NCERT Solutions for Class 11 Chemistry Chapter 8 Redox Reactions

Class 11 Chemistry NCERT Solutions Chapter 8

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

Answer: (a) In Kl₃, 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, we must consider its structure, K⁺[I —I <— I]^–. Here, 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. Thus, the O.N. of three I atoms, atoms in Kl₃ are 0, 0 and -1 respectively.

Question 3. Justify that the following reactions are redox reactions:
(a) CuO(s) + H₂(g) —–> Cu(s) + H₂0(g)
(b) Fe₂O₃(s) +3CO(g) —-> 2Fe(s) + 3CO₂(g)
(c) 4BCl₃(g) +3LiAlH₄(s) ——> 2B₂H₆(g) + 3LiCl(s) + 3AlCl₃(s)
(d) 2K(s) +F₂(g)——> 2K+F^–(s)
Answer:

Here, O.N. of B decreases from +3 in BrCl₃to -3 in B₂H₆ while that of H increases from -1 in LiAlH₄to +1 in B₂H₆. Therefore, BCl₃ is reduced while LiAlH₄ is oxidised. Further, H is added to BCl₃ but is removed from LiAlH₄, therefore, BC13 is reduced while LiAlH₄ is oxidised. Thus, it is a redox reaction.
Here, each K atom as 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 a redox reaction.
By chemical bonding, C₂ is attached to three H-atoms (less electronegative than carbon) and one CH₂OH group (more electronegative than carbon), therefore,

Thus, there is no fallacy about the O.N. of N in N0₃^–whether one calculates by conventional method or by chemical bonding method.

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.
Answer: (a) Hg(II)Cl₂, (b) Ni(II)SO₄, (c)S_n(IV)O₂ (d) T₁₂(I)SO₄, (e) Fe₂(III)(S0₄)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.
Answer:

Question 8. While sulphur dioxide and hydrogen peroxide can act as an oxidising as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants. Why?
Answer:  (i) In S0₂ , O.N. of S is +4. In principle, S can have a minimum O.N. of -2 and maximum of +6. Therefore, S in S0₂ can either decrease or increase its O.N. and hence can act both as an oxidising as well as a reducing agent.
(ii) In H₂O₂, the O.N. of O is -1. In principle, O can have a minimum O.N. of -2 and maximum of zero (+1 is possible in O₂F₂and +2 in OF₂). Therefore, O in H₂O₂ can either decrease its O.N. from -1 to -2 or can increase its O.N. from -1 to zero. Therefore, H₂O₂ acts both as an oxidising as well as a reducing agent.
(iii) In O₃, the O.N. of O is zero. It can only decrease its O.N. from zero to -1 or -2, but cannot increase to +2. Therefore, O₃ acts only as an oxidant.
(iv) In HNO₃, O.N. of N is +5 which is maximum. Therefore, it can only decrease its O.N. and hence it acts as an oxidant only.

Question 9. Consider the reactions:
(a) 6CO₂(g) 6H₂O(l) ———> C₆H₁₂O₆(s) + 6O₆(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) + 6H₂O(l) + 6O₂(g)
(b) O₃(g) + H₂O₂ (l) ———–> H₂O(l) + O₂(g) + O₂(g)
Also suggest a technique to investigate the path of above (a) and (b) redox reactions.
Answer:

Question 10. The compound AgF₂ is unstable. However, if formed, the compound acts as a very strong oxidising agent. Why?
Answer: In AgF₂ oxidation state of Ag is +2 which is very 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.

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.
Answer: (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 the oxidation state of C is +2. If, however, excess of O₂ is used, the initially formed CO gets oxidised to CO₂ in which oxidation state of C is + 4.

Question 12. How do you account 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 to an inorganic mixture containing chloride, we get colourless pungent smelling gas HCl, but if the mixture contains bromide then we get red vapour of bromine. Why?
Answer: (a) Toluene can be oxidised to benzoic acid in acidic, basic and neutral media according to the following redox equations:

In the laboratory, benzoic acid is usually prepared by alkaline KMnO₄ oxidation of toluene. However, in industry alcoholic KMnO₄ is preferred over acidic or alkaline KMnO₄ because of the following reasons:
(i) The cost of adding an acid or the base is avoided because in the neutral medium, the base (OH- ions) are produced in the reaction itself.
(ii) Since reactions occur faster in homogeneous medium than in heterogeneous medium, therefore, alcohol helps in mixing the two reactants, i.e., KMnO₄ (due to its polar nature) and toluene (because of its being an organic compound).
(b) When cone. H₂S0₄ is added to an inorganic mixture containing chloride, a pungent smelling gas HCl is produced because a stronger acid displaces a weaker acid from its salt.

