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NEET Chemistry MCQ
Redox Reactions And Electro Chemistry Mcq
Quiz 3
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Q.1
For electrochemical cell M|M+ || X- |X, Eo(M+/M)=0.44 Vand Eo(X/X-)=0.33 VFrom this data, one can deduce that [ IIT 2000]
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a) M + X → M+ + X- is the spontaneous reaction
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b) M+ + X- → M + X is the spontaneous reaction
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c)Ecell=0.77 V
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d)Ecell=-0.77 V
Explanation
M|M+ || X- |X Cell reaction should be M + X →M+ + X- Eocell=Eocathode - Eoanode Eocell=0.33 - 0.44=-0.11 V ∴ Reverse cell reaction should be spontaneous i.e M+ + X- → M + X is the spontaneous reaction Answer:(b)
Q.2
The molar conductivity is maximum for the solution of concentration
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a) 0.004 M
0%
b) 0.002 M
0%
c) 0.005 M
0%
d) 0.001 M
Explanation
Molar conductivity is inversely proportional to molarity Answer: (d)
Q.3
The emf of the cell Ag|Ag+( 0.1M)|| Ag+( 1M) | Ag at 298K is
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a)0.59 V
0%
b) 5.9 V
0%
c)0.059 V
0%
d)0.0059 V
Explanation
This is a example of concentration cell Anode and cathode are of same material Thus Ecello=0 in Nernst's equation Thus Answer: (c)
Q.4
resistance of a conductivity cell filled with solution of an electrolyte of concentration 0.1M is 100 Ω. the conductivity of this solution is 1.29 Sm-Resistance of the same cell when filled with 0.2M of the same solution is 520Ω. The molar conductivity of 0.02M solution of the electrolyte will be [ AIEEE 2006]
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a) 124 × 10-4S m2 mol-1
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b) 1240 × 10-4S m2 mol-1
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c)1.24 × 10-4S m2 mol-1
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d)12.4 × 10-4S m2 mol-1
Explanation
from the formula for conductivity κNow R=520 Ω for 0.2M, C=0.02molar conductivityAnswer: (a)
Q.5
The Eo (M3+ / M2+) values for Cr, Mn, Fe and Co are -0.41, +1.57, +0.77 and +1.97 respectively . For which one of these metals the change in oxidation state from +2 to +3 is easiest? [ AIEEE 2004]
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a) Cr
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b) Co
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c)Fe
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d)Mn
Explanation
Oxidation is easiest for metal whose reduction potential is lower Answer:(a)
Q.6
For the redox reaction Zn + Cu2+(0.1M) → Zn2+(1M) + Cu taking place in cell, Eocell is 1.1 V. Ecell for the cell will be [ AIEEE 2003]
0%
a) 2.14 V
0%
b) 1.8 V
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c) 1.07 V
0%
d) 0.82 V
Explanation
From Nernst equation Answer: (c)
Q.7
The half cell reaction for the corrosion are 2H+ + ½ O2 → H2O ; Eo=-1.23 V Fe2+ + 2e- → Fe ; Eo=- 0.44 V Find ΔGo ( in kJ) for the over all reaction [ IIT 2005]
0%
a)-76
0%
b) -322
0%
c)-161
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d)-152
Explanation
The reduction potential of two half reactions suggest that reduction of H+ and oxidation of Fe take place, so Eocell=Eocathode - EoanodeEocell=1.23 - ( -0.44)=1.67 V Applying ΔGo=-nFEo ΔGo=-2 × 96500 × 1.67=-322310J=-322 kJ Answer: (b)
Q.8
Standard reduction electrode potential of three metals A, B and C are respectively +0.05V, -3.0V and -1.2V. The reducing powers of ..[ AIEEE 2003]
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a) B > C > A
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b) A > B > C
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c)C > B > A
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d)A > C > B
Explanation
Lower the reduction potential more is the reducing powerAnswer: (a)
Q.9
Calculate the volume of H2 gas at NTP obtained by passing 4 amperes through acidified H2O for 30 minutes
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a) 0.0836 L
0%
b) 0.0432 L
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c)0.1672 L
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d)0.836 L
Explanation
Q=4 × 30 ×60 C=7200 C H2O → H2 + (1/2) O2 2H+ + 2e- → H2 ( 1 mole) or 22.4 L at NTP Thus 2F=2 × 96500 C produce 22.4 L at NTP ∴ 7200 C will give H2= Answer:(d)
Q.10
Consider the following Eo values Eo ( Fe3+/ Fe2+ )=+0.77 V , Eo ( Sn2+/ Sn )=-0.14V Under standard conditions, the potential for the reaction Sn + 2Fe3+ → 2Fe2+ + Sn2+ is .. [ AIEEE 2004]
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a) 1.68 V
0%
b) 0.63 V
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c) 0.91 V
