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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 701. |
Redox reactions play a pivotal role in chemistry and biology.The values of standard reduction potential `E^(@)` of two half cell reaction decide which way the reaction is expected to proceed. A simple exaple is Danie cell in which zinc goes in to solution and copper gets deposited Given below are set of half cell reaction (acidic medium) along with their `E^(@)` values. `I_(2)+2e^(-)rarr2I^(-) " "E^(@)=0.54` `CI_(2)+2e^(-)rarr2CI^(-) " "E^(@)=0.54` `Mn^(3+)+e^(-)rarrMn^(2+) " "E^(@)=1.36` `Fe^(3+)+e^(-)rarrMn^(2+)" "E^(@)=0.77` `O_(2)+4H^(+)e^(-)rarr2h_(2)O" "E^(@)=1.23` Using these data , obtain the correct explanation for the following question. Sodium fusion extact obvtined from aniline. On treatment with iron (II suphate and `H_(2)SO_(4))` in presence of air givers a prussian blue precipitte.The blue colour is due to the formation ofA. `Fe_(4)[Fe(CN)_(6)]^_(3)`B. `Fe_(3)[Fe(CN)_(6)]^_(2)`C. `Fe_(4)[Fe(CN)_(6)]^_(2)`D. `Fe_(3)[Fe(CN)_(6)]^_(3)` |
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Answer» Correct Answer - A `Na+C+NrarrNaCN` (Sodium fusion extract) `Fe^(2+)+6CN^(-)rarr[Fe(CN)_(6)]^(4-)` In presence of air, `Fe^(2+)` ions get oxidised to `Fe^(3+)` ions. `(Fe^(2+)rarrFe^(3+)+e^(-)]xx4,E^(@)=-0.77V)` `(O_(2)+4H^(+)+4e^(-)rarr2H_(2)O,E^(@)=+1.23 V)/(4Fe^(2+)+4H^(+)+O_(2)rarr4Fe^(3+)+2H_(2)O)` `Fe^(3+)` ions then combine with `[Fe(CN)_(6)]^(4-)` ion to form ferric ferrocyanide which has prussian blue colour `4Fe^(3+)+3[Fe(CN)_(6)]^(4-)rarrFe_(4)[Fe(CN)_(6)]_(3)` |
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| 702. |
Oxidation numbers of P in ` PO_4^(3-)`, of S in ` SO_4^(2-)`, and that of `Cr` in ` Cr_2O_7^(2-)` are respectively ,A. `-3`, `+6`, `+6`B. `+5`, `+3`, `+6`C. `+3`, `+6`, `+5`D. `+5`, `+6`, `+6` |
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Answer» Correct Answer - D Oxidation no. in each case. `PO_(4)^(3-)` `x+4(-2)= -3x=+5` `SO_(4)^(2-)` `x+(-2xx4)= -2` `x-8= -2` `x = -2+8x= +6` `Cr_(2)O_(7)^(2-)` `2x+(-2xx7)= -2` `2x-14= -2` `2x= -2+14` `2x=12` `x=+6` |
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| 703. |
The O.N of P in `Ba(H_(2)PO_(2))_(2)` is:A. `+3`B. `+2`C. `+1`D. `-1` |
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Answer» Correct Answer - C `Ba(H_(2)overset(x)PO_(2))_(2)` +2 + 4 (+1) + 2 x + 4 (-2)=0 2x-2=0 or x=+1 |
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| 704. |
Oxidation no. of `P` in `H_(4)P_(2)O_(5)`, `H_(4)P_(2)O_(6)`, and `H_(4)P_(2)O_(7)` are respectivelyA. `+3`, `+4`, `+5`B. `+4`, `+3`, `+5`C. `+3`, `+5`, `+4`D. `+5`, `+3`, `+4` |
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Answer» Correct Answer - A `H_(4)P_(2)O_(5) : 4xx1+2xxa-5xx2=0` `a=+3` `H_(4)P_(2)O_(6) : 4xx1+2xxa-6xx2=0` `a=+4` `H_(4)P_(2)O_(7) : 4xx1+2xxa-7xx2=0` `a=+5` |
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| 705. |
Which of the following has the highest value of `E_("rod")^(@)` ?A. `P_(4)`B. `Cl_(2)`C. `I_(2)`D. `F_(2)` |
| Answer» Correct Answer - D | |
| 706. |
which of the following is a redox reaction ?A. `NaCl + KNO_(3) to NaNO_(3) + KCl`B. `CaC_(2)O_(4) + 2HCl to CaCl_(2) + H_(2)C_(2)O_(4)`C. `Ca(OH)_(2) + 2NH_(4)Cl to CaCl_(2) + 2NH_(3) + 2H_(2)O`D. `2K[Ag(CN)_(2)] + Zn to 2Ag + K_(2)[Zn(CN)_(4)]` |
