CBSE · NCERT · Class 11 Chemistry · Chapter 7

NCERT Solutions: Class 11 Chemistry Chapter 7 - Redox Reactions

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Chapter-wise NCERT intext questions and exercise answers for Redox Reactions, grounded in the official textbook.

Questions are taken verbatim from the NCERT textbook; answers were grounded against the chapter's content during generation. Items needing review are marked.
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Q.7.4Fluorine reacts with ice and results in the change: H2O(s) + F2(g) → HF(g) + HOF(g) Justify that this reaction is a redox reaction.v
Solution

In F2, fluorine has oxidation number 0. In HF and HOF, fluorine is -1, so fluorine is reduced. In H2O, oxygen is -2. In HOF, H is +1 and F is -1, so oxygen is 0. Oxygen therefore increases from -2 to 0 and is oxidised. Since oxidation and reduction occur together, the reaction is redox.

Answer:

It is a redox reaction because fluorine is reduced and oxygen in water is oxidised.

Q.7.5Calculate the oxidation number of sulphur, chromium and nitrogen in H2SO5, Cr2O7^2- and NO3-. Suggest structure of these compounds. Count for the fallacy.v
Solution

Apply oxidation-number sum → charge. In Cr2O7^2-, 2Cr + 7(-2) = -2, so Cr = +6. In NO3-, N + 3(-2) = -1, so N = +5. For H2SO5, the structure is HO-SO2-OOH, containing one peroxide linkage. The two peroxide oxygens are -1 each, while the other three oxygens are -2. Thus 2(+1) + S + 3(-2) + 2(-1) = 0, giving S = +6. If all five oxygens are incorrectly counted as -2, sulphur would appear as +8, which is the fallacy.

Answer:

S in H2SO5 = +6, Cr in Cr2O7^2- = +6, and N in NO3- = +5. The fallacy in H2SO5 is avoided by recognising one peroxide O-O bond.

Q.7.6Write 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) oxidev
Solution

Balance cation and anion charges: Hg2+ with two Cl- gives HgCl2; Ni2+ with SO4^2- gives NiSO4; Sn4+ with O2- gives SnO2; two Tl+ ions with SO4^2- give Tl2SO4; two Fe3+ ions with three SO4^2- ions give Fe2(SO4)3; two Cr3+ ions with three O2- ions give Cr2O3.

Answer:

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

Q.7.7Suggest a list of the substances where carbon can exhibit oxidation states from -4 to +4 and nitrogen from -3 to +5.v
Solution

The examples are chosen so that standard oxidation-number rules give the required sequence. Carbon becomes progressively more positive as H atoms in methane are replaced by electronegative Cl atoms. Nitrogen examples range from hydrides, where N is negative, through elemental N2, to oxides and oxoacids, where N is positive.

Answer:

Carbon: CH4 (-4), CH3Cl (-2), CH2Cl2 (0), CHCl3 (+2), CCl4 (+4). Nitrogen: NH3 (-3), N2H4 (-2), NH2OH (-1), N2 (0), N2O (+1), NO (+2), HNO2 (+3), NO2 (+4), HNO3 (+5).

Q.7.8While 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 ?v
Solution

Sulphur in SO2 is +4, so it can be oxidised to +6 or reduced to lower states. Oxygen in H2O2 is -1, so it can be oxidised to O2 (0) or reduced to H2O (-2). In O3, oxygen is 0 and readily accepts electrons to form oxide or water species. In HNO3, nitrogen is +5, its maximum common oxidation state, so it cannot be further oxidised and acts as an oxidising agent.

Answer:

SO2 and H2O2 contain elements in intermediate oxidation states, while O in O3 and N in HNO3 are at high oxidation states and are therefore reduced rather than oxidised.

