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All questions of Redox Reactions for EmSAT Achieve Exam
- a)I, II and III
- b)I and II
- c)II and III
- d)Only I
Correct answer is 'D'. Can you explain this answer?
|
Nabanita Deshpande answered |
Which is chlorate (I) ion?
a) 
b)
c)
d)
Correct answer is option 'B'. Can you explain this answer?
|
Pooja Shah answered |
- ClO3- : A very reactive inorganic anion.
- The term chlorate can also be used to describe any compound containing the chlorate ion, normally chlorate salts.
- Example: Potassium chlorate, KClO3
The difference in the oxidation number of the two types of sulphur atoms in Na2S4O6 is .........[HTJEE2011]
Correct answer is '7'. Can you explain this answer?
The difference in the oxidation number of the two types of sulphur atoms in Na2S4O6 is .........
[HTJEE2011]
|
Hrishikesh Sengupta answered |
Difference in oxidation number = 6 - (-1) = 7
Which of the following is not an example of redox reaction?- a)BaCl2 + H2SO4 ⎯→ BaSO4 + 2HCl
- b)Fe2O3 + 3CO ⎯→ 2Fe + 3CO2
- c)2K + F2 ⎯→ 2KF
- d)CuO + H2 ⎯→ Cu + H2O
Correct answer is option 'A'. Can you explain this answer?
Which of the following is not an example of redox reaction?
a)
BaCl2 + H2SO4 ⎯→ BaSO4 + 2HCl
b)
Fe2O3 + 3CO ⎯→ 2Fe + 3CO2
c)
2K + F2 ⎯→ 2KF
d)
CuO + H2 ⎯→ Cu + H2O
|
Raghav Bansal answered |
a) BaCl2 + H2SO4 → BaSO4 + 2HCl is not a redox reaction, as there is no change in the oxidation state of any element.
It is an example of double displacement reactions.
It is an example of double displacement reactions.
The oxidation half reaction for following reaction is
Fe2+(aq) + Cr2O72-(aq) → Fe3+ (aq) + Cr3+(aq)- a)Fe3+(aq) → Fe2+ (aq)
- b)Cr2O72-(aq) → Cr3+(aq)
- c)Cr3+(aq) → Cr2O72-(aq)
- d)Fe2+ (aq) → Fe3+(aq)
Correct answer is option 'D'. Can you explain this answer?
The oxidation half reaction for following reaction is
Fe2+(aq) + Cr2O72-(aq) → Fe3+ (aq) + Cr3+(aq)
Fe2+(aq) + Cr2O72-(aq) → Fe3+ (aq) + Cr3+(aq)
a)
Fe3+(aq) → Fe2+ (aq)
b)
Cr2O72-(aq) → Cr3+(aq)
c)
Cr3+(aq) → Cr2O72-(aq)
d)
Fe2+ (aq) → Fe3+(aq)
|
Geetika Shah answered |
Oxidation half reaction for a reaction is that reaction which gives us the reactant and product formed after the oxidation of the reactant. In this case Fe+2 oxidizes itself to Fe+3 and so the oxidation of Fe+2 is oxidation half reaction. Option d correct.
In the following reaction, 
, oxidation number of- a)N changes from -2 to +2
- b)N changes from -2 to 0
- c)N in
changes from -3 to 0 and that in 
changes from +3 to 0 - d)N does not change
Correct answer is option 'C'. Can you explain this answer?
In the following reaction, , oxidation number of
, oxidation number ofa)
N changes from -2 to +2
b)
N changes from -2 to 0
c)
N in changes from -3 to 0 and that in changes from +3 to 0
changes from -3 to 0 and that in changes from +3 to 0d)
N does not change
|
Learners Habitat answered |
In NH4NO2, the oxidation number of N in NH4+ is -3, and of N in NO2- is +3.
Oxidation numbers of P in PO4−3, of S in SO42− and that of Cr in Cr2O72− are respectively,- a) +5, +6 and +6
- b)+3, +6 and +5
- c)+5, +3 and +6
- d)-3, +6 and +6
Correct answer is option 'A'. Can you explain this answer?
