MP Board Class 11th Chemistry Solutions Chapter 8 Redox Reactions

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MP Board Class 11th Chemistry Solutions Chapter 8 Redox Reactions

MP Board Class 11 Chemistry Redox Reactions Textbook Questions and Answers

Question 1.
Assign oxidation number to the underlined elements in each of the following species:

Solution.
(a) Let the oxidation number of P be x. The oxidation number of each atom is shown above its symbol,

Sum of oxidation numbers of various atoms in NaH₂PO₄ is zero,
hence 1(+ 1) + 2(+ 1) + 1(x) + 4(- 2) = 0 or x – 5 = 0 or x = + 5
Thus, the oxidation number of P in NaH₂PO₄ = + 5.
(b) In the similar way as discussed, the sum of oxidation numbers in will be,
1(+ 1) + 1(+ 1) + x + 4(- 2) = 0 or x = + 6 Thus, the oxidation number of S in NaH₂PO₄ = + 6.
+ 1 x -2
(c) The sum of oxidation numbers in, will be
4(+ 1) + 2(x) + 7(- 2) = 0 or x = + 5 Thus, the oxidation number of P in H₄P₂O₇ = + 5.
(d) The sum of oxidation numbers in , will be
2(+ 1) + l(x) + 4(- 2) = 0 or x = + 6.
Thus, the oxidation number of Mn in K₂MnO₄ = + 6.
(e) The oxidation number of Ca is + 2. Thus, the sum of oxidation numbers in will be
+2+ 2 x = 0 or x = -1 or x = – 1 Thus, oxidation number of oxygen in CaO₂ = – 1.
(f) In NaBH₄, H is present as hydride ion. Therefore, its oxidation number is – 1. Thus,
for , the sum of oxidation numbers is 1(+ 1) + x + 4(- 1) = 0 or x = + 3
Thus, the oxidation number of B in NaBH₄ = + 3.
(g) The sum of oxidation numbers in, will be given as
2(+ 1) + 2(x) + 7(- 2) = 0 or x = + 6
Thus, the oxidation number of S in Na₂S₂O₇ = + 6.
(h) The sum of oxidation numbers in,

+ 1 + 3 + 2x + 8(- 2) + 12(2 x 1 – 2) = 0 or x = + 6
Alternatively, since H₂O is a neutral molecule, therefore, sum of oxidation numbers of all the atoms in H20 may be taken as zero. As such water molecules can be ignored while computing the oxidation number of S. Accordingly,
+ l + 3 + 2x-16 = 0 or x = + 6 Thus, the oxidation number of S in KAl(SO₄)₂.12 H₂O = + 6.

Question 2.
What are the oxidation numbers of the underlined elements in each of the following and how do you rationalize your results?

Solution.
(a) In KI₃, since the oxidation number of K is + 1, therefore, the average oxidation number of iodine = – 1/3. But the oxidation number cannot be fractional. Therefore, in such cases we must consider its structure, ( ). InI₃^– ion, 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 forming the coordinate bond is -1.
(b) Conventional method. The average oxidation number in is calculated as:
2(+ 1) + 4x + 6(- 2) = 0 or x = + 2.5.
But it is wrong because all the four S atoms cannot be in the same oxidation state, in the compound.
Thus, in such cases the oxidation number is calculated by chemical bonding method, i.e., the oxidation number is computed on the basis of its structure. The structure of H₂S₄O₆ is as follows:

The O.N. of each of the S-atoms linked with each other in the middle is zero while that of the remaining two S-atoms is + 5.
(c) Conventional method. The average oxidation number of Fe in will be 3x + 4(- 2) = 0
or x = 8/3.
Fe₃O₄ is a mixed oxide therefore its oxidation number is calculated by Stoichiometry method. Accordingly, Fe₃O₄ shall be written as:
Fe has O.N. of + 2 and + 3.

(d) Conventional method. The average oxidation number of carbon in C₂H₅OH or
will be
2x + 6(+ 1) + 1(- 2) = 0 or x = – 2.

On the basis of chemical bonding method as carbon C-2 is attached to three H-atoms (less electronegative than carbon) img and one CH₂OH group (more electrone- gative than carbon), therefore, O.N. of C-2 will be = 3(+ 1) + x + 1(- 1) = 0 or x = – 2. The carbon, C-l is, however, attached to one OH group (O.N. – 1) and one CH₃ group (O.N. = + 1), therefore, O.N. of C-l = + 1 + 2(+ 1) + x + 1(- 1) = 0 or x = – 2.

(e) Conventional method. The oxidation number of carbon in CH₃COOH or img will be
2x + 4 – 4 = 0 or x = 0.
Chemical bonding method. The carbon C-2 is attached to three H-atoms (less electronegative than carbon) img and one -COOH group (more electronegative than carbon), therefore, O.N. of C-2 = 3(+ 1) + a: + 1(- 1) = 0 or x = -2
The carbon, C-l is, however, attached to one oxygen atom by a double bond, one OH group (O.N. = – 1) and one CH₃ group (O.N. = + 1), therefore, O.N. of C-l = + 1 + x + 1(- 2)+ 1(— 1) = 0 or x + 2.

Question 3.
Try all possible approaches to justify that the following reactions are redox reactions.
(a) CuO(s) + H₂(g) → Cu(s) + H₂O(g)
(b) Fe₂O₃ (s) + 3CO(g) → 2Fe(s) + 3CO₂(g)
(c) 4BCl₃(g) + 3LiAlH₄(s) → 2B₂H₆(g) + 3LiCl(s) + 3A1Cl₃(S)
(d) 2K(s) + F₂(g) → 2K⁺F^–(s)
(e) 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g).
Solution.
(a) The chemical equation with oxidation number of each element shown on its symbol is written as:

Here, O is removed from CuO, therefore, it is reduced to Cu while O is added to H₂ to form H₂O₂, therefore hydrogen is oxidised. Further, O.N. of Cu decreases from + 2 in CuO to 0 in Cu but that of H increases from 0 in H₂ to + 1 in H20. Therefore, CuO is . reduced to Cu but H₂ is oxidised to H₂O. Thus, this is a redox reaction.

(b) The chemical equation of the reaction is:

Here O.N. of Fe decreases from + 3 in Fe₂O₃ to 0 in Fe while that of C increases from + 2 in CO to + 4 in CO₂. Further, oxygen is removed from Fe₂O₃ and added to CO, therefore, Fe₂O₃ is reduced while CO is oxidised. Thus, this is a redox reaction.