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d) 1.4 V
Explanation
Since reduction potential of Fe3+/ Fe2+ is grater will act as cathode Eo=Eocathode - Eoanode Eo=0.77 - ( -0.14)=0.91V Answer: (c)
Q.11
The standard reduction potential at 298K for the following half reactions are given against each, which willa act as best reducing agent Zn2+ + 2e- → Zn ; -0.762 V Cr3+ + 3e- → Zn ; -0.740 V 2H+ + 2e- → H2 ; -0.00 VFe3+ + e- → Fe2+ ; -0.762 V
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a)Zn
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b) Cr
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c)H2
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d)Fe2+
Explanation
Smaller the reduction potential stronger the reducing agentAnswer: (a)
Q.12
Standard electrode potential (Eo) for OCl-/Cl- and Cl- / (1/2) Cl2 are respectively 0.94 V and -1.36 V. The Eo value for OCl-/ Cl2 will be
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a) -0.42 V
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b) -2.20 V
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c)0.52 V
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d)1.04 V
Explanation
Half reactions areCl- → (1/2)Cl2 + e-; Eo=-1.36 VOCl- + H2O + e- → Cl- + 2OH- ; Eo=+ 0.94V Total reaction OCl- + H2O → (1/2)Cl2 + 2OH- Eo=(-1.36 + 0.94) V=-0.42 VAnswer: (a)
Q.13
The oxidation state of chromium in Cr(CO)6 is ..[AIIMS 1993]
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a)0
0%
b) +2
0%
c)-2
0%
d)+6
Explanation
Oxidation number of metal in metal carbonyls is zeroAnswer: (a)
Q.14
Given The value of standard electrode potential for the changeFe3+ + e- → Fe2+ will be.. [ AIEEE 2009]
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a) -0.072 V
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b) 0.85 V
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c) 0.770 V
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d) -0.270 V
Explanation
Given Fe3+ + 3e- → Fe ; E10=-0.036 V Fe2+ + 2e- → Fe ; E20=-0.439 VRequired equation is Fe3+ + e- → Fe2+ ; E30=?Applying ΔG3o=ΔG1o - ΔG2o(-n3E3o)=(-n1E1o) - ((-n2E2o) E3o=3E1o - 2E2oE3o=3 × ( -0.036) - 2 × (-0.439) E3o=0.77 VAnswer: (c)
Q.15
The unit of molar conductivity is
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a)Ω cm2 mol-1
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b) Ω-1 cm2 mol-1
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c)Ωcm2 mol-1
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d)Ω-1 cm-2 mol-1
Explanation
Answer: (b)
Q.16
In corrosion of iron
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a) An electrochemical ( galvanic cell) is formed in which Fe acts as anode and cathode where O2 is reduced
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b) Electrons flow from anode to cathode through the metal while ions flow through the water droplets
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c)Dissolved O2 oxidises Fe2+ to Fe3+ before it is deposited as rust ( Fe2 . H2O)
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d)All of the above takes place
Explanation
Corrosion is an electrochemical process in which Fe2+ are oxidised to Fe3+Answer: (d)
Q.17
A standard hydrogen electrode has zero electrode potential because [ IIT 1997]
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a) Hydrogen is easier to oxidise
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b) This electrode potential is assumed to be zero
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c)Hydrogen atom has only electron
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d)hydrogen is the lightest element
Explanation
Answer:(b)
Q.18
The oxidation potential for the half cell reactions are Zn → Zn+ +2e- ; Eo=+0.76 V Fe → Fe+ +2e- ; Eo=+0.41 V The emf of cell reaction Fe+ + Zn → Zn+ + Fe is [IIT 1981]
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a) -0.35 V
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b) +0.35 V
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c) +1.17 V
0%
d) 0.117 V
Explanation
Oxidation potential is given Since Zn → Zn+ +2e- oxidation potential is more will act as anode Eocell=Eanode - Ecathode Eocell=0.76 -0.41=0.35 V Answer: (b)
Q.19
Given The potential of cell Cr|Cr3+(0.1M) || Fr2+(0.01M)| Fe is [ AIEEE 2008]
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a)-0.26 V
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b) 0.26 V
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c)0.339 V
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d)-0.339 V
Explanation
Since reduction potentials are given Eocell=Eocathode - EoanodeEocell=-0.42-(-0.72)=0.3According to Nernst equationAnswer: (b)
Q.20