Question 15. Justify-giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic add is the best reductant.
Answer: Halogens have a strong tendency to accept electrons. Therefore, they are strong oxidising agents. Their relative oxidising power is, however, measured in terms of their electrode potentials. Since the electrode potentials of halogens decrease in the order: F₂ (+2.87V) > Cl₂ (+1.36V) > Br₂ (+1.09V) > I₂ (+0.54V), therefore, their oxidising power decreases in the same order.
This is evident from the observation that F₂ oxidises Cl^– to Cl₂, Br^–to Br₂, I – to I₂ ; Cl₂ oxidises Br^–to Br₂ and F to I₂ but not F^– to F₂. Br₂, however, oxidises F to I₂ but not F^–  to F₂ , and Cl^–   to Cl₂.
F₂(g) + 2Cr(aq) ———–> 2F^–(aq) + Cl₂(g); F2(g) + 2Br^–(aq) ———-> 2F^–(aq) + Br₂ (Z)
F₂(g) + 2I^–(aq) ———-> 2F^–(aq) + I₂(s); Cl2 (g) + 2Br^–(aq) ————> 2Cl^–(aq) + Br₂ (Z)
Cl₂(g) + 2I^–(aq) ———–> 2Cl^–  (aq) + I₂(s) and Br₂ (Z) + 2F ———> 2Br^– (aq) + I₂(s)
Thus, F₂ is the best oxidant.
Conversely, halide ions have a tendency to lose electrons and hence can act as reducing agents. Since the electrode potentials of halide ions decreases in the order: I^–(-0.54 V) > Br^– (-1.09 V) > Cl^–(-1.36 V) > I₂ (-2.87 V), therefore, the reducing power of the halide ions or their corresponding hydrohalic acids decreases in the same order: HI > HBr > HCl > HF. Thus, hydroiodic acid is the best reductant. This is supported by the following reactions. For example, HI and HBr reduce H₂S0₄ to S0₂ while HCl and HF do not.
2HBr + H₂S0₄ —–> Br₂+ S0₂ + 2H₂O; 2HI + H₂S0₄ ——> I₂ + S0₂ + 2H₂O
Further F reduces Cu²⁺ to Cu⁺ but Br does not.
2Cu²⁺(aq) + 4I^–(aq) >Cu₂I₂(s) + I₂(aq); Cu2+(aq) + 2Br^–> No reaction.
Thus, HI is a stronger reductant than HB_r.
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(aq) + 2H₂O
MnO₂ (s) + 4HF(l) ———–> No reaction.
Thus, the reducing character of hydrohalic acids decreases in the order: HI > HBr > HCl > HF.

Question 17. Consider the reactions:
(a) H₃P0₂(aq) + 4AgNO₃(aq) + 2H₂O(l) ————->H₃PO₄(aq) + 4Ag(s) + 4HNO₃(aq)
(b) H₃P0₂(aq) + 2CuS0₄(aq) + 2H₂O(l) ————->H₃P0₄(aq) + 2Cu(s) + H₂S0₄(aq)
(c) C₆H₅CHO(l) + 2[Ag(NH₃)₂]⁺(aq) + 30H^–(aq)———–> C₆H₅COO^–(aq) + 2Ag(s) + 4NH₃(aq) + 2H₂0(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?
Answer: Reactions (a) and (b) indicate that H₃P0₂ (hypophosphorous acid) is a reducing agent and thus reduces both AgNO₃ and CuS0₄ to Ag and Cu respectively. Conversely, both AgNO₃ and CuS0₄ act as oxidising agent and thus oxidise H₃P0₂to H₃P0₄ (orthophosphoric acid) 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 strong deoxidising agent than Cu²⁺ ion.