(c) The chemical equation along with oxidation number of each element is:

Here, O.N. of Br decreases from + 3 in BrCl₃ to – 3 in B₂H₆ while that of H increases from – 1 in LiAlH₄ to + 1 in B₂H₆. Therefore, BCl₃ is reduced while LiAlH₄ is oxidised. Further, H is added to BrCl₃ but is removed from LiAlH₄, therefore, BCl₃ is reduced while LiAlH₄ is oxidised. Thus, it is a redox reaction.

(d) The chemical equation along with oxidation number of each element will be written as:

Each K atom has lost one electron to form K⁺ while F₂ has gained two electrons to form two F^– ions. Therefore, K is oxidised while F₂ is reduced. Thus, it is redox reaction.
(e) In the equation,

O.N. of N increases from -3 in NH₃ to + 2 in NO while that of O decreases from O in O₂ to-2 inNO or H₂O.
Therefore, NH₃ is oxidised while O₂ is reduced.
Further H has been removed from NH₃ but added to O₂. Therefore, NH₃ has been oxidised while O₂ is reduced. Thus, this is a redox reaction.

Question 4.
Fluorine reacts with ice and results in the change:

Justify that this reaction is a redox reaction. If oxygen of HOF disproportionates at room temperature, then what reaction is possible? Solution.
In the given reaction O.N. of F₂ changes from zero to – 1 in HF and HOF whereas O.N. of oxygen changes from – 2 in H₂O to zero in HOF. Thus, F₂ is reduced, whereas oxygen is oxidised and, therefore, it is a redox reaction.
HOF is a highly unstable molecule and hence decomposes to form O₂ and HF

If oxygen of HOF disproportionates, then in the disproportionation reaction oxygen must have three oxidation states. In simple words, the oxidation state of oxygen in one of the products should be lower than O in HOF whereas in the other product it should be higher. The oxidation state of oxygen is zero in HOF. It decrease its oxidation state to – 2 if HOF gets reduced to H₂O and also it can increase its oxidation state to + 2 if HOF gets oxidised to OF₂. Therefore, the possible reaction is:

Question 5.
Calculate the oxidation number of sulphur, chromium and nitrogen in H₂SO₅, Cr₂O₇^2- and NO₃^–. Suggest structures of these three compounds. Count for the fallacy.
Solution.
(i) Oxidation number of S in H₂SO₅ On the basis of conventional method, the oxidation number of S in H₂SO₅ comes out to be + 8 as shown below.
2 (+ 1) + x + 5(- 2) = 0 or x = + 8
This is impossible because the maximum oxidation number of S cannot be more than six since it has only six electrons in the valence shell. This fallacy is overcome by calcu¬lating the O.N. of S by chemical bonding method.
The structure of H₂SO₅ along with oxidation number of different atoms is:

Hence, 2(+ 1) + 3(- 2) + x + 2(- 1) = 0 or x = + 6

(ii) Oxidation number of Cr in Cr₂O₇^2- According to conventional method, oxidation number of Cr is: x + 5 (- 2) = 0 or x = + 10. This is impossible because the maximum oxidation number of Cr cannot be more than six since it has a maximum of six electrons (3d⁵4s²) in the outer orbital configuration which can participate in bonding. This fallacy is removed by calculating oxidation number of Cr by chemical bonding method. The structure of CrO₅ is:

From the structure, the oxidation number of Cr can be calculated as follows: x + 4(- 1) + 1(- 2) = 0 or x = + 6
It should be noted that four O atoms have peroxy linkage (oxidation number = – 1) and O held by double bond has oxidation number = – 2.
(iii) Oxidation number of N in NO₃^–
According to conventional method, oxidation number of N in NO₃^– = x + 3(- 2) = — 1 or x = + 5
According to chemical bonding method, the structure of nitrate ion is:

Hence the oxidation number of nitrogen is,
x + 3(- 2) = – 1 or x = + 5
Thus, the oxidation number of N in NO₃^– whether one calculates by conventional method or by chemical bonding method is same.

Question 6.
Write formulas for the following compounds:
(a) Mercury (II) chloride
(b) Nickel (II) sulphate
(d) Thallium (I) sulphate
(e) Iron (III) Sulphate
(f) Chromium (III) oxide.
Solution:
(a) Hg(II)Cl₂
(b) Ni(II)SO₄
(c) Sn(IV)O₂
(d) Th₂(I)SO₄
(e) Fe₂(III)(SO₄)₃
(f) Cr₂(III)O₃

Question 7.
Suggest a list of substances where carbon can exhibit oxidation states from – 4 to + 4 and nitrogen from – 3 to + 5.
Solution.
Various compounds of carbon and nitrogen formed in different oxidation states are tabulated below:

Question 8.
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?
Solution.
Any substance can act as an oxidising or reducing agent if one of the elements in the given compound is present in its intermediate oxidation state. Thus, during the reaction it can either increase its oxidation number (reducing agent) or decrease its oxidation number (oxidising agent).
(i) In SO₂, oxidation number of S is + 4. In principle, S can have a minimum oxidation number of – 2 and maximum of + 6. Therefore, S in SO₂ can either decrease or increase its oxidation number and hence can act both as an oxidising as well as a reducing agent.

(ii) In H₂O₂, the oxidation number of O is – 1. In principle, O can have a minimum oxidation number of- 2 and maximum of zero (+ 2 is possible only with 0F2). Therefore, O in H₂O₂ can either decrease its oxidation number from – 1 to – 2 or can increase its oxidation number from – 1 to zero. Therefore, H₂O₂ acts both as an oxidising as well as a reducing agent.

(iii) In O₃, the oxidation number of O is zero. It can only decrease its oxidation number from zero to -1 or – 2, but cannot increase to + 2. Therefore, O₃ acts only as an oxidant.
(iv) In HNO₃, nitrogen atom is in its maximum oxidation state of + 5. Therefore, it can only decrease its oxidation number and hence can act as an oxidant only.

Question 9.
Consider the reactions:
(а) 6CO₂ (g) + 6H₂O(l) → C₆H₁₂O₆(s) + 6O₂(g)
(b) O₃(g) + H₂O₂(l) → H₂O (l) + 2O₂(g)
Why it is more appropriate to write these reactions as;
(a) 6CO₂ (g) + 12H₂O(l) → C₆H₁₂O₆(s) + 6 H₂O(l) + 6O₂(g)
(b) O₃(g) + H₂O₂(l) → H₂O (l) + O₂(g) + O₂(g)
Also suggest a technique to investigate the paths of above (a) and (b) redox reactions.
Solution.
(a) The given reaction occurs during photosynthesis. Although the mechanism of photosynthesis is very complex but the process/reaction may be visualized to occur in two steps.
(i) In the first step, H₂O decomposes to give H₂ and O₂ in presence of chlorophyll.
(ii) In step 2, the H₂ thus produced reduces CO₂to C₆H₁₂O₆ along with the formation of some H₂O molecules as shown on next page.