Oxidation numbers of P in PO4−3, of S in SO42− and that of Cr in Cr2O72− are respectively,
a)
+5, +6 and +6
b)
+3, +6 and +5
c)
+5, +3 and +6
d)
-3, +6 and +6
|
|
Raghav Bansal answered |
The correct answer is option A
(I) xPO43− ⇒ x + 4 × (−2) = −3
⇒x = −3 + 8 = +5
⇒x = +5
Oxidation number of P = +5
(II) xSO42− ⇒ x + 4 × (−2) = −2
⇒x = −2 + 8
⇒x = +6
Oxidation number of S=+6
(III) xCr2O72− ⇒2x + 7 × (−2) = −2
⇒2x =−2+14
⇒2x=12
⇒x= 12/2 = +6
(I) xPO43− ⇒ x + 4 × (−2) = −3
⇒x = −3 + 8 = +5
⇒x = +5
Oxidation number of P = +5
(II) xSO42− ⇒ x + 4 × (−2) = −2
⇒x = −2 + 8
⇒x = +6
Oxidation number of S=+6
(III) xCr2O72− ⇒2x + 7 × (−2) = −2
⇒2x =−2+14
⇒2x=12
⇒x= 12/2 = +6
Direction (Q. Nos. 17) Choice the correct combination of elements and column I and coloumn II are given as option (a), (b), (c) and (d), out of which ONE option is correct.
Match the compounds/ions having underlined atoms of different oxidation number (in Column I) with values (in Column II).
- a)a
- b)b
- c)c
- d)d
Correct answer is option 'B'. Can you explain this answer?
Direction (Q. Nos. 17) Choice the correct combination of elements and column I and coloumn II are given as option (a), (b), (c) and (d), out of which ONE option is correct.
Match the compounds/ions having underlined atoms of different oxidation number (in Column I) with values (in Column II).
a)
a
b)
b
c)
c
d)
d
|
|
Suresh Iyer answered |
(i) CaOCI2 has
Oxidation number = + 6
Oxidation number = - 2
(ii) → (p,u)
(iii) → (q.v)
Oxidation number = + 6
Oxidation number = - 2
(ii) → (p,u)
(iii) → (q.v)
In Daniel cell, oxidation takes place at- a)Any of the two electrodes
- b)Depends on the salts and their solutions
- c)Anode
- d)Cathode
Correct answer is option 'C'. Can you explain this answer?
In Daniel cell, oxidation takes place at
a)
Any of the two electrodes
b)
Depends on the salts and their solutions
c)
Anode
d)
Cathode
|
|
Rohit Shah answered |
The solutions in which the electrodes are immersed are called electrolytes. The chemical reaction that takes place in a galvanic cell is the redox reaction. One electrode acts as anode in which oxidation takes place and the other acts as the cathode in which reduction takes place.
Hydrogen is prepared from H2O by adding- a)Ag, which acts as reducing agent
- b)Ca, which acts as reducing agent
- c)Au, which acts as oxidising agent
- d)AI, which acts as oxidising agent
Correct answer is option 'B'. Can you explain this answer?
Hydrogen is prepared from H2O by adding
a)
Ag, which acts as reducing agent
b)
Ca, which acts as reducing agent
c)
Au, which acts as oxidising agent
d)
AI, which acts as oxidising agent
|
|
Saumya Dey answered |
Preparation of Hydrogen from Water
- Hydrogen gas can be prepared from water by using a reducing agent, which reduces water to hydrogen gas and also gets oxidized in the process.
- Calcium (Ca) is a good reducing agent and can be used to prepare hydrogen gas from water.
- When calcium is added to water, it reacts with water to form calcium hydroxide and hydrogen gas.
- The chemical equation for the reaction is:
Ca + 2H2O → Ca(OH)2 + H2
- Calcium has a strong affinity for oxygen and readily reacts with water to form calcium hydroxide, liberating hydrogen gas.
- This reaction is exothermic and produces a considerable amount of heat, which can be used to heat water or other substances.
- The liberated hydrogen gas can be collected by upward displacement of air or by using a gas syringe or gas jar.
- The purity of hydrogen gas prepared by this method is relatively low, as it may contain impurities like calcium hydroxide, unreacted water, and other gases like nitrogen and oxygen.
- Therefore, additional purification steps may be required to obtain pure hydrogen gas.
Advantages of using Calcium as a reducing agent
- Calcium is a readily available and inexpensive reducing agent.
- It reacts readily with water, producing a large amount of hydrogen gas.
- The reaction is exothermic and produces heat, which can be utilized in other processes.
- Calcium is a relatively safe reducing agent, as it does not react violently with water or other substances.
- The by-products of the reaction, calcium hydroxide, and hydrogen gas, are non-toxic and can be disposed of safely.
- Hydrogen gas can be prepared from water by using a reducing agent, which reduces water to hydrogen gas and also gets oxidized in the process.
