NEET Chemistry Questions: Redox Reactions & Electrochemisty

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Calculate the energy obtainable from a lead storage battery in which 0.1 mol lead is consumed. Assume a constant concentration of 10.0 M H2 SO4 





(A) PbO2 + 4H+ + SO42- + 2e-PbSO4 + 2H2O; Eº = 1.70Pb + SO42-PbSO4+ 2e-;       Eº = 0.3PbO2 + Pb + 4H+ + 2SO42- 2PbSO4 + 2H2O ; Eº = 2.01

E=E°-0.05922log1H+4SO42-2=2.01-0.05922log1204 102=2.22 V

Now, (0.100 mol Pb)2 mol e-mol Pb 95500 Cmol e-=19300 C

Energy = qE = (19300 C) (2.22 V) = 42.8 kJ.

Adiponitrile is manufactured electrolytically from acrylonitrile CH2= CHCN  CN  (CH2 )4  CH

How many kg of adiponitrile (molecular mass = 108) is produced in 9.65 hr using a current of 3750 A with 80% efficiency ?





(B) 2CH2= CHCN + 2H+ + 2e- CN  (CH2)4 CN

m=1082×10-3×3000×9.65×360096500=58.32 kg

A certain metal salt solutions is electrolysed in series with a silver coulometer. The weights of silver and the metal deposited are 0.5094 g and 0.2653 g. Calculate the valency of the metal if its atomic weight is nearly that of silver. 





(B) Evidently 0.50940.2653=Equivalent weight of AgEquivalent weight of metal0.2

valency ratio =Valency of metalValency of Ag =2

The reduction potential of a half-cell consisting of a Pt electrode immersed in 1.5 M Fe2+and 0.015 M Fe3+ solution at 25°C is EAg+, Ag°=+0.770 v.





(A) E=0.77-0.0591 logFe2+Fe3+=0.77-0.059 log1.50.015=0.77-0.118=0.652 V

In a concentration cell, Zn/Zn2+ (1.0 M) || Zn2+ (0.15 M)/Zn as the cell discharges,





(B). In the above representation of the cell, the reaction proceeds to the left; 1.0 M solution is used up.

In a half-cell containing [Tl3+] = 0.1 M and [Tl+ ] = 0.01 M, the cell potential is – 1.2496 V for the reaction Tl+ Tl3+ + 2e-  The standard reduction potential of the Tl+ /Tl3+ couple at 25°C is





(D) For the oxidation reaction, 

Tl+Tl3++2e-E=E°-RTnFlogoxired-1.2496 = E°-0.0592log0.010.01                  E°=1.220 V

The standard reduction potentials for Mn3+/Mn2+ and MnO2 /Mn3+ couples in acid medium at 25°C are 1.50 V and 1.00 V respectively. The following reaction:

2Mn3+ + 2H2O  Mn2+ + MnO2 + 4H+ is





(C) Mn3++e-Mn2+;E°=1.50 VMn3++2H2OMnO2+4H++e-; E°=-1.0 V

Eº for the above reaction=1.50 V + (– 1.0)V = 0.50 V

Since Eº is positive,Gº is negative and the reaction is spontaneous.

If a 100 mL solution of 0.1M HBr is titrated using a very concentrated solution of NaOH, then the conductivity (specific conductance) of the solution at the equivalence point will be (assume volume change is negligible due to addition of NaOH). Calculate your answer in terms of 10-1Sm-1,

Given °Na+=8×10-3Sm2mol-1 °Br-=4×10-3S m2 mol-1





(B) At the equivalence point the concentrations will be [Br-] = 100 m3, [Na+] = 100 m3

Therefore, 

Ktotal=kBr- +kNa+=1.2 Sm-1         =12×10-1Sm-1

The volume of gases liberated at STP when a charge of 2F is passed through aqueous solution of sodium phosphate, is :





(C) At anode 2H2O O2 + 4H+ + 4e-

At cathode 2H2O + 2e-2OH-+ H2

After passage of 2F, one mole of H2 and 1/2 mole of O2 would be produced. The total volume is 33.6L.