Question 18. Balance the following redox reactions by ion-electron method.
(a) MnO₄^–(aq) +I^–(aq) ———>Mn0₂(s) + I₂ (s) (in basic medium)
(b) MnO₄^–(aq) + S02(g) ——-> Mn²⁺(aq) +H₂S0₄^–(in acidic solution)
(c) H₂O₂(aq) + Fe2+(aq) ———-> Fe³+(aq) + H₂O(l) (in acidic solution)
(d) Cr₂O₇^2-(aq) + S0₂ (g)——> Cr³⁺(aq) + SO₄^2-(aq) (in acidic solution)
Answer: (a) Do it yourself.
(b) The balanced half reaction equations are:
Oxidation half equation:
S0₂(g) + 2H₂O(l) ——–> HS0₄^– (aq) + 3H⁺(aq) +2e^–                        …(i)
Reduction half equation:
MnO₄^–(aq) + 8H⁺(aq) + 5e^– ——–> Mn²⁺(aq) + 4H₂O(l) ………..(ii)
Multiply Eq. (i) by 3 and Eq. (ii) by 2 and add, we have,
2MnO₄^–(aq) + 5S0₂(g) + 2H₂0(l) + H⁺(aq) ————> 2Mn²⁺(aq) + 5HSO₄^–(aq)
(c) Oxidation half equation: Fe²⁺(aq) ———> Fe³⁺(aq) + e^– …(i)
Reduction half equation: H₂O₂(aq) + 2H⁺(aq) + 2e^– ———> 2H₂O(l) …(ii)
Multiply Eq. (i) by 2 and add it to Eq. (ii), we have,
H₂O₂(aq) +2Fe²⁺(aq) +2H⁺(aq) ——-> 2Fe³⁺(aq) + 2H₂O(l)
(d) Following the procedure detailed on page 8/23, the balanced half reaction equations are:
Oxidation half equation:
SO₂(g) + 2H₂O(l) ————> SO₄^2-(aq) + 4H⁺(aq) + 2 e^–  …(i)
Reduction half equation:
Cr₂O₇2^–(aq) + 14H⁺(aq) + 6e^– ————> 2Cr³⁺(aq) + 7H₂0(l) …(ii)
Multiply Eq. (i) by 3 and add it to Eq. (ii), we have,
Cr₂O₇2^–(aq) + 3SO₂(q) + 2H⁺(aq) ————> 2Cr³⁺(aq) + 3SO₄^2-(aq) + H₂0(l)

Question 19. Balance the following equation in basic medium by ion electron method and oxidation number method and identify the oxidising agent and the reducing agent.
(a) P₄(s) + OH^–(aq) ———> PH₃(g) + H₂PO₂^–(aq)
(b) N₂H₄(l) + ClO^–(aq) ——–> NO(g) + CV(aq)
(c) Cl₂O₇(g) + H₂O₂(aq) ———-> ClO₂^–(aq) + O₂(g) + H⁺
P₄ acts both as an oxidising as well as a reducing agent.
Answer:

Question 20. Write Jour informations about the reaction:
(CN)₂(g) + 2OH^–(aq) —–> CN^–(aq) + CNO^–(aq) + H₂O(l)
Answer: Let x be the O.N. of C.
O.N. of C in cyanogen, (CN)₂ = 2 (x – 3) = 0 or x = +3 O.N. of C in cyanide ion, CN- = x – 3 = -1 or x = +2 O.N. of C in cyanate ion, CNO =x-3-2 = -lora: = +4 The four information about the reaction are:
(i) The reaction involves decomposition of cyanogen, (CN)₂ in the alkaline medium to cyanide ion, CN and cyanate ion,CNO^–.
(ii) The O.N. of C decreases from +3 in (CN)₂ to +2 in CN^–ion and increases from +3 in(CN)₂ to +4 in CNO^– ion. Thus, cyanogen is simultaneously reduced to cyanide ion and oxidised to cyanate ion.
(iii) It is an example of a redox reaction in general and a disproportionation reaction in particular.
(iv) Cyanogen is a pseudohalogen (behaves like halogens) while cyanide ion is a pseudohalide ion (behaves like halide ion).