- Calcium (Ca) is a good reducing agent and can be used to prepare hydrogen gas from water.
- When calcium is added to water, it reacts with water to form calcium hydroxide and hydrogen gas.
- The chemical equation for the reaction is:
Ca + 2H2O → Ca(OH)2 + H2
- Calcium has a strong affinity for oxygen and readily reacts with water to form calcium hydroxide, liberating hydrogen gas.
- This reaction is exothermic and produces a considerable amount of heat, which can be used to heat water or other substances.
- The liberated hydrogen gas can be collected by upward displacement of air or by using a gas syringe or gas jar.
- The purity of hydrogen gas prepared by this method is relatively low, as it may contain impurities like calcium hydroxide, unreacted water, and other gases like nitrogen and oxygen.
- Therefore, additional purification steps may be required to obtain pure hydrogen gas.
Advantages of using Calcium as a reducing agent
- Calcium is a readily available and inexpensive reducing agent.
- It reacts readily with water, producing a large amount of hydrogen gas.
- The reaction is exothermic and produces heat, which can be utilized in other processes.
- Calcium is a relatively safe reducing agent, as it does not react violently with water or other substances.
- The by-products of the reaction, calcium hydroxide, and hydrogen gas, are non-toxic and can be disposed of safely.
in basic medium. 1 mole of KMnO4 is reduced by
The reduction potential values of M, N and O are +2.46, -1.13 and -3.13 V respectively. Which of the following order is correct regarding their reducing property?- a)M>N>O
- b)O>M>N
- c)M>O>N
- d)O>N>M
Correct answer is option 'D'. Can you explain this answer?
The reduction potential values of M, N and O are +2.46, -1.13 and -3.13 V respectively. Which of the following order is correct regarding their reducing property?
a)
M>N>O
b)
O>M>N
c)
M>O>N
d)
O>N>M
|
|
Geetika Shah answered |
Reduction potential means the tendency to reduce itself, i.e SO if we need reducing agent, then we need elements having more oxidising potential or reversing reduction potential order.
The order of reduction potential - O < N < M
So the order of reducing agent = M < N < O
The order of reduction potential - O < N < M
So the order of reducing agent = M < N < O
The complex [Fe(H2O)5NO]2+ is formed in the ring-test for nitrate ion 
when freshly prepared FeSO4 solution is added to aqueous solution of 
followed by the addition of conc. H2SO4. NO exists as NO+ (nitrosyl).Q. Magnetic moment 
of Fe in the ring is - a)zero BM
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
The complex [Fe(H2O)5NO]2+ is formed in the ring-test for nitrate ion when freshly prepared FeSO4 solution is added to aqueous solution of followed by the addition of conc. H2SO4. NO exists as NO+ (nitrosyl).
when freshly prepared FeSO4 solution is added to aqueous solution of followed by the addition of conc. H2SO4. NO exists as NO+ (nitrosyl).Q. Magnetic moment of Fe in the ring is
of Fe in the ring is a)
zero BM
b)
c)
d)
|
Knowledge Hub answered |
Fe2+ is added asFeSO4
Fe+ is formed by charge transfer from NO to Fe2+
Fe+ has three unpaired electrons (N).
Fe+ is formed by charge transfer from NO to Fe2+
Fe+ has three unpaired electrons (N).
I− reduces IO3- and I2 and itself oxidised to I2 in acidic medium. Thus, final reaction is- a)
- b)
- c)
- d)None of the above is correct
Correct answer is option 'A'. Can you explain this answer?
I− reduces IO3- and I2 and itself oxidised to I2 in acidic medium. Thus, final reaction is
a)
b)
c)
d)
None of the above is correct
|
|
Pooja Shah answered |
To balance oxidation number, cross-multiply by change in oxidation number
For the redox reactionMnO4– + C2O42- + H+ → Mn2+ + CO2 + H2O
The correct coefficients of the reactants for the balanced reaction are:- a)MnO4– C2O42- H+ 16 5 2
- b)MnO4– C2O42- H+ 2 16 5
- c)MnO4– C2O42- H+ 2 5 16
- d)MnO4– C2O42- H+ 5 16 2
Correct answer is option 'C'. Can you explain this answer?