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Answer» Correct Answer - D (a) `overset(+1)Naoverset(-1)(Cl) + overset(+5)K overset(-2)(NO_(3)) to overset(+1)(Na)overset(+5)Noverset(-2)(O_(3)) + overset(+1)Koverset(-1)(Cl)` (b) `overset(+2)(Ca)overset(+3)(C_(2))overset(-2)(O_(4)) + 2overset(+1)Hoverset(-1)(Cl) to overset(+2)(Ca)overset(-1)(Cl_(2)) + 2overset(+2)(Ca)overset(-1)(Cl_(2)) + 2overset(-3)overset(+1)H_(3) + 2overset(+1)(H_(2))overset(-2)O` (c) `overset(+2)(Ca)overset(-1)(OH)_(2) + 3overset(-3)(NH_(4))overset(-1)(Cl) to overset(+2)(Ca)overset(-1)(Cl_(2)) + 2overset(-3)Noverset(+1)(H_(3)) + 2overset(+1)H_(2)overset(-2)O` In all these cases, during reaction there is no change in oxidation state of ion or molecule or constituent atom, there are simple ionic reactions. (d) `2K[Ag(CN)_(2)] + Zn to 2Ag + K_(2)[Zn(CN)_(4)]` `Ag^(+) to Ag`, gain of electron, reduction `Zn to Zn^(2+)`, loss of electron, oxidation |
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| 707. |
Which one of the following statements is correct?A. Oxidation of a substance is followed by the reduction of anotherB. Reduction of a substance is followed by the reduction of anotherC. Oxidation and reduction are complementary reactionsD. It is not necessary that both oxidation and reduction should take place in the same reaction |
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Answer» Correct Answer - C Both oxidation and reduction are complementary to each other. |
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| 708. |
In which fo the following has the oxidation number of oxygen been arragned in increasing order ?A. `KO_2 lt OF_2 lt O_3 lt BaO_2`B. `BaO_2 lt KO_2 lt O_3 lt OF_2`C. `BaO_2 lt O_3 lt OF_2 lt KO_2`D. `OF_2 lt KO_2 lt BaO_2 lt O_3` |
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Answer» Correct Answer - B Let x be the oxidation number (O.N.) of O atom in a given compound: O.N. of O in ` BaO_2 rArr (+2) +2x rArr =0 rArr x=- 1` O.N of O in `KO_2 rArr (+1) + 2x =0 rArr x=- 1//2` O.N. of O in `O_3 rArr 0` (Free state or uncombined state ) O.N. of O in ` OF_2 rArr x+ 2(-1) =0 rArr x= +2` `:. Baoverset(-1)(O_(2)) lt Koverset(-1//2)(O_(2)) lt overset(0)(O_(2)) lt overset(+2)(OF_(2))` |
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| 709. |
The oxidation number of oxygen in `OF_(2)` is |
| Answer» Correct Answer - 3 | |
| 710. |
In the reaction `3Cl_(2)+6NaOHrarrNaClO_(3)+5NaCl+3H_(2)O` the element which loses as well as gains electrons isA. NaB. ClC. OD. None of these |
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Answer» Correct Answer - B In the reaction `3Cl_(2)+6NaOHrarrNaClO_(3)+5NaCl+3H_(2)O` O.N. of Cl increases from zero in `Cl_(2)` to +5 in `NaClO_(3)` and decreases from zero in `Cl_(2)` to -1 to NaCl. |
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| 711. |
In the reaction `3Cl_(2)+6NaOHrarrNaClO_(3)+5NaCl+3H_(2)O` the element which loses as well as gains electrons isA. `Na`B. `O`C. `Cl`D. None of these |
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Answer» Correct Answer - C It is a disproportionation reaction, so `Cl_(2)` undergoes both oxidation and reduction. |
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| 712. |
The oxidation number of oxygen in `OF_(2)` isA. `+1`B. `+2`C. `-1`D. `-2` |
| Answer» Correct Answer - C | |
| 713. |
A `KMnO_(4)` solution can be standarised by titration against `As_(2)O_(3(s))`. A `0.1156 g` sample of `As_(2)O_(3)` requires `27.06mL` of the `KMnO_(4(aq.))` for its titration. What is the molarity of the `KMnO_(4(aq.))` [As `=75`]? `5As_(2)O_(3)+4MnO_(4)^(-)+9H_(2)O+12H^(+)rarr10H_(2)AsO_(4)+4Mn^(2+)` |