It is desired to convert the energy derived from the combustion of propane into electrical energy, via a fuel cell. Given that the standard free energies of formation of C3H8 (g), H2O(l) and CO2 (g) are -23.5, -237.2 and -394.4 kJ mol-1 respectively, calculate the standard EMF of the propane fuel cell.





(A) For the combustion of propane

C3H8+5O23CO2g+4H2Ol the overall G° for the reaction

= 3x + 394.4 + 4x  237.2 = [23.5 + 0]= 1183.2  948.8 + 23.5 = 2108.5 kJC3H8(g) + 6H2O(l)  3CO2(g) + 20H = 20e5O2(g) + 2OH+ + 20e 10H2O(l)C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (l)

20 electrons are involved in the cell reactions

G0=-nFE0;E0=-G0nF=+21085×10320×96500=1.10 V

Na-amalgam is prepared by electrolysis of NaCl solution using liquid Hg as cathode. How long should the current of 10 amp. is passed to produce 10% Na – Hg on a cathode of 10 gm Hg. (atomic mass of Na = 23).





(A) 90 gm Hg has 10 gm Na

10 gm Hg=1090×10=109gm Naweight of Na=Mn×i×t96500109=231×10×t96500    Na++eNa  =10×965009×10+23=7.77 min

The standard reduction potential of a silver chloride electrode is 0.2 V and that of a silver electrode is 0.79 V. The maximum amount of AgCl that can dissolve in 106 L of a 0.1 M AgNO3 solution is 





(B) AgCl+1e-Ag+Cl-    E°=-0.2 VAg             Ag++1e-   E°=-0.2 VAgCl 1e-Ag+Cl-          E°=-0.2 VE°=0.059nlog K -0.59=0.0591log Ksp Ksp=10-10

Now solubility of  AgCl in 0.1 M AgNO3

SS+0.1 = 10-10 S = 10-9 mol/L

Hence 1 mole dissolves in 109 L solution hence in 106 L amount that dissolves in 1 mmol.

The k= 4.95 × 10-5S cm-1 for a 0.00099 M solution. Calculate the reciprocal of the degree of dissociation of acetic acid, if m0 for acetic acid is 400 S cm2mol-1 





(B) λM=100×4.95×10-50.00099=50 S cm2mol-1α=50400=0.125

1/.125 = 8

A hydrogen electrode X was placed in a buffer solution of sodium acetate and acetic acid in the ratio a : b and another hydrogen electrode Y was placed in a buffer solution of sodium acetate and acetic acid in the ratio b : a. If reduction potential values for two cells are found to be E1and E2 respectively w.r.t standard hydrogen electrode, the pKa value of the acid can be given as





(B) H++e1/2H2gE1=0-0.0591 log 1H+1E1=0+0.591 log H+1=-0.591 pH1E2=-0.591 pH2pH1=pka+logSaltAcid ; pH1=pka+logab     .....1pH1=pka+log ba ; pH2=pka-log ab       .........2Add 1 & 2 pH1+pH2=2 pka2pka=- E10.0591-E20.0591 ; pka=-E1+E20.118

At what Br-CO32-does the following cell have its reaction at equilibrium ?

Ag(s) | Ag2CO3 (s) | Na2CO3 (aq) | | KBr(aq) | AgBr(s) | Ag(s)

KSP = 8 × 10-12 for Ag2 CO3 and KSP = 4 × 10-13 for AgBr





(B) anode : Ag(s)Ag+aq+1e-

cathode: Ag+aq+1e-Ag

Net :  Ag+AgBr1e-Ag+Ag2CO30=0+0.0591log KSPAgBrBr-KSpAg2CO3CO32-KSPAgBrBr-=KSpAg2CO3CO32-4×10-138×10-2=Br-CO32-Br-CO32-=2×10-7