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.
Popular Questions of Class Chemistry
- Q:-
Calculate the amount of carbon dioxide that could be produced when
(i) 1 mole of carbon is burnt in air.
(ii) 1 mole of carbon is burnt in 16 g of dioxygen.
(iii) 2 moles of carbon are burnt in 16 g of dioxygen.
- Q:-
The mass of an electron is 9.1 × 10–31 kg. If its K.E. is 3.0 × 10–25 J, calculate its wavelength.
- Q:-
Calculate the wavelength of an electron moving with a velocity of 2.05 × 107 ms–1.
- 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:-
Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass.
- Q:-
In a reaction A + B2 → AB2 Identify the limiting reagent, if any, in the following reaction mixtures.
(i) 300 atoms of A + 200 molecules of B
(ii) 2 mol A + 3 mol B
(iii) 100 atoms of A + 100 molecules of B
(iv) 5 mol A + 2.5 mol B
(v) 2.5 mol A + 5 mol B
- Q:-
Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction,
CaCO3(s) + 2 HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l)
What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl?
- Q:-
Density of a gas is found to be 5.46 g/dm3 at 27 °C at 2 bar pressure. What will be its density at STP?
- Q:-
A sample of drinking water was found to be severely contaminated with chloroform, CHCl3, supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass).
(i) Express this in percent by mass.
(ii) Determine the molality of chloroform in the water sample.
- Q:-
Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol–1
Recently Viewed Questions of Class Chemistry
- 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:-
Addition of HBr to propene yields 2-bromopropane, while in the presence of benzoyl peroxide, the same reaction yields 1-bromopropane. Explain and give mechanism.
- Q:-
Calculate the temperature of 4.0 mol of a gas occupying 5 dm3 at 3.32 bar.
(R = 0.083 bar dm3 K–1 mol–1).
- Q:-
What are hybridisation states of each carbon atom in the following compounds ?
(i) CH2=C=O,
(ii) CH3CH=CH2,
(iii) (CH3)2CO,
(iv) CH2=CHCN,
(v) C6H6
- Q:-
Use molecular orbital theory to explain why the Be2 molecule does not exist.
- Q:-
Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction:
CH4 (g) + H2O (g) ↔ CO (g) + 3H2 (g)
(a) Write as expression for Kp for the above reaction.
(b) How will the values of Kp and composition of equilibrium mixture be affected by
(i) increasing the pressure
(ii) increasing the temperature
(iii) using a catalyst ?
- Q:-
Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.
- Q:-
An alkene 'A' on ozonolysis gives a mixture of ethanal and pentan-3-one. Write structure and IUPAC name of 'A'.
- Q:-
Explain the difference in properties of diamond and graphite on the basis of their structures.
- Q:-
Find the oxidation state of sodium in Na2O2.
- NCERT Chapter