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Answer» Meq.of `As_(2)O_(3)=` Meq.of `KMnO_(4)` `(As_(2)^(3+)rarr2As^(5+)rarr4e)` `(0.1156)/(198//4)xx1000=Mxx5xx27.08` `(Mn^(7+)+5erarrMn^(2+))` `therefore M=0.0172` `therefore M_(MnO_(4)^(-))=0.0172M` |
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| 714. |
The oxidation number of oxygen in `OF_(2)` isA. `+2`B. `-2`C. `+1`D. `-1` |
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Answer» Correct Answer - A `overset(+2_-1xx2)(OF_(2)):x-2=0impliesx=2` `:.` Oxidation number of `O = +2` |
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| 715. |
A particular acid-rain water has `SO_(3)^(2-)`. If a `25.00 mL` sample of this water requires `34.08 mL` of `0.01964M KMnO_(4)` for its titration, what is the molarity of `SO_(3)^(2-)` in acid-rain? `2MnO_(4)^(-)+5SO_(3)^(2-)+6H^(+)rarr5SO_(4)^(2)+2Mn^(2+)+3H_(2)O` |
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Answer» Meq.of. `SO_(3)^(2-)="Meq.of" KMnO_(4)` `therefore Nxx25=34.08xx0.01964xx5` `N=0.1339` `therefore M=(0.1339)/(2)=0.0669` `S^(4+)rarrS^(6+)+2e` `N_(SO_(3)^(2-))=0.1339N` `M_(SO_(3)^(2-))=(0.1339)/(2)=0.0669M` |
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| 716. |
An oxidation process involvesA. Increase in oxidation numberB. Decrease in oxidation numberC. Both decrease and increase in oxidation numberD. No change in oxidation number |
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Answer» Correct Answer - A The statement is self-explanotory. |
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| 717. |
A solution containing `1.984 g` of crystalline `NA_(2)SO_(2)O_(3).xH_(2)O` in water required `40ml` of `N//5` lodine solution for complete reaction .Calculate the value of `x` |
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Answer» `2S_(2)O_(3)^(2-)rarrS_(4)O_(6)^(2-)+2e` `2e+I_(2)rarr2I^(-)` Also Meq.of `I_(2)="Meq.of" Na_(2)S_(2)O_(3).xH_(2)O` `40xx(1)/(2)=(1.984)/(M//1)xx1000` `therefore M_(Na_(2)S_(2)O_(3).xH_(2)O)=248` `2xx23+2xx32+16xx3+18x=248` or `x=5` |
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| 718. |
Barium permangnate solution (20mL, 0.1 M) is mixed with 0.1 N `I^(-)` giving precipitate fo `IO_(3)^(-)` and `MnO_(2)` . Resulting solution is filtered and titrated against `Mo^(3+)`, giving `MoO_(2)^(2+)` and `Mn^(2+)`. Which required 0.5 M, 10mL acidified `Mo^(3+)` .Select the correct option (s).A. Volume of `I^(-)` solution taken is 30mLB. Volume of `I^(-)` solution taken is 50mLC. Per mole `Mn^(2+)` formed, 4 moles of `H^(+)` are consumedD. Per mole `IO_(3)^(-)` formed, 2 moles of `MnO_(4)^(-)` are consumed |
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Answer» Correct Answer - a,d |
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| 719. |
Which of the following are redox reactions?A. `Zn+2HClrarrZnCl_(2)+H_(2)`B. `Al(OH)_(3)+3HClrarrAlCl_(3)+3H_(2)O`C. `Ag^(o+)+I^(-)rarrAgl`D. Disproportionation of `Cu^(o+)` in aqueous solution. |
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Answer» Correct Answer - A::D (a) and (b) |
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| 720. |
In balancing the half reaction `CN^(ө)rarrCNO^(ө)`(skeltan) The number of electrosn that must be added isA. `1` on the rightB. `0`C. `1` on the leftD. `2` on the right |
