NCERT Solutions for Class 11 Chemistry Chapter 8 Redox Reactions includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 exams must go through NCERT Solutions for Class 11 Chemistry Chapter 8 Redox Reactions. Going through the solutions provided on this page will help you to know how to approach and solve the problems.
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NCERT Solutions for Class 11 Chemistry Chapter 8 Redox Reactions
NCERT Solutions for Class 11 Chemistry Chapter 8 Redox Reactions are been solved by expert teachers of CBSETuts.com. All the solutions given in this page are solved based on CBSE Syllabus and NCERT guidelines.
Assign oxidation number to the underlined elements in each of the following species :
Let the oxidation no. of underlined element in all the given compounds = x
What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results?
(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.
Justify that the following reactions are redox reactions:
(a) CuO(s) + H2(g) ➝ Cu(5) + H2O(g)
(b) Fe2O3(s) + 3CO(g) ➝ 2Fe(s) + 3CO2(g)
(c) 4BCI3(g) + 3LiAIH4(s) ➝ 2B2H6(g) + 3LiCI(s) + 3AICI3(s)
(d) 2K(s) + F2(g) ➝ 2K+F–(s)
(e) 4NH3(g) + 5O2(g) ➝ 4NO(g) + 6H2O(g)
Fluorine reacts with ice and results in the change:
H2O(s) + F2(g) ➝ HF(g) + HOF(g)
Justify that this reaction is a redox reaction.
Calculate the oxidation number of sulphur, chromium and nitrogen in H2SO5, Cr2O72- and NO3–. Suggest structure of these compounds. Count for the fallacy.
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
Suggest a list of the substances where carbon can exhibit oxidation states from -4 to +4 and nitrogen from -3 to +5.
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?
The oxidation state of sulphur in sulphur dioxide is +4. It can be oxidised to +6 oxidation state or reduced to +2. Therefore, sulphur dioxide acts as a reducing agent as well as oxidising agent. Similarly, the oxidation state of oxygen in hydrogen peroxide is -1. It can be oxidised to O2 (zero oxidation state) or reduced to H2O or OH– (-2 oxidation state) and therefore, acts as reducing as well as oxidising agents.
However, both ozone and nitric acid can only decrease their oxidation number and therefore, act only as oxidising agents.
Consider the reactions :
(a) 6CO2(g) + 6H2O(l) ➝ C6H12O6(ag) + 6O2(g)
(b) O3(g) + H2O2(l) ➝ H2O(l) + 2O2(g)
Also suggest a technique to investigate the path of the above (a) and (b) redox reactions.
The compound AgF2 is unstable compound. However, if formed, the compound acts as a very strong oxidising agent. Why ?
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 oxidising agent.
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.
(i) C is a reducing agent while O2 is an oxidising 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 oxidised to CO2 in which oxidation state of C is + 4.
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 colourless pungent smelling gas HCI, but if the mixture contains bromide then we get red vapour of bromine. Why?
(a) In neutral medium, KMnO4 acts as an oxidant as follows :
MnO4– + 2H2O + 3e– ➝ MnO2 + 40H–
In laboratory, alkaline KMnO4 is used to oxidise toluene to benzoic acid.
Identify the substance oxidised, reduced, oxidising agent and reducing agent for each of the following reactions:
Consider the reactions:
Why does the same reductant, thiosulphate react differently with iodine and bromine?
Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.
The halogens (X2) have strong electron accepting tendency and have positive standard oxidation potential values. They are therefore, powerful oxidising agents. The decreasing order of oxidising powers of halogens is :
Why does the following reaction occur ?
Consider the reactions :
Balance the following redox reactions by ion – electron method:
Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.
What sorts of informations can you draw from the following reaction?
(CN)2(g) + 2OH–(aq) ➝ CN–(aq) + CNO–(aq) + H2O(l)
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.
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.
(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.
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.
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.
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?
The reaction involved in the manufacturing process is :
Using the standard electrode potentials, predict if the reaction between the following is feasible:
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.
(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 oxidised 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 oxidised 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 SO2-4 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.
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.
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.
Given the standard electrode potentials,
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.
The given redox reaction is
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