Q.7.9Consider the reactions: (a) 6 CO2(g) + 6H2O(l) → C6 H12 O6(aq) + 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) → C6 H12 O6(aq) + 6H2O(l) + 6O2(g) (b) O3(g) + H2O2 (l) → H2O(l) + O2(g) + O2(g) Also suggest a technique to investigate the path of the above (a) and (b) redox reactions.v
Solution

In photosynthesis, the oxygen evolved as O2 comes from water, not directly from CO2, so writing water on both sides helps track the oxygen atoms. In the ozone-hydrogen peroxide reaction, the two O2 molecules are not necessarily equivalent in origin, so writing them separately is more informative. The path can be investigated using isotopic labelling, such as H2^18O or ^18O-labelled peroxide/ozone, followed by analysis of the labelled oxygen in products.

Answer:

The rewritten equations show the separate source and fate of oxygen atoms; isotope labelling can investigate the path.

Q.7.10The compound AgF2 is unstable compound. However, if formed, the compound acts as a very strong oxidising agent. Why ?v
Solution

Fluorine is highly electronegative and can stabilise Ag(II) only with difficulty. AgF2 contains Ag2+, while Ag+ is much more stable. Therefore AgF2 has a strong tendency to accept an electron and be reduced to AgF, oxidising other substances in the process.

Answer:

Because Ag is in the +2 oxidation state and is readily reduced to the more stable +1 state.

Q.7.11Whenever 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.v
Solution

For carbon, limited oxygen gives CO, where carbon is +2, while excess oxygen gives CO2, where carbon is +4: 2C + O2 → 2CO and C + O2 → CO2. For phosphorus, limited chlorine gives PCl3, where P is +3, while excess chlorine gives PCl5, where P is +5: P4 + 6Cl2 → 4PCl3 and P4 + 10Cl2 → 4PCl5. For sulphur, limited oxygen gives SO2, where S is +4, while stronger/excess oxidation gives SO3, where S is +6: S + O2 → SO2 and 2SO2 + O2 → 2SO3.

Answer:

Excess reducing agent limits oxidation and gives lower oxidation-state products; excess oxidising agent drives the element to higher oxidation-state products.

Q.7.13Identify 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) + 2 Cu2+(aq) + 5 OH- (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)v
Solution

A substance oxidised is the reducing agent; a substance reduced is the oxidising agent. In (a), Ag+ in AgBr is reduced to Ag, while hydroquinone C6H6O2 is oxidised to quinone C6H4O2. In (b) and (c), formaldehyde is oxidised to formate, while Ag+ or Cu2+ is reduced. In (d), nitrogen in hydrazine goes from -2 to 0, so N2H4 is oxidised; oxygen in H2O2 goes from -1 to -2, so H2O2 is reduced. In (e), Pb(0) is oxidised to PbSO4 containing Pb2+, while PbO2 containing Pb4+ is reduced to Pb2+.

Answer:

(a) C6H6O2 oxidised/reducing agent; AgBr reduced/oxidising agent. (b) HCHO oxidised/reducing agent; [Ag(NH3)2]+ reduced/oxidising agent. (c) HCHO oxidised/reducing agent; Cu2+ reduced/oxidising agent. (d) N2H4 oxidised/reducing agent; H2O2 reduced/oxidising agent. (e) Pb oxidised/reducing agent; PbO2 reduced/oxidising agent.

Q.7.14Consider the reactions : 2 S2O3^2- (aq) + I2(s) → S4 O6^2-(aq) + 2I- (aq) S2O3^2-(aq) + 2Br2(l) + 5 H2O(l) → 2SO4^2-(aq) + 4Br- (aq) + 10H+(aq) Why does the same reductant, thiosulphate react differently with iodine and bromine ?v
Solution

Iodine is a milder oxidant, so it oxidises thiosulphate only to tetrathionate, S4O6^2-. Bromine is a stronger oxidant and oxidises thiosulphate further to sulphate, SO4^2-. The different oxidising strengths of I2 and Br2 therefore give different oxidation products from the same reductant.

Answer:

Because bromine is a stronger oxidising agent than iodine.

Q.7.15Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.v
Solution

Fluorine can oxidise other halide ions: F2 + 2Cl- → 2F- + Cl2, F2 + 2Br- → 2F- + Br2, and F2 + 2I- → 2F- + I2. The ease of oxidation of halide ions increases down the group, so I- in HI is the strongest reducing halide. For example, 2HI + Cl2 → 2HCl + I2 and 2HI + Br2 → 2HBr + I2.