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.
Answer:

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.
Answer: (a) F. Fluorine being the most electronegative element shows only a -ve oxidation state of -1.
(b) Cs. Alkali metals because of the presence of a single electron in the valence shell, exhibit an oxidation state of +1.
(c) I. Because of the presence of seven electrons in the valence shell, I shows an oxidation state of -1 (in compounds of I with more electropositive elements such as H, Na, K, Ca, etc.) or an oxidation state of +1 compounds of I with more electronegative elements, i.e., O, F, etc.) and because of the presence of d-orbitals it also exhibits +ve oxidation states of +3, +5 and +7.
(d) Ne. It is an inert gas (with high ionization enthalpy and high positive electron gain enthalpy) 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 chlorine is removed by treating with sulphur dioxide. Present a balanced equation for the reaction for this redox change taking place in water.
Answer: The skeletal equation is:

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 show disproportionation reaction.
Answer: 

Question 26. Using the standard electrode potentials given in Table 8.1, predict if the reaction between the following is feasible:
(a) Fe³⁺(aq) and I-(aq) (b) Ag⁺   (aq) and Cu(s)
(c) Fe³⁺(aq) and Cu(s) (d)Ag(s) and Fe³⁺(aq)
(e) Br₂ (aq) and Fe³⁺ (aq).
Answer: (a) It may be noted that for oxidation reactions, i.e., Eq. (i), the sign of the electrode potential as given in Table 8.1 is reversed. To get the equation for the overall reaction, the number of electrons lost in Eq. (i) and gained in Eq. (ii) must be cancelled. To do so, Eq. (ii) is multiplied by 2 and added to Eq. (i). Further, it may be noted that whenever any half reaction equation is multiplied by any integer, its electrode potential is not multiplied by that integer. Thus,
Overall reaction: 2Fe³⁺ (aq) + 2I^–(aq) ——-> 2Fe²⁺  (aq) + I₂(s); E° = + 0.23 V
Since the EMF for the above reaction is positive, therefore, the above reaction is feasible.
(b) The possible reaction between Ag⁺(aq) and Cu(s) is Cu(s) + 2Ag⁺  (aq)—> Cu²⁺(aq) + 2Ag(s)
The above redox reaction can be split into the following two half reactions. Writing electrode potential for each half reaction from Table 8.1, we have,