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Answer» Correct Answer - D `CN^(ө)rarrCNO^(ө)+2e^(-)` `x-3=-1, x-3-2=-1` `x=2, x=4` |
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| 721. |
Which of the following compounds acts both as an oxidising as wll as a reducing agent?A. `HNO_(2)`B. `H_(2)O_(2)`C. `H_(2)S`D. `SO_(2)` |
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Answer» Correct Answer - a,b,d |
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| 722. |
A compound can be formed by the elements A,B and C having oxidation state +1,+2 and -3 respectively. Then which compound may be formed?A. ABCB. `B_(3)C_(2)`C. `A_(2)BC`D. `A_(4)BC_(2)` |
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Answer» Correct Answer - a,d |
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| 723. |
Choose the correct statement(s):A. 1 mole of `MnO_(4)^(-)` ion can oxidise 10 moless of `Fe^(2+)`ion in acidic mediumB. 1 mole of `Cr_(2)O_(7)^(2-)` ion can oxidise 12 moles of `Fe^(2+)` ion in acidic mediumC. 2 mole of `Cu_(2)S` can be oxidize by 2.6 moles of Mn `O_(4)^(-)` ion in acidic medium. `(Cu_(2)S toCu^(2+)+SO_(2))`D. 2 moles of `Cu_(2)S` can be oxidize by 8/3 moles of `Cr_(2)O_(7)^(2-)` ion in acidic medium `(Cu_(2)S toCu^(2+)+SO_(2))` |
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Answer» Correct Answer - a,b,c |
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| 724. |
Which of the following samples of reducing agents is /are chemically equivalent to 25mL of 0.2 N `KMnO_(4)` to be reduced to `Mn^(2+)` and water?A. 25mL of 0.2M `FeSO_(4)` to be oxidized to `Fe^(3+)`B. 50mL of 0.1M `H_(3)AsO_(3)` to be oxidized to `H_(3)AsO_(4)`C. 25mL of 0.1 `H_(2)O_(2)` to be oxidized to `H^(+)` and `O_(2)`D. 25mL of 0.1 M `SnCl_(2)` to be oxidized to `Sn^(4+)` |
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Answer» Correct Answer - a,c,d |
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| 725. |
In which of the following reactions does `H_(2)O_(2)` acts as a resucing agents ?A. `2FeCl_(2)+2HCl+H_(2)Orarr2FeCl_(3)+2H_(2)O`B. `Cl_(2)+H_(2)O_(2)rarr2HCl+O_(2)`C. `2HI+H_(2)O_(2)rarr2H_(2)O+I_(2)`D. `H_(2)SO_(3)+H_(2)O_(2)rarrH_(2)SO_(4)+H_(2)O` |
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Answer» Correct Answer - b |
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| 726. |
The oxidation state of chrominium in the final product formed in the reaction between `KI` and acidified potassium dichromate soluttion isA. ` +4`B. ` +6`C. ` +2`D. ` +3` |
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Answer» Correct Answer - D Always remenber that acidified potassium bechromate is always reduced to ` Cr^(3+)` ions : ` Cr_2O_7^(2-) + 14H^(+) + 6I^(-) rarr 2Cr^(3+) + 7H_2O + 3I_2` The oxidation state fo chromun in ` Cr^(3+) ` ion is ` +3`. |
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| 727. |
Classify the following redox reactions: a. `N_(2)(g)+O_(2)(g)rarr 2NO(g)` b. `2Pb(NO)_(3)(s)rarr2PbO(s)+2NO_(2)(g)+(1)/(2)O_(2)(g)` c. `NaH(s)H_(2)O(l)rarr NaOH(aq)+H_(2)(g)` d. `2NO_(2)(g)+2overset(ө)OH(aq)rarrNO_(2)^(ө)(aq)+NO_(3)^(ө)(aq)+H_(2)O(l)` |
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Answer» a. Combination redox reaction: `N_(2) and O_(2)` combines together to give `NO`( nitric oxide). b. Decomposition redox(nitric oxide): `Pb(NO_(3))_(2)`( lead nitrate) breaks into three components. c. Displacement redox reaction: Hydrogen of `H_(2)O` is displaced by `H^(ө)`( hydride ion)into `H_(2)` gas. d. Disproporationation redox reaction: `NO_(2)(+4 "oxidation state")` disproportionates into `NO_(2)^(ө)(+3"oxidation state")` and `NO_(2)^(ө)(+5"oxidation state")`. |
