NCERT 12 Chemistry Electrochemistry Chapter 3 exercises 2

Q – 3.1. 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.
Ans. Mg, Al, Zn, Fe, Cu, Ag
Q – 3.2. Given the standard electrode potentials K+/K = – 2.93 V, Ag+/Ag = 0.80 V, Hg2+2 / Hg = 0.79 V,
Mg2+/Mg = – 2.37 V, Cr2+/Cr = – 0.74 V
arrange these metals in their increasing order of reducing power.
Ans. Higher the oxidation potential, more easily it is oxidized and hence greater is the reducing power. Thus, increasing order of reducing power will be Ag < Hg < Cr < Mg < K.
Q – 3.3. Depict the galvanic cell in which the reaction Zn (s) + 2 Ag+ (aq) → Zn2+ (aq) + 2 Ag (s) takes place. Further, show (i) which of the electrodes is negatively charged?
(ii) the carriers of the current in the cell. (iii) individual reaction at each electrode.
Ans. the sell will be represented as:
Zn (s) l Zn2+ (aq) l l Ag+ (aq) l Ag (s)
Anode, i.e., zinc electrode will be negatively charged.
The current will flow from silver to copper in the external circuit.
At Anode: Zn (s) → Zn2+ (aq) + 2 e-
At Cathode: Ag+ (aq) + e → Ag
Q – 3.4. Calculate the standard cell potential of galvanic cells in which the following reactons takes place:
2 Cr (s) + 3 Cd2+ (aq) → 2 Cr3+ (aq) + 3 Cd (s)
Fe2+ (aq) + Ag+ (aq) → Fe3+ (aq) + Ag (s)
Given E0 Cr3+, Cr = – 0.74 V, E0 Cd2+,Cd = – 0.40 V,
E0 Ag+Ag = 0.80 V, E0 Fe3+, Fe2+ = 0.77 V
Also calculate ∆r G0 and equilibrium constants of the reactions.
Ans. (i) E0 cell = E0 cathode = – 0.40 V – (- 0.74 V) = + 0.34 V
∆rG0 = – nF E0 cell = – 6 mol × 96500 C mol-1 × 0.34 V
= – 196860 CV mol-1 = – 196860 J mol-1 = – 196.86 KJ mol-1
∆rG0 = 2.303 RT log K
196860 = 2.303 × 8.314 × 298 log K or log K = 34.5014
K = Antilog 34.5014 = 3.192 × 1034
E0cell = + 0.80 V – 0.77 V = + 0.03 V.
∆rG0 = nF E0cell = – (1 mol) × (96500 C mol-1) × (0.03 V)
= – 2895 CV mol-1 = – 2895 J mol-1
= – 2.895 kJ mol-1
∆rG0= – 2.303 RT log K
– 2895 = – 2.303 × 8.314 × 298 × log K
Or log K = 0.5074 or K = Antilog (0.5074) = 3.22.
Q – 3.5. Write the Nernst equation and the e.m.f. of the following cells at 298 K:
(i) Mg (s) l Mg2+ (0.001 M) l l Cu2+ (0.0001 M) l Cu (s)
(ii) Fe (s) l Fe2+ (0.001 M) l l H+ ( 1 M) l H2 (g) (1 bar) l Pt (s)
(iii) Sn (s) l Sn2+ (0.050 M) I I H+ (0.020 M) l H2 (g)
(1 bar) l Pt (s)
(iv) Pt (s) l Br2 (i) l Br – (0.010 M) l H+ (0.030 M) l H2
(g) (1 bar) l Pt (s)
given E0 Mg2+/Mg = – 2.37 V, E0 Cu2+/Cu = + 0.34 V, E0 Fe2+, Fe = – 0.44 V
E0Sn2+/Sn = – 0.14 V, E01/2 Br2 Br- = + 1.08 V.
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Thus, oxidation will occur at the hydrogen electrode and reduction on the Br2 electrode. E cell = 1.288 V.
Q – 3.6. In the button cell widely used in watches and other devices, the following reaction takes place:
Zn (s) + Ag2O + H2O (i) → Zn2+ (aq) + 2 Ag (s) + OH- (aq)
Determine E0 and ∆rG0 for the reaction.
Given Zn → Zn2+ 2 e – , E0 = 0.76 V ; Ag2O + H2O + 2 e- → 2 Ag + 2 OH – , E0 = 0.344 V.
Ans. (a) Zn is oxidized and Ag2O is reduced (as Ag+ ions change into Ag)
(b) E0 cell = E_( 〖Ag〗_2 O/Ag)^0 (red) + E_( Zn/Zn)^02+ (Ox) = 0.344 + 0.76 = 1.104 V
∆G = – n FE0 cell = – 2 × 96500 × 1.104 J = – 2.13 × 105 J
Q – 3.7. Define conductivity and molar conductivity for solution of an electrolyte. Discuss their variation with concentration.
