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NEET Chemistry MCQ
Redox Reactions And Electro Chemistry Mcq
Quiz 1
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Q.1
Measuring Zeta potential is useful in determining which property of colloidal solution? [NEET 2020]
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a) Stability of the colloidal particles
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b) Size of the colloidal particles
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c)Viscosity
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d)OSolubility
Explanation
Answer: (a)
Q.2
On electrolysis of dil. sulphuric acid using Platinum (Pt) electrode, the product obtained at anode will be ... [NEET 2020]
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a) H2S gas
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b) SO2 gas
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c) Hydrogen gas
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d) Oxygen gas
Explanation
Answer: (d)
Q.3
The number of Faradays(F) required to produce 20 g of calcium from molten CaCl2 (Atomic mass of Ca = 40 g mol-1) is [NEET 2019]
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a) 3
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b) 4
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c) 1
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d) 2
Explanation
Answer:(c)
Q.4
What is the change in oxidation number of carbon in the following reaction? CH4 (g) + 4Cl2 (g) → CClsub>4 (l)+ 4HCl(g) [NEET 2019] Answer: (a)
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a) – 4 to + 4
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b) 0 to – 4
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c) + 4 to + 4
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d) 0 to + 4
Explanation
Answer: (a)
Q.5
Which of the following reactions are disproportionation reaction? [NEET 2019] (a) 2Cu+ → Cu2+ + Cu0 (b) 3MnO42- + 4H+ → 2MnO4 + MnO2 + 2H2O (c) 2KMnO4 → K2 MnO4+ MnO2 + O2 (d) 2MnO4- + 3Mn2+ + 2H2 O +5MnO2+ 4H+ Select the correct option from the following
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a)(a) and (b) only
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b) (a), (b) and (c)
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c)(a), (c) and (d)
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d) (a) and (d) only
Explanation
Answer: (a)
Q.6
For the cell reaction 2Fe3+ (aq) + 2I- (aq) → 2Fe2+ (aq) + I2 (aq) EΘcell=0.24 V at 298 K. The standard Gibbs energy (ΔrG Θ) of the cell reaction is : [Given that Faraday constant F = 96500 C mol-1]
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a) – 46.32 kJ mol–1
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b) – 23.16 kJ mol–1
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c)46.32 kJ mol–1
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d)23.16 kJ mol–1
Explanation
Answer: (a)
Q.7
he correct order of N-compounds in its decreasing order of oxidation states is ... [NEET 2018]
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a) HNO3, NH4Cl, NO, N2 2mole -1
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b)HNO3, NO, NH4Cl, N2
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c)HNO3, NO, N2, NH4Cl
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d) NH4Cl, N2, NO, HNO3
Explanation
Answer:(c)
Q.8
Consider the change in oxidation state of Bromine corresponding to different emf values as shown in the diagram below : Then the species undergoing disproportionation is [NEET 2018]
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a) Br2
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b) BrO4-
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c) 3BrO3-
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d) HBrO
Explanation
Answer: (d)
Q.9
For the redox reaction ... [2018] MnO4- + C2O42- + H+ → Mn2+ + CO2 + H2O the correct coefficients of the reactants for the balanced equation are
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a)MnO4- = 3; C2O42- = 16; H+ = 5
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b) MnO4- = 2; C2O42- = 5; H+ = 16
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c) MnO4- = 16; C2O42- = 5; H+ = 2
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d) MnO4- = 5; C2O42- = 16; H+ = 2
Explanation
Answer: (b)
Q.10
For a cell involving one electron E&dgecell = 0.59 V at 298 K, the equilibrium constant for the cell reaction is : Given that
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a) 1.0 × 102
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b) 1.0 × 105
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c) 1.0 × 1010
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d) 1.0 × 1030
Explanation
Zn2+ is stronger reducing agent, so can not reducedAnswer: (c)
Q.11
The chemical reaction.2AgCls + H2(g) → 2HCl(aq) + 2Ag(s)taking place in a galvanic cell is represented by the notation [ AFMC 2009]