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 silver nitrate with platinum electrodes.
(iii) A dilute solution of H₂S0₄with platinum electrodes.
(iv) An aqueous solution of CuCl₂ with platinum electrodes. 
Answer: (i) In aqueous solution, AgNO₃ ionises to give Ag⁺(aq) and NO₃^– (aq) ions.
AgN0₃(aq) ——–> Ag⁺(aq) + NO₃^– (aq)
Thus, when electricity is passed, Ag⁺(aq) ions move towards the cathode while NO₃^– ions move towards the anode. In other words, at the cathode, either Ag⁺(aq) ions or H₂O molecules may be reduced. Which of these will actually get discharged would depend upon their electrode potentials which are given below:
Ag⁺(aq) +e^–———-> Ag(s); E° = +0.80 V …(i)
2H₂O(Z) + 2e^– ————> H₂(g) + 2OH^–(aq); E° = -0.83 V …(ii)
Since the electrode potential (i.e., reduction potential of Ag⁺(aq) ions is higher than that of H₂O molecules, therefore, at the cathode, it is the Ag⁺(aq) ions (rather than H₂O molecules) which are reduced.
Similarly, at the anode, either Ag metal of the anode or H₂O molecules may be oxidised. Their electrode potentials are:
Ag(s) ———–> Ag⁺(aq) + e^–; E° = -0.80 V …(iii)
2H₂O(l) ————–> 02(g) +4H+(aq)+4e^– ; E° = -1.23 V …(iv)
Since the oxidation potential of Ag is much higher than that of H₂O, therefore,
at the anode, it is the Ag of the silver anode which gets oxidised and not the H₂O molecules. It may, however, be mentioned here that the oxidation potential of N0₃^–ions is even lower than that of H₂O since more bonds are to broken during reduction of N0₃ ions than those in H₂O.
Thus, when an aqueous solution 0f AgN0₃is electrolysed, Ag from Ag anode dissolves while Ag⁺(aq) ions present in the solution get reduced and get deposited on the cathode.
(ii) If, however, electrolysis of AgN0₃ solution is carried out using platinum electrodes, instead of silver electrodes, oxidation of water occurs at the anode since Pt being a noble metal does not undergo oxidation easily. As a result, O₂ is liberated at the anode according to equation (iv).
Thus, when an aqueous solution of AgNO₃ is electrolysed using platinum electrodes, Ag⁺ ions from the solution get deposited on the cathode while 02 is liberated at the anode.
(iii) In aqueous solution, H₂S0₄ionises to give H⁺(aq) and SO₄^2-(aq) ions.
H₂S0₄(aq) ——> 2H⁺(aq) +S0₄^–(aq)
Thus, when electricity is passed, H⁺ (aq) ions move towards cathode while SO₄^2-(aq) ions move towards anode. In other wode either H⁺(aq) ions or H₂O molecules are reduced. Their electrode potentials are:2H⁺(aq)2e^– ——-> H₂(g); E° = 0.0 V
H₂O(aq) + 2e^– ——-> H₂(g) + 2OH^–((aq); E° = -0.83 V
Since the electron potential (i.e., reduction potential) of H⁺(aq) ions is higher than that of H₂O, therefore, at the cathode, it is H⁺(aq) ions (rather than H₂O molecules) which are reduced to evolve H₂ gas.
Similarly at the anode, either SO₄^2-(aq) ions or H₂O molecules are oxidised. Since the oxidation potential of SO₄ is expected to be much lower (since it involved cleavage of many bonds as compared to those in H20) than that of HjO molecules, therefore, at the anode, it is H₂O molecules (rather than SO₄^2- ions) which are oxidised to evolve O₂ gas.
From the above discussion, it follows that during electrolysis of an aqueous solution of H₂S0₄ only the electrolysis of H₂O occurs liberating H₂ at the cathode and O₂ at the anode.
(iv) In aqueous solution, CuCl₂ ionises as follows:
CuCl₂(aq) ——-> CU²⁺(aq) + 2Cl^–(aq)
On passing electricity, CU²⁺(aq) ions move towards cathode and CU²⁺(aq) ions move towards anode.
Thus, at cathode, either CU²⁺(aq) or H₂O molecules are reduced. Their electrode potentials are:
CU²⁺+ 2e^– ———> Cu(s); E° = +0.34 V
H₂O(l) + 2e^– ——–> H₂(g) + 2OH^–; E° = -0.83 V
Since the electrode potential of CU²⁺(aq) ions is much higher than that of H₂O, therefore, at the cathode, it is CU²⁺(aq) ions which are reduced and not H₂Omolecules.
Similarly, at the anode, either Cl^–(aq) ions or H₂O molecules are oxidised. Their oxidation potentials
2Cl^–(aq) ——> Cl₂(g) + 2e^–; AE° = -1.36 V
2H₂O(l) ——>O₂(g) + 4H⁺(aq) + 4e^–; ∆E° = -1.23 V
Although oxidation potential of H₂O molecules is higher than that of Cl^– ions, nevertheless, oxidation of Cl^–(aq) ions occurs in preference to H₂O since due to overvoltage much lower potential than -1.36 V is needed for the oxidation of H₂O molecules.
Thus, when an aqueous solution of CuCl₂ is electrolysed, Cu metal is liberated at the cathode while Cl₂ gas is evolved at the 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.
Answer: It is based upon the relative positions of these metals in the activity series. 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.
Answer:  Lower the electrode potential, better is the reducing agent. Since the electrode potentials increase in the oder; K⁺/K (-2.93 V), Mg²⁺/Mg (-2.37 V), Cr³⁺/Cr (-0.74 V), Hg²⁺/Hg (0.79 V), Ag⁺/Ag (0.80 V), therefore, reducing power of metals decreases in the same order, i.e., K, Mg, Cr, Hg, Ag.

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.
Answer: The given redox reaction is Zn(s) + 2Ag⁺(aq) ——————-> Zn²⁺(aq) + 2Ag(s)
Since Zn gets oxidised to Zn²⁺ ions, and Ag⁺ gets reduced to Ag metal, therefore,

MORE QUESTIONS SOLVED

NCERT Solutions for Class 11 Chemistry Chapter 8 Very Short Answer Type Questions

Question 1. Define electrochemical cell.
Answer: Electrochemical cell is a device in which the redox reaction is carried indirectly and the decrease in free energy appears as electrical energy.