For the redox reaction
MnO4– + C2O42- + H+ → Mn2+ + CO2 + H2O
The correct coefficients of the reactants for the balanced reaction are:
The correct coefficients of the reactants for the balanced reaction are:
a)
MnO4– C2O42- H+ 16 5 2
b)
MnO4– C2O42- H+ 2 16 5
c)
MnO4– C2O42- H+ 2 5 16
d)
MnO4– C2O42- H+ 5 16 2
|
|
Raghav Bansal answered |
NaHC2O4 is neutralised by NaOH and can also be oxidised by KMnO4 (in acidic medium). Equivalent weight is related to m olecular w eight (M) of NaHC2O4 in these two reactions as- a)M,M
- b)M,M/2
- c)M/2, M/5
- d)M/5, M/2
Correct answer is option 'B'. Can you explain this answer?
NaHC2O4 is neutralised by NaOH and can also be oxidised by KMnO4 (in acidic medium). Equivalent weight is related to m olecular w eight (M) of NaHC2O4 in these two reactions as
a)
M,M
b)
M,M/2
c)
M/2, M/5
d)
M/5, M/2
|
|
Debanshi Datta answered |
Thus, equivalent weight when reacted with NaOH = M
Direction (Q. Nos. 13 and 14) This section contains 2 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONE or MORE THANT ONE is correct.Q. Which of the following species has/have oxidation number of the metal as + 6 ?- a)
- b)
- c)
- d)
Correct answer is option 'A,B,C'. Can you explain this answer?
Direction (Q. Nos. 13 and 14) This section contains 2 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONE or MORE THANT ONE is correct.
Q. Which of the following species has/have oxidation number of the metal as + 6 ?
a)
b)
c)
d)
|
|
Knowledge Hub answered |
To balance the oxygen atom in the given reaction in acidic medium
Cr2O72- (aq) → Cr3+(aq) we- a)Add water (H2O) on right side
- b)Add water (H2O) on left side
- c)Add O on right side
- d)Add O on left side
Correct answer is option 'A'. Can you explain this answer?
To balance the oxygen atom in the given reaction in acidic medium
Cr2O72- (aq) → Cr3+(aq) we
Cr2O72- (aq) → Cr3+(aq) we
a)
Add water (H2O) on right side
b)
Add water (H2O) on left side
c)
Add O on right side
d)
Add O on left side
|
Supriya Senapati answered |
Yes because right side is deficient of oxygen.
Assign oxidation number to P in NaH2PO4- a)+6
- b)+1
- c)+4
- d)5.0
Correct answer is option 'D'. Can you explain this answer?
Assign oxidation number to P in NaH2PO4
a)
+6
b)
+1
c)
+4
d)
5.0
|
Raksha Nambiar answered |
Oxidation state of P in NaH2PO4
Select the set of compounds with oxidation-reduction duality.- a)Cl2, H3PO4, HCHO, HNO2
- b)Cl2, H3PO3, C6H5CHO, H2O2
- c)Br2, H3PO2, CH3CHO, H3PO4
- d)CrO2Cl2, KMnO4, SO3, CO2
Correct answer is option 'B'. Can you explain this answer?
Select the set of compounds with oxidation-reduction duality.
a)
Cl2, H3PO4, HCHO, HNO2
b)
Cl2, H3PO3, C6H5CHO, H2O2
c)
Br2, H3PO2, CH3CHO, H3PO4
d)
CrO2Cl2, KMnO4, SO3, CO2
|
Neha Sharma answered |
Com pounds having oxidising and reducing nature in given reaction are said to have oxidation-reduction duality. Such compounds are said to undergo disproportionation reaction.
(This is called Cannizzaro reaction.)
Note Such compounds have O.N. of the affected atoms intermediate of oxidation part and reduction part
The oxidation number of oxygen in most compounds is- a)-3
- b)1
- c)4
- d)-2.0
Correct answer is option 'D'. Can you explain this answer?
The oxidation number of oxygen in most compounds is
a)
-3
b)
1
c)
4
d)
-2.0
|
Niharika Nair answered |
Oxidation number of oxygen in most compounds is -2.
Explanation:
- Oxidation number is the number assigned to an atom to indicate its degree of oxidation or loss/gain of electrons.
- Oxygen is a highly electronegative element, meaning it has a strong tendency to attract electrons.
- In most compounds, oxygen has an oxidation number of -2 because it tends to gain electrons to achieve a stable octet configuration (8 valence electrons).
- For example, in water (H2O), each hydrogen atom has an oxidation number of +1 and the oxygen atom has an oxidation number of -2, which balances out the charge to zero.
- There are some exceptions to this rule, such as in peroxides where oxygen has an oxidation number of -1, and in compounds with more electronegative elements where oxygen may have a positive oxidation number.