Ans. Conductivity – the reciprocal of resistivity is known specific conductance or simply conductivity. It is denoted by K (kappa). Thus, if K is the conductivity and G is the conductance of the solution, then
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Now, if I = 1 cm and a = 1 sq. cm, then K = G. Hence,
Molar conductivity – of a solution at a dilution V is the conductance of all the ions produced from one mole of the electrolyte dissolved in V cm3 of the solution when the electrodes are one cm apart and the area of the electrodes is so large that the whole of the solution is contained between them. It is usually represented by ∧m.
Variation of conductance – in general, for weak as well as strong electrolytes, electrolytic conductance as well as equivalent conductivity and molar conductivity increase with dilution whereas the specific conductivity of an electrolytic solution decreases with dilution.
Q – 3.8. the conductivity of 0.20 M solution of KCI at 298 K is 0.0248 S cm-1. calculate its molar conductivity.
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Q – 3.9. The resistance of a conductivity cell containing 0.001 M KCI solution at 298 K is 1500 Ώ. What is the cell constant if conductivity of 0.001 M KCI solution at 298 K is 0.146 × 10-3 S cm-1 ?
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Q – 3.10. the conductivity of NaCl at 298 K has been determined at different concentrations and the result are given below:
Concentration/M     0.001       0.010       0.020        0.050        0.100
102 × k/s m-1         1.237        11.85       23.15        55.53      106.74
Calculate ∧ for all concentration and draw a p;ot between ∧ and c1/2, find the value of ∧0.
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Q – 3.11. Conductivity of 0.00241 M acetic acid is
7.896 × 10-5 S cm-1. Calculate its molar conductivity and if ∧0 for acetic acid is 390.5 S cm2 mol-1, what is its dissociation constant ?
chemistry 12 chapter 3 ex 2 qus 11
Q – 3.12. two electrolytic cells containing silver nitrate solution and copper sulphate solution are connected in series. A steady current of 2.5 ampere was passed through them till 1.078 g of Ag were deposited. How long did the current flow? What weight of copper will be deposited?
(At mass of Ag = 107.8, Cu = 63.5)
Ans. deposition of Ag, reaction is: Ag+ + e – →Ag
Thus, 1 mole, i.e., 107.8 g Ag is deposited by 1 F = 96500 C
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Q – 13 How much electricity in terms of faraday is required to produce
20.0 g of Ca from molten CaCl2
40.0 g of Al from molten Al2O3 ?
Ans. (i) Ca2+ + 2 e – → Ca
Thus, 1 mol of Ca, i.e., 40 g of Ca require electricity = F
؞ 20 g of Ca will require electricity + 1 F
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Q – 3.14. Calculate the charge in coulombs required for three oxidation of:
2 moles of H2O to O2
1 mol of FeO to Fe2O3.
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؞ Quantity of electricity required for oxidation of 1 mol of H2O = 2 F
Or Quantity of electicity required for oxidation of 1 mol H2O = 4 F
= 4 × 96500 C = 386000 C
The electrode reaction for 1 mol of FeO is:
FeO →1/2Fe2O3, i.e., Fe2+ → Fe3+ = e –
؞ Quantity of electricity required = 1 F = 96500 C.
Q – 3.15. a solution of Ni (NO3)2 is electrolysed between platinum electrodes using a current of 5.0 ampere for 20 minutes. What mass of nickel will be delosited at the cathode ? (At. Mass of Ni = 58.7)
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Q – 3.16.Three electrolytic cells A, B and C containing electrolytes ZnSO4, AgNO3 and CuSO4 respectively were connected in series. A steady current of 1.50 ampere was passed through them unit 1-45 g of Ag were deposited at the cathode of cell B. how long did the current flow ? what mass of copper and zinc were deposited ? ( At. Wts. Of Cu = 63.5, Zn = 65.3, Ag = 108)
Ans. Ag+ + e – → Ag, i.e., 108 g of Ag are deposited by 1 F = 96500 C
chemistry 12 chapter 3 ex 2 qus 16
Q = I × t or t = Q/I = 1295.6/1.50 = 863.7 s = 14 min, 24 sec.
