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NEET Chemistry MCQ
Redox Reactions And Electro Chemistry Mcq
Quiz 6
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Q.1
Among the following, identify the species with an atom in +6 oxidation state [ IIT screening test 2000]
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a) MnO4-
0%
b)Cr(CN)63-
0%
c)NiF62-
0%
d)CrO2Cl2
Explanation
Answer:(d)
Q.2
A mole of N2H4 loses ten moles of electrons to form a new compound Y. Assuming that all the nitrogen appears in the new compound, what is the oxidation state of nitrogen in Y ( there is no change in oxidation number of hydrogen)
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a) -1
0%
b) -3
0%
c) +3
0%
d) +5
Explanation
Oxidation number of Nitrogen in N2H4 is -2 Every N looses 5 electrons thus N-2 → Nx + 5e- x - 5=-2 x=+3 Answer: (c)
Q.3
In which of the following, transition metal has zero oxidation state? [ CBSE PMT 1999]
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a)[Fe(CO)5]
0%
b) NH2 . NH2
0%
c)NOClO4
0%
d)CrO5
Explanation
Oxidation number of metal carbonyls is zeroAnswer: (a)
Q.4
In standardization of Na2S2O3 using K2Cr2O7 by iodometery, equivalent weight of K2Cr2O7 is [ IIT 2000]
0%
a) (Molecular weight)/2
0%
b) (Molecular weight)/6
0%
c)(Molecular weight)/3
0%
d)same as molecular weight
Explanation
Cr2O7-2 +14H+ +65e- → 2Cr3+ + 7H2OOxidation number of Cr changed by 3 per atom but two atoms of Cr thus changed by 6 or electron exchange is 6 thus Molecular weight/6 is equivalent weightAnswer: (b)
Q.5
HNO3 acts as [ Manipal 2001]
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a) acid
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b) oxidizing agent
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c)reducing agent
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d)both a and b
Explanation
Answer:(d)
Q.6
n a reaction, 4 moles of electrons are transferred to one mole of HNOthe possible product obtained due to reduction is
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a) 0.5 mole of N2
0%
b) 0.5 mole of N2O
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c) 1 mole of NO2
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d) 1 mole of NH3
Explanation
Oxidation number of N in HNO3 is +5 in product it changed to +1 which is possible for N2O Thus two moles of HNO3 can gibe one mole of N2O ∴ from 1 mole of HNO3 produces 0.5moles of N2O Answer: (b)
Q.7
The equivalent weight of phosphoric acid(H3PO4 ) in the reaction, NaOH + H3PO4 → NaH2PO4 + H2O is [AIIMS 1999]
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a)59
0%
b) 49
0%
c)25
0%
d)98
Explanation
In this reaction oxidation state of the P is not changed thus equivalent weight=molecular weight of H3PO4=98gAnswer: (d)
Q.8
The reaction 2H2O(l) → 4H+(aq) + O2(g) +4e- is [kerala MEE]
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a) An oxidation reaction
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b) A reduction reaction
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c)A redox reaction
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d)Sololysis reaction
Explanation
Answer: (a)
Q.9
The average oxidation number of sulphur in Na2S4O6 is [ Rajasthan PMT 2002]
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a)1.5
0%
b) 2.5
0%
c)3
0%
d)2
Explanation
Formula is Oxidation number of two sulphur shown in dotted line box is zero thus oxidation number of remaining two is 2×(+1) + 2x +6×(-2)=0 x=+5 Average oxidation number is (5+5) /4=2.5 Answer:(b)
Q.10
Equivalent weight of MnO4- in acidic, basic and neutral media are in the ratio of
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a) 3:5:15
0%
b) 5:3:1
0%
c) 5:1:3
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d) 3:5:5
Explanation
In acidic medium oxidation number charges by 5 in basic medium oxidation number changes by 3+ In neutral medium oxidation number changes by 3 If M is the molecular weight of KMnO4, then ratio of molecular weight in acidic , basic and neutral medium M/5 : M/ 3: M/3 or 3 : 5 : 5 Answer: (d)
Q.11
The standard reduction potential at 298K for following half reactions are given against each Zn2+ +2e- → Zn(s) , -0.762 Cr3+ +3e- → Cr(s) , -0.740 2H+ +2e- → H2 , 0.00 Fe3+(aq) + e- → Fe2+ 0.777 Which is the strongest reducing agent? [ IIT 1981]
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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
More the negative reduction potential more stronger reducing agent more the positive oxidation potential more stronger reducing agentAnswer: (a)
Q.12
The two Pt electrodes fitted in a conductance cell are 1.5cm apart while the cross-sectional area of each electrode is 0.75cmWhat is the cell constant? [ Pb CET 1989]
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a) 1.25
0%
b) 0.5 cm
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c)2.0 cm-1
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d)0.2 cm-
Explanation
Cell constant=l/A=1.5cm/0.75cm2Cell constant=2.0cm- Answer: (c)
Q.13
What is the standard cell potential for the cellZn|Zn2+(1M)|| Cu2+(1M)|Cu?Eo for Zn2+/Zn=-0.76Eo for Cu2+/Cu=0.34 AIIMS 1980]
0%
a)-0.42V
0%
b) +0.42V
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c)+1.10V
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d)-1.10V
Explanation
potentials given are reduction potential Thus Cathod potential - anode potential +0.34-(-0.76)=+1.10V Answer:(c)
Q.14
The ionic mobility of the cation and the anion of a salt are 140 and 80 mhos respectively. The equivalent conductivity of the salt at infinite dilution will be [ Harayana CEET 1991]
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a) 160 mhos
0%
b) 280 mhos
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c) 66 mhos
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d) 220 mhos
Explanation
Λeqo=λa + λc Λeqo=140 + 80=220 mho Answer: (d)
Q.15
The number of Faradays needed to reduce 4gm equivalents of Cu2+ to Cu metal will be .. [ BHU]
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a)1
0%
b) 2
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c)1/2
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d)4
Explanation
1 gram equivalent requires 1FaradaysAnswer: (d)
Q.16