- Overall, the oxidation number of oxygen in most compounds is -2.
Explanation:
- Oxidation number is the number assigned to an atom to indicate its degree of oxidation or loss/gain of electrons.
- Oxygen is a highly electronegative element, meaning it has a strong tendency to attract electrons.
- In most compounds, oxygen has an oxidation number of -2 because it tends to gain electrons to achieve a stable octet configuration (8 valence electrons).
- For example, in water (H2O), each hydrogen atom has an oxidation number of +1 and the oxygen atom has an oxidation number of -2, which balances out the charge to zero.
- There are some exceptions to this rule, such as in peroxides where oxygen has an oxidation number of -1, and in compounds with more electronegative elements where oxygen may have a positive oxidation number.
- Overall, the oxidation number of oxygen in most compounds is -2.
Equivalent weight of H3PO2 in a reaction is found to be half of its molecular weight. It can be due to its- a)reaction of two H+ ion
- b)oxidation of H3PO3
- c)oxidation of H3PO4
- d)reduction of PH3
Correct answer is option 'B'. Can you explain this answer?
Equivalent weight of H3PO2 in a reaction is found to be half of its molecular weight. It can be due to its
a)
reaction of two H+ ion
b)
oxidation of H3PO3
c)
oxidation of H3PO4
d)
reduction of PH3
|
Anisha Chauhan answered |
An equivalent weight of H3PO2 when it disproportionates into PH3 and H3PO3 is
For disproportionation reactions, use a trick to find equivalent mass; E = E1 + E2; where E1 and E2 are equivalent masses of oxidation and reduction half-reactions of the same element.Now, find the n-factor(total change in oxidation number per molecule) and then equivalent mass = molar mass/n-factor.
So, when H3PO2 changes into PH3, the oxidation state of phosphorous changes from +1 to -3 so as n-factor is 4. Also when H3PO2 changes into H3PO3, the oxidation state of phosphorous changes from +1 to +5 so as n-factor is again 4.
Now equivalent mass of H3PO2, E =(M/4) + (M/4) = M/2
Hence, the correct option is b.
In this method, the two half equations are balanced separately and then added together to give balanced equation- a)Reluctant method
- b)Oxidizing agent method
- c)Reducing agent method
- d)Half reaction method
Correct answer is option 'D'. Can you explain this answer?
In this method, the two half equations are balanced separately and then added together to give balanced equation
a)
Reluctant method
b)
Oxidizing agent method
c)
Reducing agent method
d)
Half reaction method
|
Kavita Joshi answered |
In the ion-electron method (also called the half-reaction method), the redox equation is separated into two half-equations - one for oxidation and one for reduction. Each of these half-reactions is balanced separately and then combined to give the balanced redox equation.
In which of the following reactions oxidation number of chromium has been affected?- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
In which of the following reactions oxidation number of chromium has been affected?
a)
b)
c)
d)
|
Sankar Bose answered |
Oxidation number of Cr changes from +6 to +3.
The standard electrode potentials, K+/K= -2.93V, Ag+/Ag = 0.80V, the electrode which is negatively charged is- a)Ag+/Ag
- b)K+/K
- c)Any of the two
- d)None of them
Correct answer is option 'B'. Can you explain this answer?
The standard electrode potentials, K+/K= -2.93V, Ag+/Ag = 0.80V, the electrode which is negatively charged is
a)
Ag+/Ag
b)
K+/K
c)
Any of the two
d)
None of them
|
Supriya Senapati answered |
Since anode is negatively charged and in anode oxidation takes place, so the element having lower SRP value would act as anode. so answer is K+/K.
Intensity of blue colour increases gradually when _________________- a)copper rod is dipped in silver nitrate solution
- b)silver rod is dipped in copper nitrate solution
- c)zinc rod is dipped in silver solution
- d)copper rod is dipped in zinc rod solution
Correct answer is option 'A'. Can you explain this answer?
Intensity of blue colour increases gradually when _________________
a)
copper rod is dipped in silver nitrate solution
b)
silver rod is dipped in copper nitrate solution
c)
zinc rod is dipped in silver solution
d)
copper rod is dipped in zinc rod solution
|
EduRev NEET answered |
When a copper rod is dipped in silver nitrate solution, a redox reaction occurs between Copper and an aqueous solution of silver nitrate.
- So the intensity of blue colour increases gradually as silver deposits on the rod.