It is more appropriate to write the equation for photosynthesis as equation (iii) because the equation clearly illustrates that 12H₂Omolecules are used per molecule of carbohydrate formed and 6H₂O molecules are produced during the process.

(b) In equation (b) dioxygen gas is written twice. The purpose of writing O₂ two times suggests that O₂ is obtained from each of the two reactants as shown by equations given below:

The path of reactions (a) and (b) can be determined by tracer techniques i.e., by using different isotopes of elements. For example, using H₂O¹⁸ in reaction (a) or by using H₂O¹⁸ or O₃¹⁸ in reaction (b).

Question 10.
The compound AgF₂ is unstable. However, if formed, the compound acts as a very strong oxidising agent. Why?
Solution.
In AgF₂, oxidation state of Ag is + 2 which is very very unstable. Since Ag can exist in a stable state of + 1, 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.

Question 11.
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 oxidising agent is in excess. Justify this statement giving three illustrations.
Solution.
(i) Illustration 1. Let us consider reaction between C (reducing agent) and O₂ (oxidising agent). If excess of carbon is burnt in a limited supply of O₂, CO is formed in which the oxidation state of C is + 2. If, however, excess of O₂ is used, the initially formed CO gets oxidised to C02 in which the oxidation state of C is + 4.

(ii) Illustration 2. Reaction between P₄ (reducing agent) and Cl₂ (oxidising agent). When excess of P₄ is used, PCl₃ is formed in which the oxidation state of P is + 3. If, how¬ever, excess of Cl2 is used, the initially formed PCl₃ reacts further to form PCl₅ in which the oxidation state of P is + 5


(iii) Illustration 3. Reaction between Na (reducing agent) and O₂ (oxidising agent). When excess of Na is used, sodium oxide is formed in which the oxidation state of O is – 2. If, however, excess of O₂ is used, Na₂O₂ is formed in which the oxidation state of O is – 1 which is higher than – 2.

Question 12.
How do you account for the following observations:
(а) 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 to an inorganic mixture containing chloride, we get pungent-smelling gas HCl, but if the mixture contains bromide then we get red vapour of bromine. Why?
Solution.
(a) Toluene can be oxidised to benzoic acid in acidic, basic and neutral medium as per the following redox reactions
(i) Acidic medium

(ii) In Basic and neutral medium

(b) When cone. H₂SO₄ is added to an inorganic mixture containing chloride, a pungent smelling gas, HCl, is produced because a stronger acid can displace a weaker acid from its salt.
2NaCl + 2H₂SO₄ → 2NaHSO₄ + 2HCl
Since HCl is a very weak reducing agent, it cannot reduce H₂SO₄ to SO₂ and hence HCl is not oxidised to Cl₂ i.e., the following reaction will not be possible.
2HCl + H₂SO₄ → Cl₂ + SO₂ + 2H₂O

However, when the mixture contains bromide ion, the initially produced HBr being a stronger reducing agent reduces H₂SO₄ to SO₂ and is itself oxidised to produce red vapour of Br₂.
2NaBr + 2H₂SO₄ → 2NaHSO₄ + 2HBr
2HBr + H₂SO₄ → Br₂ + SO₂ + 2H₂O

Question 13.
Identify the substance oxidised, reduced, oxidising agent and reducing agent for each of the following reactions.
(a) 2AgBr (s) + C₆H₆O₂(aq) → 2Ag(s) + 2HBr(aq) + C₆H₄O₂(aq)
(b) HCHO(l) + 2[Ag(NH₃)₂]+(aq) + 3OH^–(aq) →
2Ag(s) + HCOO^– (aq) + 4NH₃(aq) + 2H₂O(l)
(c) HCHO(l) + 2Cu²⁺ (aq) + 5OH^–(aq) → Cu₂O(s) + HCOO^–(aq) + 3H₂O(l)
(d) N₂H₄(l) + 2H₂O₂(l) → N₂(g) + 4H₂O(l)
(e) Pb(s) + PbO₂(s) + 2H₂SO₄(aq) → 2PbSO₄(s) + 2H₂O(l)
Solution.
The list of substance oxidised, reduced, etc., for the given reactions are tabulated below :

Question 14.
Consider the reactions:
2S₂O₃^2-(aq) + I₂(s) → S₄O₆^2-(aq) + 2I^–(aq)
S₂O₃^2- (aq) + 2Br₂(l) + 5H₂O(l) → 2SO₄^2-(aq) + 4Br^–(aq) + 10H⁺(aq)
Why does the same reductant, thiosulphate react differently with iodine and bromine?
Solution.
The average O.N. of S in S₂O₃^2- is + 2 while in S₄O₆^2- it is + 2.5. The O.N. of S in SO₄^2- is + 6. Since Br₂ is a stronger oxidising agent that I₂, it oxidises S of S₂O₃^2- to a higher oxidation state of + 6 and hence forms SO₄^2- ion. I₂, however, being a weaker oxidising agent oxidises S of S₂O₃^2- ion to a lower oxidation of + 2.5 in S₄O₆^2-ion. It is because of this reason that thiosulphate reacts differently with Br₂ and I₂.

Question 15.
Justify giving reactions that among halogens, fluorine is the best oxidant and among Hydrohalic compounds, hydriodic acid is the best reductant.
Solution. The oxidising power of halogens decreases in the order: F₂ > Cl₂ > Br₂ > I₂. This is evident from the observation that F₂ oxidises Cl^– to Cl₂, Br^– to Br₂,I^– to I₂ ; Cl₂ oxidises Br^– to Br₂ and I^– to I₂ but not F to F₂.Br₂, however, oxidises I^– to I₂ but not F^– to F₂, and Cl^– to Cl₂.
F₂(g) + 2Cl^–(aq) → 2F^–(aq) + Cl₂(g);
F₂(g) + 2Br^–(aq) → 2F^–(aq) + Br₂(l)
F₂(g) + 2I ^–(aq) → 2F^–(aq) + I₂(s);
Cl₂(g) + 2Br^– (aq) → 2Cl^– (aq) + Br₂(l)
Cl₂(g) + 2I^–(aq) → 2Cl^–(aq) + I₂(s)
and Br₂( l) + 2I^– → 2Br^–(ag) + I₂(s)
Thus, F₂ is the best oxidant.