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| 728. |
`2H_(2) O_(2) (l) rarr 2H_(2)o(l) + O_(2) (g)` `100 mL` of `X` molar `H_(2)O_(2)` gives `3L` of `O_(2)` gas under the condition when 1 mole occupies `24 L`. The value of `X` isA. `2.5`B. `1`C. `0.5`D. `0.25` |
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Answer» Correct Answer - A Mole of `O_(2)` formed `= (3)/(24) = (1)/(8)` `:.` Mole of `H_(2)O_(2) = (1)/(8) xx 2 = (1)/(4)` `:. 100 xx X = (1)/(4) xx 1000 :. X = 2.5` |
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| 729. |
For the redox reaction: `Cr_(2)O_(7)^(2-)+H^(o+)+NirarrCr^(3+)+Ni^(2)+H_(2)O` The correct coefficient of the reactants for the balanced reaction are:A. `Cr_(2)O_(7)^(2-)=1, Ni=3, H^(o+)=14`B. `Cr_(2)O_(7)^(2-)=3, Ni=3, H^(o+)=12`C. `Cr_(2)O_(7)^(2-)=2, Ni=3, H^(o+)=14`D. `Cr_(2)O_(7)^(2-)=1, Ni=1, H^(o+)=16` |
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Answer» Correct Answer - A `14H^(o+)+cancel(6e^(-))+Cr_(2)O_(7)^(2-)rarr2Cr^(3+)+7H_(2)O` `2x=14=-2, 2x=6` `2x=12` `NirarrNi^(+2)+cancel(2e^(-))]xx3` `ulbar(14H^(o+)+CrO_(7)^(2-)+3Ni2Cr^(3+)+3Ni^(2+)+7H_(2)O)` |
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| 730. |
The oxidation number of `S` in `S_(8),S_(2)F_(2)`, and `H_(2)S`, respectively, areA. `0,+1` and `-2`B. `+2, +1`, and `-2`C. `0,+1` and `+2`D. `-2,+1` and ` -2` |
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Answer» Correct Answer - A `{:(S_(8):axx8=0,,:. a=0,,,),(S_(2)F_(2):axx2+2xx(-1)=0,,:. a=+1,,,),(H_(2)S:2xx1+a=0,,:. a=-2,,,):}` |
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| 731. |
Which of the following statements is`//`are correct?A. The oxidation states of `N` in `NH_(3), HN_(3)`, and `N_(2)H_(4)` are `-3, -1//3`, and `-2`, respectively.B. The oxidation state of `N` in `NO_(2), N_(2)O_(4), and `NO_(2)^(-)` are `+4, +4`, and `+3`, respectively.C. The oxidation states of `N` in `NH_(2)OH, NO`, and `HNO_(3)` are `-1, +2`, and ` +5`, respectively.D. The oxidation states of `N` in `N_(2)O` and `HCN` are `+1` and `-3`, respectively. |
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Answer» Correct Answer - A::B::C::D All statements are self-explanatory. |
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| 732. |
For the redox reaction `Cr_(2)O_(7)^(-2)+H^(+)+Ni rarr Cr^(3)+Ni^(2+)+H_(2)O` The correct coefficents of the reactions for the balanced reaction areA. (`Cr_(2)O_(7)^(2-)=1`, `Ni=3`, `H^(+)=14`)B. (`Cr_(2)O_(7)^(2-)=2`, `Ni=3`, `H^(+)=14`)C. (`Cr_(2)O_(7)^(2-)=1`, `Ni=1`, `H^(+)=16`)D. (`Cr_(2)O_(7)^(2-)=3`, `Ni=3`, `H^(+)=12`) |
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Answer» Correct Answer - A The balanced equation is `Cr_(2)O_(7)^(2-)+14H^(+)+3Ni rarr 2Cr^(3+)+7H_(2)O+3Ni^(2+)` |
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| 733. |
The oxidation number of `S` in `S_(8),S_(2)F_(2)`, and `H_(2)S`, respectively, areA. `0,+1` and `-2`B. `+2, +1` and `-2`C. `0,+1` and `+2`D. `-2, +1` and `-2` |
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Answer» Correct Answer - A `S_(8)`: Oxidation numbers of `S=0` `S_(2)F_(2): 2x-2=0impliesx=1,` Oxidation number of `S=1` `H_(2)S:2+x=0impliesx=2` , Oxidation number of `S= -2` |