Direction (Q. Nos. 16-18) This section contains 4 multiple choice questions. Each question has four
choices (a), (b), (c) and (d), out of which ONE or MORE THANT ONE is correct.Q. Sulphate (IV) ion reduces chromate (VI) to green coloured salt. Thus,- a) Cr (III) is formed
- b)Equivalent mass of chromate (VI) = M/3
- c) Equivalent mass of sulphate (IV) =M/2
- d) Oxidation product is sulphate (VI)
Correct answer is option 'A,B,C,D'. Can you explain this answer?
Direction (Q. Nos. 16-18) This section contains 4 multiple choice questions. Each question has four
choices (a), (b), (c) and (d), out of which ONE or MORE THANT ONE is correct.
choices (a), (b), (c) and (d), out of which ONE or MORE THANT ONE is correct.
Q. Sulphate (IV) ion reduces chromate (VI) to green coloured salt. Thus,
a)
Cr (III) is formed
b)
Equivalent mass of chromate (VI) = M/3
c)
Equivalent mass of sulphate (IV) =M/2
d)
Oxidation product is sulphate (VI)
|
|
Amrita Choudhary answered |
Thus, (a) is correct.
A cell is prepared by dipping a chromium rod in 1M Cr2(SO4)3 solution and an iron rod in 1M FeSO4 solution. The standard reduction potentials of Chromium and Iron electrodes are -0.75 V and -0.45 V respectively. What will be the standard EMF of the cell?- a)+0.30 V
- b)+0.47 V
- c)– 0.30 V
- d)-0.47 V
Correct answer is option 'A'. Can you explain this answer?
A cell is prepared by dipping a chromium rod in 1M Cr2(SO4)3 solution and an iron rod in 1M FeSO4 solution. The standard reduction potentials of Chromium and Iron electrodes are -0.75 V and -0.45 V respectively. What will be the standard EMF of the cell?
a)
+0.30 V
b)
+0.47 V
c)
– 0.30 V
d)
-0.47 V
|
Deadpool answered |
Reduction potential of fe is more than that of cr, so fe will reduce and act as cathode and cr will act as anode. now emf of cell is reduction potential of cathode - reduction potential of anode. so EMF= -0.45-(-0.75) = 0.3 V
Reduction is defined in terms ofI. electronation and hydrogenation
II. deelectronation and gain of oxygen
III. increase in oxidation number
IV. decrease in oxidation numberSelect the correct terms- a)I and IV
- b)I and III
- c)II and II
- d)I and II
Correct answer is option 'A'. Can you explain this answer?
Reduction is defined in terms of
I. electronation and hydrogenation
II. deelectronation and gain of oxygen
III. increase in oxidation number
IV. decrease in oxidation number
II. deelectronation and gain of oxygen
III. increase in oxidation number
IV. decrease in oxidation number
Select the correct terms
a)
I and IV
b)
I and III
c)
II and II
d)
I and II
|
Sharmila Sharma answered |
Gain of hydrogen , electron and decrease in oxidation number is reduction
The more positive the value of E0, the greater is the tendency of the species to get reduced. Using the standard electrode potential of redox couples given below find out which of the following is the strongest oxidising agent. E0values : Fe3 + / Fe2+ = +0.77; I2(s)/l- = +0.54; cu2+/ Cu = +0.34; Ag+ / Ag = +0.80V- a)Ag+
- b)Fe3+
- c)I2 (s)
- d)Cu2+
Correct answer is option 'A'. Can you explain this answer?
The more positive the value of E0, the greater is the tendency of the species to get reduced. Using the standard electrode potential of redox couples given below find out which of the following is the strongest oxidising agent.
E0values : Fe3 + / Fe2+ = +0.77; I2(s)/l- = +0.54; cu2+/ Cu = +0.34; Ag+ / Ag = +0.80V
a)
Ag+
b)
Fe3+
c)
I2 (s)
d)
Cu2+
|
Mansi Mukherjee answered |
Oxidation number of H is not always +1 . It can be -1 , 0.
Consider the elements: Cs, Ne, I and F. Identify the element(s) that exhibits only negative oxidation state- a)F
- b)I
- c)s
- d)Cs and F
Correct answer is option 'A'. Can you explain this answer?
Consider the elements: Cs, Ne, I and F. Identify the element(s) that exhibits only negative oxidation state
a)
F
b)
I
c)
s
d)
Cs and F
|
Sinjini Datta answered |
F has negative oxidation state as it is very electro negative.