Among hydrohalic acids, the reducing power decreases in the order: HI > HBr > HC1 > HF. Thus, HI and HBr reduce H₂SO₄ to SO₂ while HCl and HF do not.
2HBr + H₂SO₄ → Br₂ + SO₂ + 2H₂O
2HI + H₂SO₄ → I₂ + S02 + 2H20
Further I^– reduces Cu²⁺ to Cu⁺ but Br^– does not.
2Cu²⁺(aq) + 4I^–(aq) → Cu₂I₂(s) + I₂(aq)
Cu²⁺(aq) + 2Br^– → No reaction.

Further among HCl and HF, HCl is a stronger reducing agent than HF because HCl reduces MnO₂ to Mn²⁺ but HF does not.
MnO₂(s) + 4HCl(aq) → MnCl₂(aq) + Cl₂(g) + 2H₂O
MnO₂(s) + 4HF(l) → No reaction.
Thus, the reducing character of hydrohalic acids decreases in the order: HI > HBr > HCI > HF.

Question 16.
Why does the following reaction occur?
XeO₆^4-(aq) + 2F^–(aq) + 6H⁺(aq) → XeO₃ (s) + F₂(g) + 3H₂O(l)
What conclusion about the compound Na₄XeO₆ (of which XeO₆^4- is a part) can be drawn from the reaction?
Solution.
The balanced equation along with O.N. of the elements above their symbols will be as:

In the equation the, O.N. of Xe decreases from + 8 in XeO₆^4- to + 6 in XeO₃ while that of F increases from – 1 in F^– to 0 in F₂. Therefore, XeO₆^4- is reduced while F~ is oxidised. This reaction occurs because Na₂XeO₆^4- (or XeO₆^4-) is stronger oxidising agent than F₂.

Question 17.
Consider the reactions:
(а) H₃PO₂(aq) + 4AgNO₃(aq) + 2H₂O(l) → H₃PO₄(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) →
C₆H₅CHO(aq) + 2Ag(s) + 4NH₃(aq) + 2H₂O(l)
(d) C₆H₅CHO(l) + 2Cu²⁺(aq) + 5OH (aq) → No change observed.
What inference do you draw about the behaviour of Ag⁺ and Cu²⁺ from these reactions?
Solution.
Reactions (a) and (b) indicate that H₃PO₂ (hypophosphorous acid) is a reducing agent and thus reduces both AgNO₃ and CuSO₄ to Ag and Cu respectively. Conversely, both AgNO₃ and CuSO₄ act as oxidising agent and thus oxidise H₃PO₂ to H₃PO₄. Reaction (c) suggests that [Ag(NH₃)₂]⁺ oxidises C₆H₅CHO (benzaldehyde) to C₆H₅COO^– (benzoate ion) but reaction (d) indicates that Cu²⁺ ions cannot oxidise C₆H₅CHO to C₆H₅COO^–. Therefore, from the above reactions, we conclude that Ag⁺ ion is a stronger oxidising agent than Cu²⁺ ion.

Question 18.
Balance the following redox reactions by ion-electron method.
(а) MnO₄^–(aq) + I^–(aq) → MnO₂(s) + I₂(s) (in basic medium)
(b) MnO₄^– (aq) + SO₂(g) → Mn²⁺(aq) + HSO₄^– (in acidic solution)
(c) H₂O₂(aq) + Fe²⁺(aq) → Fe³⁺(aq) + H₂O(l) (in acidic solution)
(d) Cr₂O₇^2-(aq) + SO₂(g) → Cr³⁺(aq) + SO₄^2-(aq) (in acidic solution)
Solution.
(a) Step 1. The Skeletal equation of the reaction is:

Step 2. Writing the two half-reactions:
Oxidation half reaction, I^–(aq) →

Step 3. Balancing of atoms in the half equation 2I^–(aq) → I₂(s)
In the other equation the step is not required because Mn is balanced.

Step 4. Balancing of charges in the half-reactions by adding electrons
2I^–(ag) → I₂(s) + 2e^–

Step 5. Balancing of oxygen atoms and H atoms
MnO₄^– (aq) + 3e^– → MnO₂(s) + 2H₂O(l)
To balance H atoms, we add 4H⁺ ions on the left
MnO₄^–+ 4H⁺(aq) → MnO₂(s) + 2H₂O(l)

As the reaction is taking place in basic solution, therefore for the 4H⁺ ions we add 4 OH^– ions to both sides of the equation.
MnO₄^–(aq) + 4H+(aq) + 4OH^–(aq) → MnO₂(s) + 2H₂O(l) + 4OH^–(aq)
As H⁺ and OH^– ions form water, therefore the resultant equation is
MnO₄^–(aq) + 2H₂O(l) + 3e^– → MnO₂(s) + 4OH^–(aq)

Step 6.
(a) Balancing of electrons in the two half equations
2I^–(aq) → I₂(s) + 2e^– ………………………………………..(i)
MnO₄^–(aq) + 2H₂O(l) + 3e^– → MnO₂(s) + 4OH^–(aq) …………………………………………(ii)
Multiply (i) with 3 and (ii) with 2 and add
6I^–(aq) + 2MnO₄^–(aq) + 4H₂O(l) → 3I₂(s) + 2MnO₂(s) + 8OH^–(aq)

(b) By applying the procedure as discussed in the equations above Oxidation half equation will be:
SO₂(g) + 2H₂O(l) →HSO₄^–(aq) + 3H⁺ (aq) + 2e^– ……………………………….(i)
The reduction half equation will be:
MnO₄^–(aq) + 8H⁺(ag) + 5e^–→ Mn²⁺(aq) + 4H₂O(l) ……………………………………………………..(ii)
Multiply equation (i) by 5 and equation (ii) by 2 and adding the equations, we get,
2MnO₄^–(aq) + 5SO₂(g) + 2H₂O(l) + H⁺(aq) → 2Mn²⁺ (aq) + 5HSO₄^–(aq)

(c) By applying the first 5 steps as discussed above, the oxidation half equation will be:
Fe²⁺(aq) → Fe³⁺(aq) + e^– ……………..(i)
The reduction half equation will be:
H₂O₂(aq) + 2H⁺(ag) + 2e^– → 2H₂O(l) …………………(ii)
Multiply equation (i) by 2 and adding it to equation (ii) we get
H₂O₂(aq) + 2Fe²⁺ (aq) + 2H⁺ (aq) → 2Fe³⁺ (aq) + 2H₂O(l)

(d) Following the procedure discussed in details and applying 5 steps the balanced half-reaction equations are:
Oxidation half equation:
SO₂(g) + 2H₂O(l) → SO₄^2-(aq) + 4H⁺(aq) + 2e^– ……………………(i)

Reduction half equation:
Cr₂O₇^2-(ag) + 14H⁺(aq) + 6e^– → 2Cr³⁺ (aq) + 7H₂O(l) …………………………………………………(ii)
Multiplying equation (i) by 3 and adding it to equation (ii), we get
Cr₂O₇^2-(aq) + 3SO₂(g)+2H⁺(aq) → 2Cr³⁺ (aq) + 3SO₄^2- (aq) + H₂O(l).

