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NEET Chemistry MCQ
Redox Reactions And Electro Chemistry Mcq
Quiz 2
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Q.1
In the standardization of Na2S2O3 using K2Cr2O7 by iodomentry, the equivalent weight of K2Cr2O7 is [ IIT 2001]
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a) (Molecular weight)/2
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b) (molecular weight)/6
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c) (Molecular weight )/3
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d) Same as molecular weight
Explanation
K2Cr2O7 acts as oxidising agent in presence of dil H2SO4 Cr2O72- +14H+ +6e- → 2Cr3+ + 7H2O Equivalent wt=Molecular weight / number of electrons gain=M/6 Answer: (b)
Q.2
Oxidation state of Cl in CaOCl2 is
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a)0
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b) +1
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c)-1
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d)+1, -1
Explanation
Oxidation number of Cl in Cl-1=-1 Oxidation number of Cl in ClO+1 Thus Cl has +1 and -1 oxidation statesAnswer: (d)
Q.3
Amount of oxalic acid present in a solution can be determined by its titration with KMnO4 solution in presence of H2SOThe titration gives unsatisfactory result when carried out in the presence of HCl, because HCl ..[ AIEEE 2008]
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a) oxidises oxalic acid to carbon dioxide and water
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b) gets oxidised by oxalic acid to chlorine
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c)furnishes H+ ions in addition to those from oxalic acid
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d)reduces permanganate to Mn2+
Explanation
Titration can not be done in presence of HCl because KMno4 being strong oxidising agent oxidises HCl to Cl2 and get itself reduced to Mn2+. So actual amount of oxalic acid in solution cannot be determinedAnswer: (d)
Q.4
When copper is treated with a certain concentration of nitric acid, nitric oxide and nitrogen dioxide are liberated in equal volumes according to the equation xCu + yHNO3 → xCu(NO3)2 + 2NO + (y/2)H2O The coefficient x and y are
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a) 2 and 3
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b) 1 and 6
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c)1 and 3
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d)3 and 8
Explanation
Out of given options only x=3 and y=8 makes equation balanced HNO3 is dilute, nitric oxide instead of nitrogen dioxide 3Cu + 8HNo3 → 3Cu(NO3)2 + 2NO + 4H2 O Answer:(d)
Q.5
KI and CuSO4 solution when mixed give
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a) CuI2 + K2SO4
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b) Cu2I2 + K2SO4
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c) Cu2I2 + K2SO4 + I2
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d) CuI2 + K2SO4 + I2
Explanation
4KI + 2CuSO4 → Cu2I2 + K2SO4 + I2 Answer: (c)
Q.6
Equivalent mass of oxidising agent in the reaction SO2 + 2H2S → 3S +2H2O is
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a)32
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b) 64
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c)16
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d)8
Explanation
SO2 is acting as oxidising agent Equivalent mass of SO2=Molecular wiight / Change in oxidation number=64/4=16Answer: (c)
Q.7
Number of moles of KMnO4 required to produce one mole of Fr(C2O4) in acidic medium is
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a) 0.6
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b) 0.4
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c)0.2
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d)1.67
Explanation
MnO4- +8H+ +5e- → Mn2+ +5H2OFe(C2O4) → Fe2+ + C2O4- Fe2+ → Fe3+ + e- C2O42- → 2CO2 + 2e- We can see that one mole of KMnO4 accepts 5 electrons where as one mole of Fe(C2O40 loses 3 electrons ∴ No of moles of KMnO4 required to oxidise one mole of Fe(C2O4)=3/5=0.6 moleAnswer: (a)
Q.8
In the reaction 3CuO + 2NH3 → N2 + 3H2O + 3Cu the change of NH3 to N2 involvs
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a) Loss of 6 electrons per molecule of N2
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b) Loss of 3 electrons per molecule of N2
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c)Gain of 6 electrons per molecule of N2
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d)Gain of 3 electrons per molecule of N2
Explanation
Loss of 6 electrons for 1 molecule of N2 Answer:(a)
Q.9
White P racts with caustic soda, the products are PH3 and NaH2POthis reaction is an example of ..[ IIT 1990]
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a) Oxidation
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b) Reduction
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c) Disproportionation
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d) Neutralisation
Explanation
P under goes oxidation as well as reduction reaction inthe ssame reaction. thus it is a Disproportionation reaction Answer: (c)
Q.10
When KMnO4 acts as oxidizing agent and ultimately forms Mn42-, MnO2O3, Mn>2+ then the number of electrons transferred in each case respectively is ..[ AIEEE 2002]
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a)4, 3, 1, 5
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b) 1, 5, 3, 7