Question 2. What is a redox couple?
Answer: A redox couple consists of oxidised and reduced form of the same substance taking part in the oxidation or reduction half reaction.

Question 3. Define oxidation in terms of electronic concept.
Answer: Oxidation involves loss of one or more electrons by a species during a reaction.

Question 4. What is the source of electrical energy in a galvanic cell?
Answer: In a galvanic cell due to redox reaction released energy gets converted into the electrical energy.

Question 5. What is the oxidation state of Ni in Ni (CO)₄?
Answer:  Zero.

Question 6. What is the oxidation number of P in H₃P0₄?
Answer: 

Question 7. What is salt bridge?
Answer:  It is a U-shaped tube filled with agar-agar containing inert electrolyte like KCl or KNO₃ which does not react with solutions.

Question 8. What is meant by reducing agent? Name the best reducing agent.
Answer: A species which loses electrons as a result of oxidation is a reducing agent. Li (Lithium).

Question 9. Why is standard hydrogen electrode called reversible electrode?
Answer: A standard hydrogen electrode is called reversible electrode because it can react both as anode as well as cathode in an electrochemical cell.

Question 10. What is a disproportionation reaction ? Give one example.
Answer: In a disproportionation reaction an element in one oxidation state is simultaneously oxidised and reduced.
For example,

Question 15. Define EMF of cell.
Answer: EMF of a cell is the difference in the electrode potentials of the two electrodes in a cell when no current flows through the cell.

Question 16. Calculate the oxidation number of Cr in [Cr (H₂O)₆]³⁺ ion.
Answer:  H₂O is a neutral molecule O.N of H₂O = 0
Write the oxidation number of Cr above its symbol and that of H₂O above its formula,

Question 18. How will you identify cathode and anode in electrochemical cell ?
Answer: At cathode there is gain of electrons.
At anode there is loss of electrons.
In electrochemical cell anode is written on L.H.S while cathode is written on R.H.S.

Question 19.  Identify the oxidant and the reductant in the following reaction. 
N₂H₄(g) + ClO₄(aq) ———–> NO(g) + Cr(aq)
Answer:  N₂H₄is reducing agent i.e., reductant whereas Cl0₃^–is oxidising agent i.e., oxidant.

Question 20. What is oxidation number of Fe in [Fe(CO)₅] ?
Answer: x + 5 (0) =0 , x = 0

Question 21. In the reaction .
M₄O₂ + 4HCI ————-> M₄Cl₂ + Cl₂ + 2H20
which species is oxidised.
Answer: HCl gets oxidised.

NCERT Solutions for Class 11 Chemistry Chapter 8  Short Answer Type Questions

Question 1. What is meant by electrochemical series? What are characteristics of electrochemical series?
Answer:  Electrochemical series is the series of elements in which elements are arranged in decreasing order of their reduction potential.
Reducing power goes on increasing whereas oxidising power goes on dcreasing down the series.

Question 2. What is standard hydrogen electrode? For what purpose it is used? What are signs of oxidation potential and reduction potential decided by using SHE (Standard hydrogen electrode)?
Answer:  Standard hydrogen electrode is used as reference electrode. Its electrode potential is taken as 0.000 volt. Hydrogen electrode can be made. If we use a piece of platinum coated with finely divided black containing hydrogen gas absorbed in it. Platinum black catalyses the reaction and equilibrium is attained faster. When the given electrode acts as anode SHE, we give -ve sign to its reduction potential and +ve sign to its oxidation potential.

Answer: (a) Ag⁺  is reduced, C₆H₆O₂ is oxidised.Ag⁺  is oxidising agent whereas C₆H₆O₂ is reducing agent.
(b) HCHO is oxidised, Ag⁺   is reduced.Ag⁺  is oxidising agent whereas HCHO is reducing agent.
(c) N₂H₄is getting oxidised it is reducing agent. H₂O₂ is getting reduced it acts as an oxidising agent.

Question 10. (a) Calculate the oxidation number of
(i) C in CH₃COOH (ii) S in S₂O₈^-2
(b) Give one example of disproportionation reaction.
Answer:

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