Question 19.
Balance the following equations in basic medium by ion-electron method and oxidation number method and identify the oxidising agent and the reducing agent.
(а) P₄(s) + OH^– (aq) → PH₃(g) + HPO₂^–(aq)
(b) N₂H₄(l) + CIO₃^–(aq) → NO(g) + Cl^–(aq)
(c) Cl₂O₇(g) + H₂O₂(aq) → CIO₂^–(aq) + O₂(g) + H⁺
Solution.

P₄ acts both as an oxidising as well as reducing agent.
Oxidation number method
Total decrease in O.N. of P₄ in PH₃ = 3 × 4 = 12
Total increase in O.N. of P₄ in H₂PO₂^– =1×4 = 4
(i) To balance increase/decrease in O.N. multiply PH₃ by 1 and H₂PO₂^– by 3. After doing so we get the equation,
P₄(s) + OH^–(ag) → PH₃(g) + 3H₂PO₂^–(aq)

(ii) The oxygen atoms are balanced by multiplying OH^– by 6 and we get the equation,
P₄(s) + 6OH^–(aq) → PH₃(g) + 3H₂PO₂^–(aq)

(iii) To balance H atoms, 3H₂O are added to L.H.S. and 3 OH^– to the R.H.S. Thus, the equation is,
P₄(s) + 6OH^–(aq) + 3H₂O(l) → PH₃(g) + 3H₂PO₂^–(aq) + 3OH^–(ag)
or The correct balanced equation is:
P₄(s) + 3OH^–(aq) + 3H₂O(l) → PH₃(g) + 3H₂PO₂^–(ag) ………………………………. (i)
Ion electron method. The two half-reactions are:
Oxidation half-reaction:
P₄(s) → H₂PO₂^– …………………………… (ii)
Balancing P atoms, we have,

Balance O.N. by adding electrons,
P₄(s) → 4H₂PO₂^–(ag) + 4e^–
Balance charge by adding OH- ions
P₄(s)+ 8OH^–(ag) → 4H₂PO₂^–(ag) + 4e^– ………………… (iii)
O and H get automatically balanced. Thus, equation (iii) represents the balanced oxidation half reaction.
Reduction half reaction:

Balancing P atoms, we have,
P₄(s) →4PH₃ (g)
Balance O.N. by adding electrons,
P₄(s) + 12e^–→ 4PH₃(g)
Balance charge by adding OH- ions,
P₄(s) + 12e^– →4PH₃(g) + 12OH^–(ag)
Balance O atoms by adding 12H₂O to L.H.S. of above equation.
P₄(s) + 12H₂O(l) + 12e^– → 4PH₃(g) + 12OH^–(ag) …………………………………………… (v)

To cancel out electrons, multiply equation (iii) by 3 and add it to equation (v), we have,
4P₄(s) + 24OH^–(ag) + 12H₂O(l) → 4PH₃(aq)
+ 12H₂PO₂^–(aq) + 12 H₂O(l) + 12OH^–(ag)
or P₄(g) + 3 OH^–(aq) + 3H₂O(l) → PH₃(ag) + 12H₂PO₂^–(aq) ……………………………………………………….. (vi)
Thus, equation (vi) represents the correct balanced equation.

Therefore, N₂H₄ acts as the reducing agent while CIO₃^– acts as the oxidising agent.
Oxidation number method
Total increase in oxidation number of N = 2×4 = 8
Total decreases in oxidation number of Cl = 1 × 6 = 6
Therefore, to balance increase/decrease in oxidation number multiply N₂H₄ by 3 anc CIO₃^–by 4, we have,
3N₂H₄(l) + 4CIO₃^–(ag) → NO(g) + Cl^–(aq)
To balance N and Cl atoms, multiply NO by 6 and Cl^– by 4, we have,
3N₂H₄(l) + 4CIO₃^–(aq) → 6NO(g) + 4Cl^–( aq)
Balance O atoms by adding 6H₂O,
3N₂H₄(l) + 4CIO₃^– (aq) → 6NO(g) + 4Cl^–( aq) + 6H₂O(l) …………………………………(i)
H atoms get automatically balanced and thus equation (Z) represents the correct balanced equation.
Ion electron method. Two half-reactions are:
Oxidation half equation:

Balance N atoms,
N₂H₄(l) → 2NO(g)
Balance O.N. by adding electrons,
N₂H₄(l) → 2NO(g) + 8e^–
Balance charge by adding OH- ions,
N₂H₄(l) + 8OH^–(aq) → 2NO(g) + 8e^–
Balance O atoms by adding 6H₂O,
N₂H₄(l) + 8OH^–(aq) → 2NO(g) + 6H₂O(l) + 8e^– ………………………………………(iii)
Thus, equation (ii) represents the correct balanced oxidation half equation.
Reduction half equation:

Balance oxidation number by adding electrons,
CIO₃^– (aq)+ 6e^– → Cl^–(aq)
Balance charge by adding OH ions,
CIO₃^– (aq) + 6e^–→ Cl^–(aq) + 6OH^–(aq)
Balance O atoms by adding 3H₂O,
CIO₃^– (aq) + 3H₂O(l) + 6e^–→ Cl^–(aq)+ 6OH^–(aq) ………………………………….. (iii)
Thus, equation (iii) represents the correct balanced reduction half equation.
To cancel out electrons gained and lost, multiply equation (ii) by 3 and equation (iii) by 4 and add, we have,
3N₂H₄(l)+ 4CIO₃^–(aq) → 6NO(g) + 4Cl^–(aq) + 6H₂O(l)
This represents the correct balanced equation.