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c)1, 3, 4, 5
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d)3, 5, 7, 1
Explanation
Answer: (c)
Q.11
the chemical that undergoes self oxidation and self reduction in the same reaction is [ Ramil Nadu CET 2001]
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a) Bezyl alcohol
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b) Acetone
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c)Formaldehyde
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d)Acetic acid
Explanation
2HCHO → CH3OH + HCOOHAnswer: (c)
Q.12
What is the oxidation state of P in Ba(H2PO2)2
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a) +3
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b) +2
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c)+1
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d)-1
Explanation
Ba(H2PO2)2, let oxidation state of P=x, then+2 + 4(1) + 2(x) + 4(-2)=0x=+1 Answer:(c)
Q.13
For decolouration of 1 mole of acidified KMnO4 the moles of H2O2 required is
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a) 1/2
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b) 3/2
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c) 5/2
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d) 7/2
Explanation
(i) 2KMNO4 + 3H2SO4 → K2SO4 + H2O (ii) H2O2 + O → H2O + O2 Thus, 1 mole KMNO4 requires 5/2 moles of H2O2 for decolouration Answer: (c)
Q.14
The equivalent conductance of two strong electrolytes at infinite dilution in H2O ( Where ions moving freely through a solution)at 25°C are given below ΛCH3COONa=91.0 S cm2 / equiv. Λ HCl=426.2 S cm2 / equiv. What additional information/ quantity one needs to calculate Λ of an aqueous solution of acetic acid? [ AIEEE 2007]
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a)Λ of chloroacetic acid ( ClCH2COOH)
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b) Λ of NaCl
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c)Λ of CH3COOK
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d)The limiting equivalent conductance of H+ ( λ H+)
Explanation
According to Kohlrausch's law, the molar conductivity at infinite dilution ( Λ) for weak electrolyte, CH3COOH is Λ CH3COOH=ΛCH3COONa + Λ HCl - Λ NaCl Thus value of Λ NaCl should also be known Answer: (b)
Q.15
Several blocks of magnesium are fixed to the bottom of a ship to [ AIEEE 2003]
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a) Keep away the sharks
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b) MAke the ship lighter
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c)Prevent action of water and salt
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d)prevent puncturing by under sea-rocks
Explanation
Magnesium with lower reduction potential prevents rusting of iron in the atmosphere of water and saltAnswer: (c)
Q.16
In the electrolytic cell, flow of electrons is from .. [ IIT 2003]
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a) Cathode to anode in solution
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b) Cathode to anode through external supply
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c)Cathode to anode through internal supply
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d)Anode to cathode through internal supply
Explanation
In an electrolytic cell flow of electrons takes place from anode to cathode through external circuit and from cathode to anode through solution Answer:(a)
Q.17
Charge required to liberate 11.5 g sodium is Na+ +e- → Na 23 g of sodium requires 1 F charge for liberation for 11.5 g of sodium requires=11.5/23=0.5 F
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a) 96500 coulombs
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b) 1.5 F
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c) 0.1 F
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d) 0.5 F
Explanation
Na+ +e- → Na 23 g of sodium requires 1 F charge for liberation for 11.5 g of sodium requires=11.5/23=0.5 F Answer: (d)
Q.18
electrolysis of dilute aqueous NaCl solution was carried out by passing 10 mili ampere current. The time required to liberate 0.01 mol of H2 gas at the cathode is ( 1 Faraday=96500 C mol-1){IIT 2008]
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a)9.650 × 104 sec
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b) 19.3 ×104 sec
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c)28.95 × 104 sec
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d)38.6 × 104 sec
Explanation
Q=I × t Q=10 × 10-3 × t ---(1) 2 H2O + 2e- → H2 + 2OH- 0.1 mole of H2 is liberated by 0.02 Faraday of charge i.e Q=0.02 × 96500 .... (2) From (1) and (2) we get 10 × 10-3 × t=0.02 × 96500 t=19.3 × 104 secAnswer: (b)
Q.19
Reaction that takes place at graphite anode in dry cell is
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a) Zn2+ + 2e- → Zn
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b) Zn → Zn2+ + 2e-
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c)Mn2+ + 2e- → Mn
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d) Mn → Mn2+ + 2e-
Explanation
Answer: (b)
Q.20
The molar conductivities ΛNaOAc and ΛHCl at infinite dilution in water at 25°C are 91.0 and 426.2 S cm2/mol respectively. To calculate ΛHOAc, the additional value required is [ AIEEE 2006]
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a) ΛH2O
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b)ΛKCl
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c)ΛNaOH
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d)ΛNaCl
Explanation