Answer:

F2 is the best oxidant because it is reduced most readily to F-. HI is the best reductant because I- is oxidised most readily among halide ions.

Q.7.16Why does the following reaction occur ? XeO6^4- (aq) + 2F- (aq) + 6H+(aq) → XeO3(g)+ F2(g) + 3H2O(l) What conclusion about the compound Na4XeO6 (of which XeO6^4- is a part) can be drawn from the reaction.v
Solution

In XeO6^4-, xenon is +8. In XeO3, xenon is +6, so xenon is reduced. Fluoride ion changes from -1 in F- to 0 in F2, so it is oxidised. Since XeO6^4- oxidises F- to F2, the compound Na4XeO6 containing this ion behaves as a strong oxidising agent.

Answer:

The reaction occurs because Xe(VIII) in XeO6^4- is reduced and F- is oxidised; Na4XeO6 is a strong oxidising agent.

Q.7.17Consider the reactions: (a) H3PO2(aq) + 4 AgNO3(aq) + 2 H2O(l) → H3PO4(aq) + 4Ag(s) + 4HNO3(aq) (b) H3PO2(aq) + 2CuSO4(aq) + 2 H2O(l) → H3PO4(aq) + 2Cu(s) + H2SO4(aq) (c) C6H5CHO(l) + 2[Ag (NH3)2]+(aq) + 3OH- (aq) → C6H5COO-(aq) + 2Ag(s) + 4NH3 (aq) + 2 H2O(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 ?v
Solution

Both Ag+ and Cu2+ oxidise the stronger reducing agent H3PO2 to H3PO4, being reduced to Ag and Cu respectively. However, benzaldehyde reduces Ag+ to Ag but does not reduce Cu2+ under the stated conditions. Therefore Ag+ has the greater oxidising tendency and is the stronger oxidising agent here.

Answer:

Ag+ is a stronger oxidising agent than Cu2+ under these conditions.

Q.7.18Balance the following redox reactions by ion – electron method (a) MnO4- (aq) + I- (aq) → MnO2 (s) + I2(s) (in basic medium) (b) MnO4- (aq) + SO2 (g) → Mn2+ (aq) + HSO4- (aq) (in acidic solution) (c) H2O2 (aq) + Fe2+ (aq) → Fe3+ (aq) + H2O (l) (in acidic solution) (d) Cr2O7^2- + SO2(g) → Cr3+ (aq) + SO4^2- (aq) (in acidic solution)v
Solution

Each equation is balanced by equating electron gain and loss, then balancing O and H with H2O, H+ or OH- according to the medium. Charge and atom checks confirm: (a) both sides charge -8; (b) both sides charge -1; (c) both sides charge +6; (d) both sides charge 0.

Answer:

(a) 2MnO4- + 6I- + 4H2O → 2MnO2 + 3I2 + 8OH-; (b) 2MnO4- + 5SO2 + H+ + 2H2O → 2Mn2+ + 5HSO4-; (c) H2O2 + 2Fe2+ + 2H+ → 2Fe3+ + 2H2O; (d) Cr2O7^2- + 3SO2 + 2H+ → 2Cr3+ + 3SO4^2- + H2O.

Q.7.20What sorts of informations can you draw from the following reaction ? (CN)2(g) + 2OH- (aq) → CN-(aq) + CNO-(aq) + H2O(l)v
Solution

In (CN)2, carbon is in the +3 oxidation state. In CN-, carbon is +2, so one carbon is reduced. In CNO-, carbon is +4, so the other carbon is oxidised. The same reactant is both oxidised and reduced, giving cyanide and cyanate in basic solution; hence it is disproportionation.

Answer:

It is a disproportionation reaction of cyanogen in alkaline medium.