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a)Pts| H2(g).1 bar| 1 M KCl(aq)| AgCl(s)| Ag(s)
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b)Pts| H2(g).1 bar| 1 M HCl(aq)| 1M Ag+(aq)| Ag(s)
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c)Pts| H2(g).1 bar| 1 M HCl(aq)| AgCl(s)| Ag(s)
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d)Pts| H2(g).1 bar| 1 M HCl(aq)| Ag(s)| AgCl(s)
Explanation
Oxidation potential of Hydrogen is more than Ag hence Hydrogen will act as Anode and Ag as cathode Answer:(b)
Q.12
Among the following compound, the oxidation state of Mn is highest in ..[ AFMC 2010]
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a) KMnO4
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b) K2MnO4
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c) MnO2
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d) Mn2O3
Explanation
Oxidation state of Mn is highest in KMnO4 Oxidation states are as follows- KMnO4=+7 K2MnO4=+6 MnO2=+4Mn2O3=+3Answer: (a)
Q.13
In alkaline condition KMhO0 reacts as follows: 2KMnO4 + 2KOH → 2K2MnO4 + H2O + O The equivalent weight of KMnO4 would be ( At. mass of K=39, Mn=55, O=16)
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a)79.0
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b) 31.6
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c)52.7
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d)158.0
Explanation
Change in oxidation state=1 Thus, equivalent weight of KMNO4 is equal to its molecular weight 158Answer: (d)
Q.14
How many electrons are involved in oxidation by KMnO4 in basic medium
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a) 1
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b) 2
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c)5
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d)3
Explanation
MnO4- +2H2 +3e- →MnO2 + 4OH- Oxidation number of Mn is +7 it becomes +4 thus electron involved is three Answer: (d)
Q.15
Which of the following chemical reactions depicts the oxidising behavior of H2SO4 [ AIEEE 2006]
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a) 2HI + H2SO4 → I2 + SO2 + 2H2O
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b)Ca(OH)2 + H2SO4 → CaSO4 + 2H2O
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c)NaCl + H2SO4 → NaHSO4 + HCl
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d)2PCl5 + H2SO4 → 2POCl3 +2HCl + SO2Cl2
Explanation
H2SO4 is reduced to SO4and thus acting as oxidising agent Answer:(a)
Q.16
The reaction 3ClO- → Cl3- + 2Cl- is an example of ..[ IIT 2001]
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a) Oxidation reaction
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b) Reduction reaction
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c) disproportion reaction
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d) decomposition reaction
Explanation
In given reaction chlorine is getting oxidised as well as reduced thus this is disproportion reaction Answer: (c)
Q.17
Cr2O72- + X + H+ → Cr3+ + H2O + oxidised product of X X in above reaction can not be
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a)C2O42-
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b)SO42-
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c)S2-
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d)Fe2+
Explanation
In the given reaction X is oxidised and hence acting as reducing agent. Out of given options SO4- can't act as reducing agent, because sulphur is present in its highest oxidation state +6Answer: (b)
Q.18
Which one of the following leads to redox reaction
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a) AgNO3 + HCl
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b) KOH + HCl
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c)KI + Cl2
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d)NH3 + HCl
Explanation
Answer: (c)
Q.19
On reduction of KMnO4 by oxalic acid in acidic medium, the oxidation number of Mn changes. What is the magnitude of this charge? [ AIEEE 2007]
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a) From 7 to 2
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b) From 6 to 2
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c)From 5 to 2
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d)From 7 to 4
Explanation
Answer:(a)
Q.20
Number of moles of K2Cr2O7 reduced by one mole of Sn2+ ions is [ Hariyana CEET 2000]
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a) 1/3
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b) 3
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c) 1/6
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d) 6
Explanation
Cr2O72- + 3Sn2+ +14H+ → 2Cr3+ + 3Sn4+ + 7H2O Thus 1 mole of Sn2+ reduces (1/3) mole of K2Cr2O7 Answer: (a)
Q.21
In which of the following changes there is a transfer of five electrons?
