Class 11 Chemistry - Chapter Redox Reactions NCERT Solutions | Predict the products of electrolysis in
Predict 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.
(i) AgNO3 ionizes in aqueous solutions to form Ag+ and NO-3 ions.
On electrolysis, either Ag+ ions or H2O molecules can be reduced at the cathode. But the reduction potential of Ag+ ions is higher than that of H2O.
Ag+(aq) + e- → Ag(s) ; E° = +0.80V
2H2O(l) + 2e- → H2(g) + 2OH-(aq) ; E° = -0.83V
Hence, Ag+ ions are reduced at the cathode. Similarly, Ag metal or H2O molecules can be oxidized at the anode. But the oxidation potential of Ag is higher than that of H2O molecules.
Ag(s) → Ag+(aq) + e- ; E° = -0.80V
2H2O(l) → O2(g) + 4 H+(aq) + 4e- ; E° = -1.23V
Therefore, Ag metal gets oxidized at the anode.
(ii) Pt cannot be oxidized easily. Hence, at the anode, oxidation of water occurs to liberate O2. At the cathode, Ag+ ions are reduced and get deposited.
(iii) H2SO4 ionizes in aqueous solutions to give H+ and SO2-4 ions.
H2SO4(aq) → 2H+(aq) + SO2-4(aq)
On electrolysis, either of H+ ions or H2O molecules can get reduced at the cathode. But the reduction potential of H+ ions is higher than that of H2O molecules.
2H+(aq) + 2e- → H2(g) ; E° = 0.0V
2H2O(aq) + 2e- → H2g + 2OH-(aq) ; E° = -0.83V
Hence, at the cathode, H+ ions are reduced to liberate H2 gas.
On the other hand, at the anode, either of SO2-4 ions or H2O molecules can get oxidized. But the oxidation of SO2-4 involves breaking of more bonds than that of H2O molecules.Hence, ions have a lower oxidation potential than H2O. Thus, H2O is oxidized at the anode to liberate O2 molecules.
(iv) In aqueous solutions, CuCl2 ionizes to give Cu2+ and Cl- ions as:
CuCl2(aq) → Cu2+(aq) + 2Cl-(aq)
On electrolysis, either of Cu2+ ions or H2O molecules can get reduced at the cathode. But the reduction potential of Cu2+ is more than that of H2O molecules.
Cu2+(aq) + 2e- → Cu(aq) ; E° = +0.34V
H2O(l) + 2e- → H2(g) + 2OH- ; E° = -0.83V
Hence, Cu2+ ions are reduced at the cathode and get deposited.
Similarly, at the anode, either of Cl- or H2O is oxidized. The oxidation potential of H2O is higher than that of Cl-.
2Cl-(aq) → Cl2(g) + 2e- ; E° = -1.36V
2H2O(l) → O2(g) + 4H+(aq) + 4e- ; E° = -1.23V
But oxidation of H2O molecules occurs at a lower electrode potential than that of Cl- ions because of over-voltage (extra voltage required to liberate gas). As a result, Cl- ions are oxidized at the anode to liberate Cl2 gas.
More Questions From Class 11 Chemistry - Chapter Redox Reactions
- Q:-
Balance 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) + Fe 2+ (aq) → Fe3+ (aq) + H2O (l) (in acidic solution)
(d) Cr2O7 2– + SO2(g) → Cr3+ (aq) + SO42– (aq) (in acidic solution)
- Q:-
Assign oxidation number to the underlined elements in each of the following species:
(a) NaH2PO4
(b) NaHSO4
(c) H4P2O7
(d) K2MnO4
(e) CaO2
(f) NaBH4
(g) H2S2O7
(h) KAl(SO4)2.12 H2O
- Q:-
What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results ?
(a) KI3
(b) H2S4O6
(c) Fe3O4
(d) CH3CH2OH
(e) CH3COOH
- Q:-
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) + 3 AlCl3 (s)
(d) 2K(s) + F2(g) → 2K+F– (s)
(e) 4 NH3(g) + 5 O2(g) → 4NO(g) + 6H2O(g)
- Q:-
Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.
(a) P4(s) + OH – (aq) → PH3(g) + HPO2 – (aq)
(b) N2H4(l) + ClO3 – (aq) → NO(g) + Cl–(g)
(c) Cl2O7 (g) + H2O2(aq) → ClO – 2(aq) + O2(g) + H + (aq)
- Q:-
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.
- Q:-
Write the formulae 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 - Q:-
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?
- Q:-
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) + 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)
- Q:-
The compound AgF2 is an unstable compound. However, if formed, the compound acts as a very strong oxidizing agent. Why?
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Recently Viewed Questions of Class 11 Chemistry
- Q:-
Predict which of the following reaction will have appreciable concentration of reactants and products:
a) Cl2 (g) ↔ 2Cl (g) Kc = 5 ×10–39
b) Cl2 (g) + 2NO (g) ↔ 2NOCl (g) Kc = 3.7 × 108
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- Q:-
For the following equilibrium, Kc = 6.3 × 1014 at 1000 K
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Nitric oxide reacts with Br2 and gives nitrosyl bromide as per reaction given below:
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