The correct order of N-compounds in its decreasing order of oxidation states is- a)HNO3, NO, N2, NH4Cl
- b)HNO3, NO, NH4Cl, N2
- c)HNO3, NH4Cl, NO, N2
- d)NH4Cl, N2, NO, HN03
Correct answer is option 'A'. Can you explain this answer?
The correct order of N-compounds in its decreasing order of oxidation states is
a)
HNO3, NO, N2, NH4Cl
b)
HNO3, NO, NH4Cl, N2
c)
HNO3, NH4Cl, NO, N2
d)
NH4Cl, N2, NO, HN03
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Infinity Academy answered |
To determine the decreasing order of oxidation states of nitrogen in the given compounds, we need to find the oxidation state of nitrogen in each compound:
- HNO3 (Nitric acid): Oxidation state of nitrogen: +5
- NO (Nitric oxide): Oxidation state of nitrogen: +2
- N2 (Dinitrogen): Oxidation state of nitrogen: 0
- NH4Cl (Ammonium chloride): Oxidation state of nitrogen: -3
Now, let's arrange these compounds in decreasing order of oxidation states:
- HNO3: +5
- NO: +2
- N2: 0
- NH4Cl: -3
So, the correct order in decreasing oxidation state is:
HNO3, NO, N2, NH4Cl
The process in which the strength of an unknown solution is calculated using a known standard solution.- a)Titration
- b)Oxidation
- c)Reduction
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
The process in which the strength of an unknown solution is calculated using a known standard solution.
a)
Titration
b)
Oxidation
c)
Reduction
d)
None of these
|
|
Nandini Iyer answered |
A titration is a technique used to work out the concentration of an unknown solution based on its chemical reaction with a solution of known concentration. The process usually involves adding the known solution (the titrant) to a known quantity of the unknown solution (the analyte) until the reaction is complete.
Which of the following is true as per metal activity series?- a)Zn>Cu>Ag
- b)Zn<Cu<Ag
- c)Zn>Ag>Cu
- d)Zn<Ag<Cu
Correct answer is option 'A'. Can you explain this answer?
Which of the following is true as per metal activity series?
a)
Zn>Cu>Ag
b)
Zn<Cu<Ag
c)
Zn>Ag>Cu
d)
Zn<Ag<Cu
|
|
EduRev NEET answered |
Metal activity series or electrochemical series is a series in the decreasing order of metals which are active during a chemical reaction comparatively with each other.
- Here, Zinc’s activity is greater than Copper’s activity and Copper’s activity is greater than that of silver.
When a zinc rod is kept in a copper nitrate solution what happens?- a)zinc is deposited on copper
- b)copper is deposited in the beaker
- c)zinc is deposited in the beaker
- d)copper is deposited on zinc
Correct answer is option 'D'. Can you explain this answer?
When a zinc rod is kept in a copper nitrate solution what happens?
a)
zinc is deposited on copper
b)
copper is deposited in the beaker
c)
zinc is deposited in the beaker
d)
copper is deposited on zinc
|
|
EduRev NEET answered |
When zinc is placed in copper nitrate solution the intensity of the blue colour is produced and copper iron is deposited on zinc.
- This is a Redox reaction between zinc and an aqueous solution of copper nitrate occurring in a beaker.
A standard hydrogen electrode has zero electrode potential because- a)Hydrogen is the lightest element
- b)Hydrogen atom has only one electron.
- c)Hydrogen is easiest to oxidize
- d)The electrode potential is assumed to be zero
Correct answer is option 'D'. Can you explain this answer?
A standard hydrogen electrode has zero electrode potential because
a)
Hydrogen is the lightest element
b)
Hydrogen atom has only one electron.
c)
Hydrogen is easiest to oxidize
d)
The electrode potential is assumed to be zero
|
|
Rohan Singh answered |
This electrode potential is assumed to be zero
The standard potential of hydrogen electrode is assumed to be zero at all temperature.
The Standard hydrogen electrode (abbreviated SHE), is a redox electrode which forms the basis of the thermodynamic scale of oxidation-reduction potentials. Its absolute electrode potential is estimated to be 4.44 � 0.02 V at 25 �C, but to form a basis for comparison with all other electrode reactions, hydrogen's standard electrode potential (E0) is declared to be zero volts at all temperatures.
In the following reaction,
One equivalent of H2S (g)will reduce- a)1 Mole SO2
- b)0.5 Mole SO2
- c)0.25 Mole SO2
- d)2 Mole SO2
Correct answer is option 'C'. Can you explain this answer?