The conductivity of a saturated solution of BaSO4 is 3.06 × 10- ohm- cm- and its equivalent conductance is 1.53 ohm- cm2 equiv-. the Ksp of BaSO4 will be ..[haryana CEET 1996]
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a) 4 × 10-12
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b) 2.5 ×10-9
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c)2.5 × 10-13
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d)4 × 10-6
Explanation
Ba(SO4 → Ba2+ + So2- If S is solubility then Ksp=S2Answer: (d)
Q.17
Conductivity of a solution is directly proportional to ..[KCET 1984]
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a) dilution
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b) number of ions
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c)current density
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d)volume of the solution
Explanation
Answer:(b)
Q.18
The equivalent conductivity of 0.1M weak acid is 100times less than that at infinite dilution. The degree of dissociation is [ Tamil nadu CET2001]
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a) 100
0%
b) 10
0%
c) 0.01
0%
d) 0.001
Explanation
Answer: (c)
Q.19
the number of electrons involved in the reaction when a faraday of electricity is passed through an electrolyte in solution is [ NCERT 1982]
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a)12 × 1046
0%
b) 96540
0%
c)8 ×1016
0%
d)6 × 1023
Explanation
One Faraday=charge on 1 mole of electrons Answer: (d)
Q.20
What weight of copper will be deposited by passing 2 F of electricity through a cupric salt ( At wt. Cu=63.5] [NCERT 1973]
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a) 2.0 g
0%
b) 3.175 g
0%
c)63.5 g
0%
d)127.0 g
Explanation
Cupric salt have copper ion charge of +2 thus Two moles of charge deposited one mole of copper=63.5gAnswer: (c)
Q.21
Which one of the following will increase the voltage of the cellSn(s) + 2Ag+(aq) → Sn2+(aq) + 2Ag(s)? [ Pb CET 1985]
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a)Increase in the size of the silver ros
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b) Increase in the concentration of Sn2+ ions
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c)Increase in the concentration of Ag+ ions
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d)None of the above
Explanation
According to Nernst equationBy increasing concentration of Ag+ cell potential will increase Answer:(c)
Q.22
Specific conductance of 0.1M sodium chloride solution is 1.06×10-2 ohm-1 cm-Its molar conductance in ohm-1cm2 mol-1 is [ kerala MEE 2001]
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a) 1.06 ×102
0%
b) 1.06 ×103
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c) 1.06 ×104
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d) 5.3 ×102
Explanation
Use formula Answer: (a)
Q.23
In electrolysis of dil H2SO4 using platinum electrodes.. [ DPMT 1983]
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a)H2 is evolved at cathode
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b) SO2 is produced at anode
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c)O2 is obtained at cathode
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d)SO2 is produced at cathode
Explanation
H2 is evolved at cathode and O2 is obtained at anodeAnswer: (a)
Q.24
When the same quantity of electricity is passed through the solutions of different electrolytes in series, the amount of products obtained are proportional to their.. [ JIPMER 1994]
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a) Atomic weights
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b) Chemical equivalents
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c)gram molecular volume
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d)gram atomic ions
Explanation
According to Feraday's second lawAnswer: (b)
Q.25
2.5 F of electricity is passed through a solution of CuSOthe number of gram gram equivalents of copper deposited on the cathode will be [ DPMT 1982]
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a) 1
0%
b) 2
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c)2.5
0%
d)1.25
Explanation
1 F=1 gram equivalent of Cu 2 F=2.5 gram equivalent Answer:(c)
Q.26
Molar conductivity of a solution is 1.26×102 ohm- cm2 mol-. Its molarity is 0.Its specific conductivity will be ..[manipal PMT 2002]
0%
a) 1.26 ×10-5
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b) 1.26 ×10-3
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c) 1.26 ×10-4
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d) 0.0063
Explanation
Use formula Answer: (b)
Q.27
Faraday's laws of electrolysis are related to [ IIT 1983]
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a)atomic number of cation
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b) atomic number of anion
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c)equivalent weight of the electrolyte
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d)speed of the cation
Explanation
Answer: (c)
Q.28
The one of that is good conductor of electricity in the following list of solids is [ NCERT 1984]
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a) Sodium chloride
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b) graphite
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c)diamond
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d)sulphur
Explanation
Answer: (b)
Q.29
Equivalent conductance of NaCl, HCl and CH3COONA at infinite dilution are 126.45, 426.16 and 91 ohm-1cm2 respectively. The equivalent conductance of CH3COOH at infinite dilution would be [ CBSE PMT 1997]
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a) 101.38 ohm-1cm2
0%
b) 253.62ohm-1cm2
0%
c)390.71ohm-1cm2
0%
d)678.90ohm-1cm2
Explanation
Λo(CH3COOH)=Λo(CH3COONa) + Λo (HCL) - Λo(NaCl)Λo(CH3COOH)=91 +426.16 -126.45=390.71 ohm-cm2 Answer:(c)
Q.30
The resistance of 1N solution of acetic acid is 250Ω, when measured in a cell of cell constant 1.15cm-. The equivalent conductance ( in ohm-Cm2 equiv-1) of 1N acetic acid is ..[ CBSE PMT 1994]
0%
a) 4.6
0%
b) 9.2
0%
c) 18.4
0%
d) 0.023
Explanation
κ=G × cell constant κ=cell constant / R=1.15 /250=0.0046 Answer: (a)
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