NaOAc + HCl → HOAc + NaCl From reaction ΛHOAc=ΛNaOAc + ΛHCl - ΛNaCl Thus to calculate the value of ΛHoAC one should know the value of ΛNacl along with other values Answer:(d)
Q.21
Given the data at 25°C Ag + I - → AgI + e-; Eo=0.152 V Ag → Ag+ + e-; Eo=-0.800 VWhat is the value of logKsp for AgI [ AIEEE 2006]
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a)-8.12
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b) +8.612
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c)-37.83
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d)-16.13
Explanation
AgI + e- → Ag + I - ; Eo=-0.152 V Ag → Ag+ + e-; Eo=-0.800 VThus AgI → Ag+ + I-, Eo=-0.952 VFrom the formula for Ksp Answer: (d)
Q.22
The standard electrode reduction potentials of Zn, Ag and Cu are -0.76, 0.80 and 0.34 volt respectively, then
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a) Ag can oxidise Zn and Cu
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b) Ag can reduce Zn+ and Cu+
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c)Zn can reduce Ag+ and Cu2+
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d)Cu and oxidise Zn and Ag
Explanation
Lower the value of reduction potential stronger will be the reducing agentAnswer: (c)
Q.23
A depolarizer used in dry cell is
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a) Ammonium chloride
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b) Sodium carbonate
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c)Lead sulphate
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d)Manganese dioxide
Explanation
Answer:(d)
Q.24
The standard reduction potential forFe2+ / Fe and Sn2+ / Sn electrodes are -0.44 and -0.14 volts respectively. For the cell reaction Fe2+ + Sn → Fe + Sn2+ , the standard emf is [ IIT 1988]
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a) +0.30 V
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b) -0.58 V
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c) +0.58 V
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d) -0.30 V
Explanation
Higher value of standard reduction potential of Sn2+ / Sn shows that reduction will take take place on Sn electrode Act as cathode While Fe2+ / Fe will act as anode Now Eocell=Eocathode - Eoanode Eocell=-0.14 - ( -0.44 )=+0.30 v Answer: (a)
Q.25
In the electrolysis of water, one faraday of electrical energy would evolve
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a)8 g oxygen
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b) one mole of oxygen
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c)one g atom of oxygen
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d)22.4 litres of oxygen
Explanation
2 H2O → O2 +4H+ +4e-∴ 4 faraday produces 32 g of oxygen∴ 1 Faraday produces (32/4)=8 g of oxygen Answer: (a)
Q.26
The cell, Zn|Zn2+ ( 1M) || Cu2+ (1M) | Cu Ecello=1.10 V Was allowed to be completely discharged at 298K. The relative concentration of Zn2+ to Cu2+ is [AIEEE 2007]
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a) 9.65 × 104
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b) antilog(24.08)
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c)37.3
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d)1037.3
Explanation
Zn + Cu2+ → Zn2+ + Cu From Nernst equation When the cell is completely discharged Ecell=0 Answer: (d)
Q.27
The standard emf of a cell involving one electron chage is found to be 0.591V at 25°C. The equilibrium constant of the reaction is ( F=96500 C mol-1, R=8.314 JK- mol-)
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a) 1.0 ×101
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b) 1.0 ×1030
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c)1.0 ×1010
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d)1.0 ×105
Explanation
At equilibrium Eocell=0 thus from Nerns equation Answer:(c)
Q.28
In the cell that utilises the reaction Zn + 2H+ → Zn2+ + H2 addition of H2SO4 to cathode compartment, will [AIEEE 2004]
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a) Lower the E and shift equilibrium to left
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b) Increase the E and shift equilibrium to left
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c) Increases the E and shift equilibrium to the right
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d) Lower the E and Shift equilibrium to right
Explanation
Zn + 2H+ → Zn2+ + H2 Addition of H2SO4 ( H+ ions) shifts the equation to right and E increases Answer: (c)
Q.29
In a hydrogen - oxygen fuel cell, combustion of hydrogen occurs to [ AIEEE 2004]
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a)Generate heat
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b) Remove absorbed oxygen from electrode surfaces
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c)Produce high purity water
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d)Create potential difference between two electrodes
Explanation
In H2 - O2 fuel cell, combustion of H2 creates potential difference between two electrodesAnswer: (d)
Q.30
The e.m.f of cellZn|Zn+ ( 0.01M) || Fe+ ( 0.001M)| Fe at298K is 0.257 then the value of equilibrium constant for the cell reaction is [ IIT 2004]
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a)
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b)
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c)
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d)
Explanation
From Nernst equation We know that at equilibrium Ecell=0 thus Answer: (b)
0 h : 0 m : 1 s
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