A gas X at 1 atm pressure is bubbled through a solution containing a mixture of 1M Y- ions and 1M Z- ions at 125°C. If reduction potential of Z > Y; X > Z then
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a) Y will oxidise Z and not X
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b) Y will oxidise X and not Z
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c)Y will oxidise X and Z
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d)Y will reduce both X and Z
Explanation
Reduction potential of X > Z > Y As reducing potential of Y is lower, it will get oxidised by both X and Z that means it will reduce both X and Z Answer: (d)
Q.21
Best way to prevent rusting of iron is by
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a) Making iron cathode
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b) Putting it in saline water
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c)Both of these
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d)None of these
Explanation
Corrosion takes place on anode. By making iron, cathode, rusting of iron can be prevented Answer:(a)
Q.22
Saturated solution of KNO3 is used to make salt bridge because [ IIT 2001]
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a) Velocity of K+ is greater than that of NO3-
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b) Velocity of NO3- is greater than that of K+
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c) Velocity of both K+ and NO3-
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d) KNO3 is highly soluble in water
Explanation
Ionic mobility of both K+ and NO3- are nearly equal Answer: (c)
Q.23
Which of the following is not a redox reaction? [ AIIMS 1993]
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a)CaCO3 → CaO +CO2
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b) O2 + 2H2 → 2H2O
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c)Na + H2O → NaOH + ½ H2
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d)MnCl3 → MnCl2 + ½Cl2
Explanation
in option 'a' Oxidation number of Ca, O and C do not change so it is not a redox reactionAnswer: (a)
Q.24
Which substance serves as reducing agent in the following reaction?14H+ Cr2O7- + 3Ni → 2Cr3+ + 7H2O + 3Ni2+ [ CBSE PMT 1994]
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a) H2O
0%
b) Ni
0%
c)H+
0%
d)Cr2O72-
Explanation
Oxidation number of Ni changes from 0 to +2 and hence Ni acts as a reducing agentAnswer: (b)
Q.25
Nitric oxide acts as a reducing agent in the reaction [ DEC 1994]
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a) 4NH3 + 5O2 → 4NO + 6H2O
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b) 2NO + 3I2 +4 H2O → 2NO3- + 6I- + 8H+
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c)2NO + H2SO3 → N2O + H2SO4
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d)2NO + H2S → N2O + S + H2O
Explanation
In option 'b' reaction oxidation number of Nitrogen changes from +2 to +5 in NO3- thus acts as reducing agent Answer:(b)
Q.26
In which of the following reaction, hydrogen is acting as an oxidising agent? [ MPPMT 1987]
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a) With iodine to give hydrogen iodide
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b) With lithium to give lithium hydride
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c) With nitrogen to give ammonia
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d) With sulphur to give hydrogen sulphide
Explanation
in option 'b' hydrogen oxidation number reduces to -1 from 0, thus hydrogen acts as oxidizing agent Answer: (b)
Q.27
When SO2 is passed through solution of potassium iodate, the oxidation state of iodine changes from .. [ Roorkee 1990]
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a)+5 to 0
0%
b) +5 to -1
0%
c)-5 to 0
0%
d)-7 to -1
Explanation
3KIO3 + 3SO2 → 3KI + 3SO2 During this reaction oxidation number of iodine changes from +5 in KIO3 to -1 in KIAnswer: (b)
Q.28
M is the molecular weight of KMnOthe equivalent weight of KMnO4 when it is converted into K2MnO4 is .. [ AFMC 1994]
0%
a) M
0%
b) M/3
0%
c)M/5
0%
d)M/7
Explanation
Since only one electron is exchanged thus Equivalent weight=molecular weight / exchange of electrons==M/1=MAnswer: (a)
Q.29
In the reaction, 3Br2 + 6CO32- + 3H2O → 5Br- + BrO3- + 6HCO3- [ MP PMT 1995, Pb CET 1996]
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a) Bromine is oxidised and carbonate is reduced
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b) Bromine is reduced and water is oxidised
0%
c)Bromine is neither reduced nor oxidised
0%
d)Bromine is both reduced and oxidised
Explanation
Answer:(d)
Q.30
Which of the following elements has least oxidation number [ Rajasthan PMT 1998]
0%
a) Ni(CN)4
0%
b) Ni(CO)4
0%
c) Fe2O3
0%
d) SF4
Explanation
Answer: (b)
0 h : 0 m : 1 s
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