Thus, Cl₂O₇(g) acts an oxidising agent while H₂O₂(aq) as the reducing agent. Oxidation number method
Total decrease in oxidation number of Cl₂O₇ = 4× 2 = 8 Total increase in oxidation number of H₂O₂ = 2 × 1 = 2
∴ To balance increase/decrease in O.N. multiply H₂O₂ and O₂ by 4, we have,
Cl₂O₇(g) + 4H₂O₂(aq) → CIO₂^–(aq) + 4O₂(g)
To balance Cl atoms, multiply CIO₂^– by 2, we have,
Cl₂O₇(g) + 4H₂O₂(aq) → 2CIO₂^–(aq) + 4O₂(g)
To balance O atoms, add 3H₂O to R.H.S., we have,
Cl₂O₇(g) + 4H₂O₂(aq) → 2CIO₂^–(aq) + 4O₂(g) + 3H₂O(l)
To balance H atoms, add 2H₂O to R.H.S., and 2OH^– to L.H.S., we have,
Cl₂O₇(g) + 4H₂O₂(aq) + 2OH^–(g) → 2CIO₂^–(aq) + 4O₂(g) + 5H₂O
This represents the balanced redox equation.
Ion electron method. Two half-reactions are:
Oxidation half equation:

Balance O.N. by adding electrons,
H₂O₂(aq) → O₂(g) + 2e^–
Balance charge by adding 0H~ ions,
H₂O₂(aq) + 2OH^–(aq) → O₂(g) + 2e^–
Balance O atoms by adding H₂O,
H₂O₂(aq) + 2OH^–(aq) → O₂(g) + 2H₂O(l) + 2e^– ……………………………… (i)
Reduction half equation:

Balance Cl atoms;
Cl₂O₇(g) → 2CIO₂^–(aq)
Balance oxidation number by adding electrons,
2CIO₂^–(aq) + 8e^– → 2CIO₂^–(aq)
Add OH ions to balance charge,
Cl₂O₇(g) + 8e^– → 2CIO₂^–(aq) + 6OH^–
Balance O atoms by adding 3H20 to L.H.S., we have,
Cl₂O₇(g) + 3H₂O(l) + 8e^– → 2CIO₂^–(aq) + 6OH^–(aq) …………………………………………….. (ii)

To cancel out electrons, multiply equation (t) by 4 and add it to equation (it), we have,
4H₂O₂(aq) + 8OH^–(aq) + Cl₂O₇(g) + 3H₂O(l) →
2CIO₂^–(aq) + 6OH^–(aq) + 4O₂(g) + 8H₂O(l)
or Cl₂O₇(g) + 4₂O₂(aq) + 2HO^–(aq) → 2CIO₂^–(aq) + 4O₂(g) + 5H₂O(l).

Question 20.
Write four informations about the reaction:
(CN)₂(g) + 2OH^–(aq) → CN^–(aq) + CNO (aq) + H₂O(l)
Solution.

1. It is a disproportionate reaction.
2. Cyanogen (CN)₂ gets simultaneously reduced to CN^– ion as well as oxidised to cyanate ion.
3. Oxidation number of N in (CN)₂ is – 3 while that in CN^– is – 2 and in CNO^– is – 5.
4. The reaction occurs in a basic medium.

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.
The skeletal equation is:
Mn³⁺ (aq) → Mn²⁺(aq) + MnO₂ (s) + H⁺(aq)
Oxidation half equation:

Balance oxidation number by adding electrons,
Mn³⁺ (aq) → MnO₂ (s) + e^–
Balance charge by adding H⁺ ions,

Balance O atoms by adding H₂O,

Reduction half equation:
Mn³⁺ (aq) → Mn²⁺
Balance oxidation number by adding electron,
Mn³⁺ (aq)+ e^– → Mn²⁺(aq) ……………………………………………… (ii)
Adding equation (i) and equation (ii), the balanced equation for the disproportionation reaction is
2Mn³⁺ (aq) + 2H₂O(l) → MnO₂ (s) + Mn²⁺(aq) + 4H⁺(aq).

Question 22.
Consider the elements: Cs, Ne, I, F
(a) Identify the element that exhibits -ve oxidation state.
(b) Identify the element that exhibits +ve oxidation state.
(c) Identify the element that exhibits both +ve and -ve oxidation states.
(d) Identify the element which neither exhibits -ve nor +ve oxidation state.
Solution.
(a) Fluorine (F) being most electronegative element shows only a -ve oxidation state of- 1.
(b) Cesium (Cs) is a highly electropositive alkali metal and has a single electron in the valence shell. Therefore it can exhibit an oxidation state of + 1.
(c) Iodine (I) because of the presence of seven electrons in the valence shell, shows an oxidation state of – I and because of the presence of d-orbitals it also exhibits +ve oxidation states of + 1, + 3, + 5 and + 7.
(d) Neon (Ne) is an inert gas and hence it neither exhibits -ve nor +ve 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 the reaction for this redox change taking place in water.
Solution.
The skeletal equation for the process would be:
Cl₂(aq) + SO₂(aq) + H₂O(l) → 2Cl^–(aq) + SO₄^2-(aq)
Reduction half equation:
Cl₂(aq) → Cl^–(aq)
Balance Cl atoms,

Balance oxidation number by adding electrons:
Cl₂(aq)+ 2e^– → 2Cl^–(aq) ……………………………………………. (i)
Oxidation half equation:

Balance oxidation number by adding electrons:
SO₂(aq) → SO₄^2-(aq) + 2e^–
Balance charge by adding H⁺ ions:
SO₂(aq) → SO₄^2-(aq) + 4H⁺(aq) + 2e^–
Balance O atoms by adding 2H₂O,
SO₂(aq) + 2H₂O(l) → SO₄^2-(aq) + 4H⁺(aq) + 2e^– ……………………………………………. (ii)
Adding equation (i) and equation (ii), we get balanced redox reaction as,
Cl₂(aq) + SO₂(aq) + 2H₂O(l) → 2Cl^–(aq) + SO₄^2-(aq) +4H⁺(aq)

Question 24.
Refer to the periodic table given in your NCERT book and now answer the following questions:
(a) Select the possible non-metals that can show disproportionation reaction. (6) Select three metals that show disproportionation reaction.
Solution.
(a) The non-metals are: phosphorus, chlorine and sulphur. The disproportionation reactions shown by them are given as follows:
(i) P₄(s) + 3OH^–(aq) + 3H₂O(l) → PH₃(g) + 3H₂PO₂^–(aq)
(ii) Cl₂(aq) + 3OH^–(aq) → Cl^–(aq) + CIO^–(aq) + H₂O(l)
(iii) S₈(s) + 12OH^–(aq) → 4S²(aq) + 2S₂O₃^2-(aq) + 6H₂O(l)

(b) The metals are: copper, gallium and indium etc. The disproportionation reactions given by the metal ions Cu⁺, Ga⁺ and In⁺ are given below:
(i) 2Cu⁺(aq) → Cu²⁺(aq) + Cu(s)
(ii) 3Ga⁺(aq) → Ga³⁺(aq) + 2Ga(s)
(iit) 3In⁺(aq) → In³⁺(aq) + 2In(s)

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 10.0 g of ammonia and 20.0 g of oxygen?
Solution.
The balanced equation for the reaction is:

(I) Identifying limiting reagent
Here, 160 g of O₂ will react with NH₃ = 68 g
∴ 20 g of O₂ will react with NH₃ =\(\frac{68}{160}\) × 20 = 8.5 g
Thus,O₂ is the limiting reagent, accordingly the amount of nitric oxide in the reaction would depend upon the amount of 02 taken and not on the amount of NH₃ taken.
(II) Calculating amount of NO formed
160 g of O₂ produce NO = 120 g
∴ 20 g of 02 will produce NO =\(\frac{120}{160}\) × 20 = 15 g.