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a)MnO4- → Mn2+
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b) CrO4- → Cr3+
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c)MnO4- → MnO2
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d)Cr2O72- → 2Cr3+
Explanation
Answer: (a)
Q.22
he colour of K2Cr2O7 changes from red orange to lemon yell on treatment with aqueous KOH because of
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a) Reduction of Cr(VI) to Cr(III)
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b) Formation of chromium ion to chromate
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c)Conversion of dichromate ion to chromate
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d)Oxidation of potassium hydroxide to potassium peroxide
Explanation
In basic medium K2Cr2,/O7 + 2 KOH → 2K2CrO4 + H2O Answer: (c)
Q.23
Cr2O72- → Cr3+ eq.wt of Cr2O72- is
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a)mol.wt/6
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b) mol.wt/3
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c)mol.wt./4
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d)mol.wt./8
Explanation
∴ Equivalent wt. of Cr2O72-= Molecular weight / no. number of electrons gain=M/6 Answer:(a)
Q.24
What is the equivalent mass of IO4- when it is converted into I2 in acid medium
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a) M/6
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b) M/7
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c) M/5
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d) M/4
Explanation
Equivalent weight of IO4-= Molecular mass/ gain in number of electrons=M/7 Answer: (b)
Q.25
xMnO4-+ yHSO3- → 2Mn2+ + 2H2O + OH-+ zSO42- In this reaction value of x, y and z are
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a)2, 5, 5
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b) 5, 2, 9
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c)3, 5, 5
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d)2, 6, 6
Explanation
Out of given options x=2, y=5 and z=5 makes the equation perfectly balancedAnswer: (a)
Q.26
The standard reduction potential of Cu2+ / Cu and Cu2+ / Cu+ are 0.337 and 0.153 respectively. the standard electrode potential of Cu+ / Cu half cell is ... [ IIT 1997]
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a) 0.184 V
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b) 0.827 V
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c)0.521 V
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d)0.490 v
Explanation
Cu2+ + 2e- → Cu ; n1=2 and ΔG1Cu2+ + e- → Cu+ ; n2=1 and ΔG2Subtracting (2) from (1) we getCu+ + e- →Cu ; n3=1 and ΔG3 Now G3=G1 - G2 As ΔG=-nEF -n3FE3=-n1FE1 - ( -n2FE2) E3=2E1 - E2 E3=2 × 0.337 - 0.152=0521 V Thus E3 i.e E Cu+ / Cu=0.521 V Answer:(c)
Q.27
A current of 2amp when passed for 5 hours through a moltan salt deposits 22.2 g of metal of atomic massThe oxidation state of the metal in the metal salt is
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a) +1
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b) +2
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c) +3
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d) +4
Explanation
Q=I × t Q=2 × 5 × 3600=36000 coulomb Answer: (c)
Q.28
The compound that can work both as an oxidising as well as a reducing agent is [ CPMT 1986]
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a) KMnO4
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b) H2O2
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c) Fe2(SO4)3
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d) K2Cr2O7
Explanation
The oxidation number of O in H2O2 is -1. It can either increase to zero or decrease to -2 in H2O. therefore H2O2 can act as an oxidising as well as a reducing agent Answer: (b)
Q.29
For the redox reaction. the correct coefficient of the reactants for the balanced reaction are
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a) MnO4-=2; C2O4-=5; H+=16
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b) MnO4-=16; C2O4-=3; H+=12
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c)MnO4-=15; C2O4-=16; H+=12
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d)MnO4-=2; C2O4-=16; H+=5
Explanation
Let us balance the given equation by oxidation number method Step(i) Skeletal equationStep(ii)Balance all elements except H and O and find total oxidation number,Identify change in oxidation numberStep (iii)Equalising total increase and decrease in oxidation numberTotal increase in oxidation number=2 Total decrease in oxidation number=5 Step(iv) Balance charge on both side by adding H+ By adding 15H+ on left side of the equation gets balanced for charge Step(v) Balance H and by adding H2O from balanced equation option (a) is correctAnswer: (a)
Q.30
In acidic, dichromatic ion oxides ferrous ion to ferric ion. If the gram molecular weight of potassium dichromate is 294 grams, its gram equivalent weight is ... grams
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a) 294
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b) 127
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c)49
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d)24.5
Explanation
In acidic medium K2Cr2O7 acts as strong oxidising agent and self gets reduced to Cr3+ Equivalent weight=Molecular weight/6=294/6=49 Answer:(c)
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