NEET Chemistry Questions: Redox Reactions & Electrochemisty

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How much time is required for the complete decomposition of 2 moles of water using a current of 2 ampere-





(B) 2OH-H2O+12O2+2e-Anode2F=1 mol H2O decomposed; so for 2 mol of H2O, 4F of electricity required.Q=i×ti=2amp.t=4×965002=193000sec=53.61 hours.

A certain current liberates 0.504 g of hydrogen in 2 hours. How many gram of copper can be liberated by the same current flowing for the same time in aqueous CuSO4 solution :





(B) Eq. of H2 = Eq. of Cu0.1541=w63.5/2wCu=16 g

Calculate the maximum work that can be obtained from the Daniell cell given below -

Zn(s) | Zn2+ (aq) || Cu2+ (aq) | Cu(s).

Given that EZn2+/Zn°=-0.76 V and ECu2+/Cu°=+0.34 V





(A) Cell reaction is : Zn(s) + Cu2+(aq)Cu(s) + Zn2+(aq)

Here n=2

Ecell= Ecathode- Eanode

                          (On the basis of reduction potential)

= + 0.34 – (– 0.76) = 1.10 V

We know that : Wmax =G =  nFE

=  (2 mol) × (96500 C mol) × (1.10 V) =  212300 J

or Wmax =  212300 J.

The metal that cannot be produced on reduction of its oxide by aluminium is :





(A) EOP of K > EOP of Al.

We have taken a saturated solution of AgBr, Ksp of AgBr is 12 × 10-14. If 10-7mole of AgNO3 are added to 1 litre of this solution then the conductivity of this solution in terms of 10-7 Sm-1 units will be

[λ°Ag+=4×10-3Sm2mol-1 ;λ°Br-=6×10-3Sm2mol-1,λ°NO3-=5×10-3Sm2mol-1]





(A). The solubility of AgBr in presence of 10-7 molar  AgNO3 is 3 × 10-7M.

Therefore [Br-] = 3 × 10-4 m3, [Ag+] = 4 × 10-4 m3 and NO3-=10-4m3

Therefore ktotal = kBr-+kAg+ + kNO3- = 55 Sm-1

The standard reduction potentials, E, for the half reactions are as Zn = Zn2++ 2e; E° = + 0.76 Vand Fe = Fe2+ + 2e ; E° = + 0.41 V; the e.m.f. for the cell reaction, Fe2++ Zn = Zn2++ Fe  is-





(B). Since oxidation potential of Zn is higher than Fe so it will act as anode simultaneously Fe will act as cathode so EMF of cell will be Eº (ox.) of Zn + Eº (red) of Fe.

E = 0.76 + ( - 0.41) =  + 0.35 V

The calomel electrode is reversible with respect to-





(D). Since half cell reaction;

2e + Hg2Cl2(aq.)  2Hgl + 2Cl-

Zn amalgam is prepared by electrolysis of aqueous ZnCl2 using Hg cathode (9 gm). How much current is to be passed through ZnCl2 solution for 1000 seconds to prepare a Zn Amalgam with 25% Zn by wt. (Zn = 65.4)





C). Let x gm of Zn deposit on 9 gm of Hg

% of Zn in Amalgam =x9+x×100=25   x=3gm

Eq. of Zn= 3×265.4; Current=665.4×965001000=8.85 amp.

The standard oxidation potentials of Cu/Cu2+and Cu+ /Cu2+ are  0.34V and  0.16 Vrespectively. The standard electrode potential of Cu+ /Cu would be :





(B). Reactions Cu2++ 2e- Cu ; G° =  nFE°

          Cu+  Cu2++ e- ;G°I =2F × 0.34 = 0.68F

G°II = F × 0.16 = 0.16F

Adding, we get 

 Cu+ + e-Cu

G°III=G°I + G°II= 0.52F =  FE° E° = 0.52V

 

Acidified water is electrolysed using an inert electrode. The volume of gases liberated at STP is 0.168L. The quantity of charge passed through the acidified water would be:





(C) 2H2O2x2H2gx+O2g3x = 0.168x = 0.056LVH2 = 2x = 0.112L, VO2 = x = 0.056L11.2L of H2 at STP  1F0.112L of H2 at STP  0.01F0.056L of O2 at STP = 0.01F

 The amount of electricity passed = 0.01F = 965C

 E0 for the reaction Fe + Zn2+ = Zn + Fe2+ is  0.35 V.The given cell reaction is- 





(B). Since EMF of the cell is negative i.e. Free energy change will be positive so cell reaction will not be feasible.

How much charge should be supplied to a cell for the electrolytic production of 245 gm NaClO4 from NaClO3 if the anode efficiency for the required reaction is 60%?





(A) ClO4- + 2H+ + 2e-ClO3- + H2O

Number of equivalents of NaClO4=24561.25=44F

No. of Faradays =4×10060=6.67F=6.43×105C

During an electrolysis of conc. H2 SO4, perdisulphuric acid (H2 S2O8 ) and O2 form in equimolar amount. The amount of H2 that will form simultaneously will be 

(2H2 SO4H2 S2O8 + 2H+ + 2e- )





(A) Anode 2H2SO4H2S2O8+2H++2e-2H2OO2+4H++4e-

Cathode 2H2OH2+2OH-2e-×3__________________________________

Net : 2H2SO4 + 8H2O H2S2O8 + O2 + 3H2 + 6H+ + 6OH-

Hence ratio of nO2 and nH2 is 1 : 3.

The specific conductivity of solution depends upon :





(C). Specific conductance is the conductance per c.c. solution

Value of m0 for  SrCl2  in water at 25°C from the following data

Conc. (mol/It)                         0.25             1

mΩ-1cm2mol-1            260              250





(A). m=m0-bC260=m0-0.5 b                  .....1250=m0-b                         ......2On solving1&2, we getm0=270Ω-1cm2mol-1