Question 26.
Using standard electrode: Predict if the reaction between as the following is feasible.
(i) Fe³⁺ (aq) and I^– (aq)
(ii) Ag+ (aq) and Cu
(iii) Fe3+ and Br~ (aq)
(iv) Ag and Fe3+ (aq)
(v) Br₂ (aq) and Fe²⁺ (aq)
Given E°I₂/I = 0.541 V, E°_Cu²⁺/Cu = 0.34 V, E°_Br2/Br^–= 1.09 V, E°_Ag⁺/Ag = 0.80 V,
Solution.
A redox reaction is feasible if its E° is positive
(i) Fe₃₊ (aq) + I^– (aq) ⇌ Fe²⁺ (aq) + \(\frac{1}{2}\) I₂
E°_Redox = E°_{Reduced species} – E°_{Oxidised species}
= E°_{Fe³⁺/Fe²⁺} – E°I₂/I
= 0.77 -0.54 = + 0.23 V (Feasible)
(ii) 2Ag⁺ (aq) + Cu ⇌ 2Ag(s) + Cu²⁺(aq)
E°_Redox = E°_Ag⁺/Ag – E° _Cu²⁺/Cu
= + 0.80 – 0.34 = + 0.46 V (Feasible)
(iii) Fe³⁺ (aq) + Br^– (aq) ⇌ Fe²⁺ (aq) + \(\frac{1}{2}\) Br₂
E°_Redox = E°_{Fe³⁺/Fe²⁺} – E° _Br2/Br^–
= 0.77 — 1.09 = – 0.32 V (Not Feasible)
(iv) Ag(s) + Fe³⁺ (aq) ⇌ Ag⁺(aq) + Fe²⁺(aq)
E°_Redox = E°_{Fe³⁺/Fe²⁺} – E°_Ag⁺/Ag
= 0.77 – 0.80 = – 0.03 V (Not Feasible)
(v) \(\frac{1}{2}\) Br₂ (aq) + Fe²⁺ (aq)⇌ Br ^–(aq)+ Fe³⁺(aq)
E°_Redox = E°_{Reduced species} – E°_{Oxidised species}
= E°_Br2/Br^– – E°_{Fe³⁺/Fe²⁺}
= + 1.09 – 0.77 = + 0.32 V (Feasible)

Question 27.
Predict the products of electrolysis in each of the following:
(i) An aqueous solution of AgNOs with silver electrodes.
(ii) An aqueous solution of AgN03 with platinum electrodes.
(iii) A dilute solution of sulphuric acid using platinum electrodes.
(iv) An aqueous solution of CuCl2 with platinum electrodes.
Solution.
(i) Anode Ag → Ag⁺ + e^–
Cathode Ag⁺ + e^– → Ag
(ii) Cathode Ag⁺ + e^–→ Ag
Anode 2OH^– → H₂O +\(\frac{1}{2}\)O₂ + 2e^–

(iii) In dilute H₂SO₄ electrolysis of
H₂O ⇌ H⁺ + OH^–
Cathode 2H⁺ + 2e^– → H₂ ↑
Anode 2OH^– → H₂O +\(\frac{1}{2}\)O₂ + 2e^–
(iv) CuCl₂ → Cu²⁺ + 2Cl^–
Cathode Cu²⁺ + 2e^– →Cu
Anode 2Cl^– → Cl₂ + 2e^–

Question 28.
Arrange the following metals in the order in which they displace each other from their salts.
Al, Cu, Fe, Mg and Zn
Solution:
This is based upon the relative positions of these metals in the activity series.
The metal placed lower in the series can displace the metals occupying a higher position present as its salt. Based upon this, the correct order is:
Mg, Al, Zn, Fe, Cu.

Question 29.
Given the standard electrode potentials
K⁺/K = – 2.93 V, Ag⁺/Ag = 0.80 V Hg²⁺/Hg = 0.79 V; Mg²⁺/Mg = – 2.37 V, Cr³⁺/Cr= – 0.74 V
Arrange these metals in increasing order of their reducing power.
Solution:
If maybe noted that lesser the E° value for an electrode, more will be reducing power. The increasing order of reducing power is:
Ag < Hg < Cr < Mg < K.

Question 30.
Depict the galvanic cell in which the reaction Zn(s) + 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag(s) takes place.
Further show:
(i) which electrode is negatively charged.
(ii) the carriers of the current in the cell.
(iii) Individual reaction at each electrode.
Solution:
The galvanic cell for the reaction is
Zn(s)|Zn²⁺(aq) || 2Ag⁺(aq) 12Ag(s)
(i) Zinc electrode (anode) is negatively charged.
(ii) Current flows from silver to zinc in outer circuit
(iii) Anode: Zn(s) → Zn²⁺(aq) + 2e^– (oxidation)
Cathode: 2Ag⁺(aq) + 2e^– → 2Ag(s) (reduction)

MP Board Class 11 Chemistry  Redox Reactions Important Questions and Answers

Question 1.
The formation of sodium chloride from gaseous sodium and gaseous chlorine is a redox process, justify.
Answer:
Na atoms get oxidised and chlorine is reduced.

Question 2.
When magnesium ribbon is burnt in air, two products are formed, magnesium oxide and magnesium nitride. Point out the oxidising and reducing agent.
Answer:
3Mg + N₂ → Mg₃N₂ ; 2Mg + O₂→ 2MgO.
Mg is reducing agent N₂ and O₂ are oxidising agents.

Question 3.
Give cathodic reaction involved in the electrolysis of acidified water.
Answer:
H₂O + e^– → OH^– + \(\frac{1}{2}\) H₂(g)

Question 4.
2A + H₂SO₄ → A₂SO₄ + H₂. Give the representation of the cell which involves the above redox reaction.
Answer:
A/A +(aq) || H⁺(aq)/H₂, Pt

Question 5.
What is meant by disproportionation? Give one example.
Answer:
A process in which same element undergoes increase as well as decrease in oxidation number.