Cu2+ + 2 e- →Cu
i.e., 2 × 96500 C deposit Cu = 63.5 g
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Q – 3.17. Predict if the reaction between the following is feasible :
Fe3+ (aq) and I- (aq) (ii) Ag+ (aq) and Cu
Fe3+ and Br-1 (aq) (iv) Ag (s) and Fe3+ (aq)
(v) Br2 (aq) and Fe2+ (aq)
Given standard electrode potentials :
= E_( 〖1/2I〗_2,I)^0 = 0.541, E_( Cu)^02+, Cu = 0.34 V,
= E_( 〖1/2Br〗_2,Br-)^0 = 1.09 V, E_( Ag)^0+, Ag = + 0.80 V ,
E_( Fe^(3+),Fe^(2+))^0 = + 0.77 V
Ans. A reaction feasible if EMF of the cell reaction is + ve
Fe3+(aq) → Fe2+ + (aq) + 1/2 I_2 i.e.,
Pt I I2 I I- (aq) I I Fe3+ (aq) I Fe2+ (aq) I Pt
؞ E0 cell = E_( Fe^(3+),Fe^(2+))^0 – E_( 〖1/2I〗_2,I)^0- = 0.77 – 0.54 = 0.23 V (Feasible)
(ii) Ag+ (aq) + Cu → Ag (s) + Cu2+ (aq),
i.e., Cu I Cu2+ (aq) I I Ag+ (aq) I Ag
E0 = E_( Ag^+,Ag^-)^0- E_( Cu)^02+, Cu = 0.80 – 0.34 = 0.46 V ( Feasible)
(iii) Fe3+(aq) + Br – (aq) → Fe2+ (aq) + 1/2 Br2 ,
E0 cell = 0.77 – 1.09 = – 0.32 V (not feasible)
(iv) Ag (s) + Fe3+ (aq) → Ag+ (aq) + Fe2+ (aq) E0 cell
= 0.77 – 0.80 = – 0.03 V (Not feasible)
(v) 1/2 Br2 (aq) + Fe2+ (aq) → Br – + Fe3+,
E0 cell = 1.09 – 0.77 = 0.32 V (Feasible)
Q – 3.18. Predict the products of electrolysis of the following :
Aqueous solution of AgNo3 with silver electrodes
An aqueous solution of AgNO3 with platinum electrodes
A dilute aqueous solution of H2SO4 with platinum electrodes
An aqueous solution of CuCl2 with platinum electrodes.
Given E0 Ag+/Ag = + 0.80 V, E0 E_( Cu)^02+/ Cu = 0.34 V
Ans. (i) electrolysis of aqueous solution of AgNO3 with silver electrodes.
AgNO3 (s) + aq → Ag+ (aq) + NO-3 + (aq)
H2O ⇌ H+ + OH-
At cathode : Ag+ ions have lower discharge potential than H+ ions. Hence, Ag+ ions will be deposited as Ag in preference to H+ ions.
Alternatively, we have standard reduction potentials as
Ag+ (aq) e- → Ag (s), E0 = + 0.80 V
H+ (aq) + e- → 1/2 H2 (g) , E0 = 0.00 V
As Ag+ ions have higher standard reduction potentials than that of H+ ions, hence Ag+ ions will be reduced more easily and deposited as Ag.
At Anode: as Ag anode is attacked by NO3- ions, Ag of the anode will dissolve to from Ag+ ions in the solution.
Ag ⟶ Ag+ + e-
Alternatively, out of the three possible oxidation reactions occurring at the anode, i.e.,
Ag ⟶ Ag+ + e-, 2 OH- ⟶ H2O + 1/2 O2 + e-
And NO-3 ⟶ NO3 + e- ,
Ag has highest oxidation potential. Hence, Ag of anode is oxidized to Ag+ ions which ass into the solution.
(ii) Electrolysis of aqueous solution of AgNo3 using platinum electrodes.
At Cathode : Same as above.
At Anode: As anode is not attackable, out of OH- and NO-3 ions, OH- ions have lower discharge potential. Hence, OH- ions will be discharged in preference to NO-3 ions, which than decompose to give out O2.
OH- (aq) ⟶ OH + e- , 4 OH ⟶ 2 H2O (I) + O2 (g)
(iii) Electrolysis of dilute H2SO4 with platinum electrodes.
H2SO4(aq) ⟶ 2 H+ (aq) + SO_4^(2-)(aq)
H2O ⇋ H+ + OH-
At Cathode: H+ + e ⟶ H, H + H ⟶ H2 (g)
At Anode: OH- ⟶ OH + e, 4 OH ⟶ 2 H2O + O2 (g)
Thus, H2 is liberated at the cathode and O2 at the anode.
(iv) Electrolysis of aqueous solution of CuCl2 with platinum electrodes
CuCl2 (s) + aq ⟶ Cu2+ (aq) + 2 Cl- (aq)
H2O ⇌ H+ + OH-
at Cathode: Cu2+ ions will be reduced in preference to H+ ions
Cu2+ + 2 e- ⟶ Cu
At Anode: Cl- ions will be oxidized in preference to OH- ions
Cl- ⟶ Cl + e , Cl + Cl ⟶ Cl2 (g)
Thus, Cu will be deposited on the cathode and Cl2 will be liberated at the anode.
Updated: September 27, 2020 — 1:12 pm

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