₃ and OH^– ions will migrate. But OH^– ions with less discharge potential will be oxidized in preference to NO₃ ions which will remain in solution.
OH- (aq) → OH + e^–; 4OH → 2H₂O(l) + O₂ (g)
Thus, as a result of electrolysis, silver is deposited on the cathode while O₂ is evolved at the anode. The solution will be acidic due to the presence of HNO₃.

(3) A dilute solution of H₂SO₄ using platinum electrodes :
On passing current, both acid and water will ionise as follows :

At cathode :
H⁺ (aq) ions will migrate to the cathode and will be reduced to H₂.
H+ (aq) + e-→ H ; H + H→ H₂ (g)
Thus, H₂ (g) will evolve at cathode.
At anode : OH ions will be released in preference to SO^2-₄ ions because their discharge potential is less. They will be oxidised as follows :
OH^– (aq) → OH + e^– ; 4OH → 2H₂O(l) + O₂ (g)
Thus, O₂ (g) will be evolved at anode. The solution will be acidic and will contain H₂SO₄.

(4) An aqueous solution of CuCl₂ using platinum electrodes :

The electrolysis proceeds in the same manner as discussed in the case of AgNO₃ solution. Both CuCl₂ and H₂O will ionise as follows :

At cathode :  Cu²⁺ ions will be reduced in preference to H⁺ ions and copper will be deposited at cathode.
Cu²⁺ (aq) + 2e^– → Cu (deposited)
At anode : Cl^– ions will be discharged in preference to OH^– ions which will remain in solution.
Cl^–→Cl^–+ e^– ; Cl + Cl → Cl₂ (g) (evolved)
Thus, Cl₂ will evolve at anode.

Question 28.
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.

Solution.

Since a metal with lower electrode potential is a stronger reducing agent, therefore, Mg can displace all the above metals from their aqueous solutions, Al can displace all metals except Mg from the aqueous solutions of their salts. Zn can displace all metals except Mg and Al from the aqueous solutions of their salts while Fe can displace only Cu from the aqueous solution of its salts. Thus, the order in which they can displace each other from the solution of their salts is Mg, Al, Zn, Fe, Cu.

Question 29.
Given the standard electrode potentials,

Solution.

Question 30.
Depict the galvanic cell in which the reaction,
Zn_(s) + 2Ag⁺_(aq) ➝ zn²⁺_(aq) + 2Ag_(s) takes place. Further show:
(i) which of the electrode is negatively charged,
(ii) the carriers of current in the cell and
(iii) individual reaction at each electrode.

Solution.
The given redox reaction is