Question 6.
A cell is set up by the two electrodes Cu/Cu²⁺ and Al/Al³⁺. What is the net cell reaction?
Answer:
2Al + 3Cu²⁺ → 2Al³⁺ + 3Cu

Question 7.
Standard electrode potentials of some elements are given as X( – 1.66 V), Y(+ 0.32 V), Z(- 3.00 V), E(+ 0.8 V) and G(- 0.76 V). Which one is the strongest oxidising agent and which one will be the strongest reducing agent?
Answer:
Strongest oxidising agent: E; Strongest reducing agent: Z.

Question 8.

Answer:
Oxidation number of Cr = + 6
Four O atoms involved in peroxide linkage have O.N. = – 1 while the-fifth O atom has O.N. = – 2

Question 9.
What is the oxidation number of Pt in [Pt (C₂H₄) Cl₃]^–?
Answer:
x+0-3=-1
∴ x = + 4.

Question 10.
In the reaction given below which species is called spectator ion?
Zn + Cu²⁺ + SO₄^2- → Zn²⁺ + Cu + SO₄^2-.
Answer:
SO₄^2- ions do not participate in reaction and thus, they are spectator ions.

Question 11.
A solution of silver nitrate was stirred with iron rod. Will it cause any change in the concentration of silver and nitrate ions?
Answer:
The concentration of Ag⁺ ions will decrease but NO₃^– ions will not have any change in concentration
2Ag⁺ + Fe → 2Ag + Fe²⁺.

Question 12.
Can we use KCl as electrolyte in the following cell Cu/Cu²⁺ || Al⁺/Al.
Answer:
KCl cannot be used in salt bridge in the given cell, because it will undergo chemical reaction with Ag⁺ ions to form a precipitate of AgCl.

Question 13.
What is oxidation numbers of C in HCN and HNC?
Answer:

Question 14.
A layer of CuO on copper vessel can be easily cleaned by dilute HCl but oxide layer of aluminium on the aluminium vessel cannot be cleaned by HCl. Give reason.
Answer:
It is because Cu falls beneath hydrogen in the electrochemical series and hence copper from Cu⁺² state can be easily reduced to Cu. On the other hand Al falls above H in electrochemical series and hence Al⁺³ cannot be reduced to Al.

Question 15.
An iron nail accidentally fell into a 1 M solution of sliver nitrate placed in glass vessel. What observations will be made?
Answer:
As reduction potential of Fe is less than Ag therefore Fe will get oxidised to Fe⁺³ whereas Ag⁺ will get reduced to Ag. In simple words the iron nail will get rusted.

Question 16.
Consider the following redox reactions that produce electricity in a galvanic cell:
(i) 2Fe³⁺ + 2Cl^– → 2Fe²⁺ + Cl₂(g)
(ii) Cd(s) + I₂ → Cd²⁺ + 2I^–
(iii) Cu²⁺ + Cr → Cu + Cr³⁺
Write the anode and cathode reactions for galvanic cell.
Answer:
(i) Anode: 2Cl^– → Cl₂ + 2e^– ;
Cathode-. Fe³⁺ +e^– → Fe²⁺
(ii) Anode: Cd → Cd²⁺ + 2e^– ;
Cathode: I₂ + 2e^– → 2I^–
(iii) Anode: Cr → Cr³⁺ + 3e^– ;
Cathode: Cu²⁺ + 2e^– → Cu

Question 17.
Draw the diagram for the galvanic cell which would have overall chemical reaction as
Zn + 2Ag⁺ → Zn²⁺ + 2Ag
Answer the following:
(i) Write the reactions occurring at each electrode.
(ii) In which directions do the electrons flow in the external circuit?
(iii) Name the salt to be taken in salt bridge.
(iv) How is the electrical neutrality maintained in the solutions of the two half cells?
(v) Label the anode and cathode.
(vi) How does the EMF change when the concentration of silver ions is decreased?
Answer:
(i) Anode: Zn → Zn²⁺ + 2e^– ;
Cathode: Ag⁺ + e^– → Ag
(ii) From Zn to Ag
(iii) KC1
(iv) By migration of ions through salt bridge (v) Zn (anode), Ag (cathode)
(vi) EMF decreases.

Question 18.
Starting with the correctly balanced half reactions, write the overall ionic reaction for the following changes:
(i) Chloride ion is oxidised to Cl₂ by MnO₄^– in acid solution
(ii) Nitrous acid (HNO₂) reduces Mn04″ in acid solution
(iii) Nitrous acid (HNO₂) oxidises I^– to I₂ in acid solution
Answer:
(i) 2MnO₄^– + 10Cl^– + 16H⁺ → 2Mn²⁺ + 8H₂O + 5Cl₂
Oxidation number of Mn changes from 7 to + 2
(ii) 5HNO₂ + 2MnO₄^–+ 6H⁺→ 2Mn²⁺ + 5HNO₃ + 3H₂O
Oxidation number of N changes from + 3 to + 5
(iii) 2HNO₂ + 2I^– + 2H⁺ → 2NO + 2H₂O + I₂
Oxidation number of N changes from + 3 to + 2

Question 19.
(a) Describe the role of oxidation number in nomenclature.
(b) State which of the following is the strongest oxidising agent and why?
(i) MnO₄^–| Mn²⁺ E° = + 1.52
(ii) Fe³⁺ | Fe²⁺ E° = + 0.76
(iii) BrO₃^– | Br₂ E° = + 1.50
(iv) Cr₂O₇^2- | Cr³⁺ E° = + 1.33
Answer:
(b) MnO₄^– is the strongest oxidising agent because it has highest reduction potential.

Question 20.
Some standard electrode potentials are given below:

(i) Can Fe (II) act as an oxidising agent
(ii) When iron is reacted with Cl₂, will it most likely form FeCl₂ or FeCl₃?
Answer:
(i) No, because its reduction potential is least
(ii) FeCl₂ is preferred.

Question 21.
A student constituted a cell by the two electrodes Zn/Zn²⁺ (1 M) and Mg/Mg²⁺ (1 M) and represented it as Zn/Zn²⁺ (1 M) | | Mg²⁺ (1 M)/Mg. Is he correct or not?
(E°_Mg⁺²/Mg = – -37 V; E°_Zn⁺²/Zn = -0.76 V) Justify.
Answer:
The electrode with lower reduction potential will act as anode and the other with higher reduction potential acts as cathode. The representation of given cell is incorrect.