In the following reaction,
One equivalent of H2S (g)will reduce
One equivalent of H2S (g)will reduce
a)
1 Mole SO2
b)
0.5 Mole SO2
c)
0.25 Mole SO2
d)
2 Mole SO2
|
|
Rithika Khanna answered |
Change in oxidation number of S of H2S = 2 units
Change in oxidation number of S of SO2 = 4 units
Then one mole H2S = 2 equivalents H2S
One mole SO2 = 4 equivalent SO2
One equivalent H23 one equivalent SO2 = 0.25 mole SO2
A metal in a compound can be displaced by another metal in the uncombined state. Which metal is a better reducing agent in such a case?
- a)Better reducing agent is the one that looses more electrons
- b)Better reducing agent is the one that looses less electrons
- c)Both are same in reducing capacity
- d)The reduced metal is a better reducing agent than the reducing metal
Correct answer is option 'A'. Can you explain this answer?
A metal in a compound can be displaced by another metal in the uncombined state. Which metal is a better reducing agent in such a case?
a)
Better reducing agent is the one that looses more electrons
b)
Better reducing agent is the one that looses less electrons
c)
Both are same in reducing capacity
d)
The reduced metal is a better reducing agent than the reducing metal
|
Bhavana Chavan answered |
Concept of Reducing Agent:
A reducing agent is a substance that loses or "donates" an electron to another substance in a redox chemical reaction. Therefore, a good reducing agent is the one that gets oxidized easily, or in other words, the one that can easily lose electrons.
Characteristics of a Good Reducing Agent:
A reducing agent is a substance that loses or "donates" an electron to another substance in a redox chemical reaction. Therefore, a good reducing agent is the one that gets oxidized easily, or in other words, the one that can easily lose electrons.
Characteristics of a Good Reducing Agent:
- Electron Loss: A better reducing agent is the one that loses more electrons. This is because by losing electrons, the reducing agent gets oxidized and in turn reduces the other substance. This is the basic principle of a redox reaction.
- Reactivity: The reactivity of the metal also determines its capacity as a reducing agent. Metals that are high in the reactivity series are good reducing agents. This is because they can easily lose electrons and get oxidized.
- Stability: Metals that are less stable are better reducing agents because they can easily lose electrons to attain a stable state.
Hence, Option A is the correct answer - a better reducing agent is the one that loses more electrons.
Consider the following experimental facts,I. When Cl2 gas is passed into Kl solution containing CHCI3, violet colour appears in CHCI3 layer.
II. When Cl2 gas is passed into KBr solution containing CHCI3, orange colour appears in CHCI3 layer.
III. When Cl2 gas is passed into a solution containing KBr, Kl and KCI, containing CHCI3, violet colour appears in CHCI3 layer.Select the correct experimental facts.- a)I and II
- b)I and III
- c)III Only
- d)II and III
Correct answer is option 'D'. Can you explain this answer?
Consider the following experimental facts,
I. When Cl2 gas is passed into Kl solution containing CHCI3, violet colour appears in CHCI3 layer.
II. When Cl2 gas is passed into KBr solution containing CHCI3, orange colour appears in CHCI3 layer.
III. When Cl2 gas is passed into a solution containing KBr, Kl and KCI, containing CHCI3, violet colour appears in CHCI3 layer.
II. When Cl2 gas is passed into KBr solution containing CHCI3, orange colour appears in CHCI3 layer.
III. When Cl2 gas is passed into a solution containing KBr, Kl and KCI, containing CHCI3, violet colour appears in CHCI3 layer.
Select the correct experimental facts.
a)
I and II
b)
I and III
c)
III Only
d)
II and III
|
Poulomi Singh answered |
Oxidising power of F2 > CI2 > Br2 > l2
Thus, (II) is true.
In experiment (III), orange colour of Br2 does not appear, as it also oxidises Kl to l2. Hence , only violet colour appears in CHCI3 layer. Thus, (III) is also true.
Thus, (II) is true.
In experiment (III), orange colour of Br2 does not appear, as it also oxidises Kl to l2. Hence , only violet colour appears in CHCI3 layer. Thus, (III) is also true.
Coefficient x, y and z are respectively
- a)a
- b)b
- c)c
- d)d
Correct answer is option 'C'. Can you explain this answer?
Coefficient x, y and z are respectively
a)
a
b)
b
c)
c
d)
d
|
Hansa Sharma answered |
Cross -multiply by change in oxidation number and balance H by H+ ions.
Thus, x = 2, y = 5 and z = 6
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