Q.7.21The Mn3+ ion is unstable in solution and undergoes disproportionation to give Mn2+, MnO2, and H+ ion. Write a balanced ionic equation for the reaction.v
Solution

One Mn3+ is reduced to Mn2+ and another Mn3+ is oxidised to Mn(IV) in MnO2. Balancing oxygen with water and hydrogen with H+ gives 2Mn3+ + 2H2O → Mn2+ + MnO2 + 4H+. Atoms and total charge are balanced.

Answer:

2Mn3+ + 2H2O → Mn2+ + MnO2 + 4H+.

Q.7.22Consider the elements : Cs, Ne, I and F (a) Identify the element that exhibits only negative oxidation state. (b) Identify the element that exhibits only postive 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.v
Solution

Fluorine is the most electronegative element and shows only -1 in compounds. Caesium is an alkali metal and shows +1. Iodine can show -1 as iodide and positive oxidation states in oxo-compounds. Neon is a noble gas and ordinarily shows neither positive nor negative oxidation states.

Answer:

(a) F; (b) Cs; (c) I; (d) Ne.

Q.7.23Chlorine 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.v
Solution

Sulphur in SO2 is oxidised from +4 to +6 in H2SO4, while chlorine is reduced from 0 in Cl2 to -1 in HCl. Balancing in water gives Cl2 + SO2 + 2H2O → 2HCl + H2SO4.

Answer:

Cl2 + SO2 + 2H2O → 2HCl + H2SO4.

Q.7.25In 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 ?v
Solution

The reaction is 4NH3 + 5O2 → 4NO + 6H2O. Moles NH3 = 10.00/17.0 = 0.588 mol. Moles O2 = 20.00/32.0 = 0.625 mol. To consume 0.588 mol NH3, O2 required = (5/4)(0.588) = 0.735 mol, so O2 is limiting. Moles NO formed = (4/5)(0.625) = 0.500 mol. Mass NO = 0.500 x 30.0 = 15.0 g.

Answer:

15.0 g of NO.

Q.7.27Predict the products of electrolysis in each of the following: (i) An aqueous solution of AgNO3 with silver electrodes (ii) An aqueous solution AgNO3 with platinum electrodes (iii) A dilute solution of H2SO4 with platinum electrodes (iv) An aqueous solution of CuCl2 with platinum electrodes.v
Solution

With silver electrodes in AgNO3, Ag+ is reduced to Ag at the cathode and Ag metal is oxidised to Ag+ at the anode. With platinum electrodes in AgNO3, Ag+ is reduced at the cathode and water is oxidised at the anode to oxygen. In dilute H2SO4 with platinum electrodes, water/H+ gives H2 at the cathode and water gives O2 at the anode. In CuCl2 with platinum electrodes, Cu2+ is reduced to Cu at the cathode and chloride ions are oxidised to Cl2 at the anode.

Answer:

(i) Ag deposits at cathode and silver anode dissolves. (ii) Ag deposits at cathode and O2 forms at anode. (iii) H2 forms at cathode and O2 at anode. (iv) Cu deposits at cathode and Cl2 forms at anode.

Q.7.28Arrange the following metals in the order in which they displace each other from the solution of their salts. Al, Cu, Fe, Mg and Zn.v
Solution

A more reactive metal displaces a less reactive metal from its salt solution. From the activity/electrochemical series, the decreasing displacement tendency is Mg, Al, Zn, Fe, Cu.

Answer:

Mg > Al > Zn > Fe > Cu.

Q.7.30Depict 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 the current in the cell, and (iii) individual reaction at each electrode.v
Solution

Zinc is oxidised, so the zinc electrode is the anode and is negatively charged in the galvanic cell. Silver ion is reduced at the cathode. The cell notation is Zn(s)|Zn2+(aq)||Ag+(aq)|Ag(s). Electrons flow through the external wire from Zn to Ag, while ions in the solutions and salt bridge complete the internal circuit.

Answer:

Cell: Zn(s)|Zn2+(aq)||Ag+(aq)|Ag(s). The zinc electrode is negative. Electrons carry current in the external circuit and ions carry current through the electrolyte/salt bridge. Anode: Zn → Zn2+ + 2e-. Cathode: Ag+ + e- → Ag.