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| 734. |
Which of the following reactions should be balanced in basic medium?A. `NH_(3)+MnO_(4)^(ө)rarrMnO_(2)+NO_(2)`B. `Cr(OH)_(2)+I_(2)rarrCr(OH)_(3)+2I^(ө)`C. `HNO_(3)+Fe^(3+)+NO_(2)`D. `H_(2)O_(2)+Fe^(3+)rarrO_(2)+Fe^(2+)` |
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Answer» Correct Answer - A::B In (a), `MnO_(4)^(ө)rarrMnO_(2)` occurs only in basic medium In (b), `Cr(OH)_(2)rarrCr(OH)_(3)` occurs only in basic medium. `(c )` and `(d)` occur only is acidic medium. |
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| 735. |
Among these, identify the species with an atom in `+6` oxidation state: .A. `CrO_(2)Cl_(2)`B. `[Cr(CN)_(6)]^(3-)`C. `MnO_(4)^(-)`D. `[NiF_(6)]^(3-)` |
| Answer» Correct Answer - A | |
| 736. |
Among the following identify the species with an atom in `+6` oxidation state.A. `MnO_(4)^(-)`B. `Cr(CN)_(6)^(3-)`C. `NiF_(6)^(2-)`D. `CrO_(2)Cl_(2)` |
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Answer» Correct Answer - D `Cr` in `CrO_(2)Cl_(2)`: `a+2xx(-2)+2xx(-1)=0` `a= +6` |
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| 737. |
Arrange the following in order of increasing O.N of iodine: |
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Answer» `overset(0)I_(2),overset(-1)HI,overset(+3)HIO+_(2),overset(+5)KIO_(3),overset(+1)ICI` The increasing order is: `HIltI_(2)ltICIlthIO_(2)ltKIO_(3)` |
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| 738. |
Among the following identify the species with an atom in `+6` oxidation state.A. `MnO_(4)^(ө)`B. `[Cr(CN)_(6)]^(3-)`C. `[NiF_(6)]^(2-)`D. `CrO_(2)Cl_(2)` |
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Answer» Correct Answer - D `overset(+7)MnO_(4)^(ө)` , b. `[overset(+3)(Cr)overset(-1xx6)((CN)_(6))]^(3-)` c. `[overset(+4)(Ni)overset(-1xx6)(F_(6))]^(2-)` , d. `overset(+6)(Cr)overset(-2xx2)(O_(2))overset(-1xx2)(Cl_(2))` |
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| 739. |
Which of the following reactions is not a redox reaction?A. `H_(2)O_(2)+KOHrarrKHO_(2)+H_(2)O`B. `Cr_(2)O_(7)^(2-)+2overset(ө)OHrarr2CrO_(4)^(2-)+H_(2)O`C. `Ca(HCO_(3))_(3)overset(Delta)rarrCaCO_(3)+CO_(2)+H_(2)O`D. `H_(2)O_(2)rarrH_(2)O+(1)/(2)O_(2)` |
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Answer» Correct Answer - A::B::C (a) Oxidation state of `K is +1` in both reactant and product. (b), Oxidation state of `Cr(+6)` does not change. In (c ), oxidation states of `Ca` and `C` and `O` do not change. In (d), the `H_(2)O_(2)` which disproportionates is both oxidising and a reducing agent. |
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| 740. |
No reaction occurs in which of the following equations?A. `I^(ө)-Fe^(2+)rarr`B. `F_(2)+2NaClrarr`C. `Cl_(2)+2NaFrarr`D. `I_(2)+2NaBerarr` |
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Answer» Correct Answer - A::C::D Reduction potential of `F_(2) gt Cl_(2) gt Br_(2) gt I_(2)`. So `F_(2)` can diplace `CI^(ө), Br^(ө)`, and `I^(ө)` but not vice versa and `Br_(2)` can displace only `I^(ө)` but not vice versa. In (a), `Fe( "not" Fe^(2+))` is a better reducing agent than `I^(ө)`. |
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| 741. |
Define reducing agent. |
| Answer» A species which loses electrons as a result of oxidtion is reducing agent | |
| 742. |
Choose correct statements (s) regarding the following reactions. `Cr_(2)O_(7)^(2-)(aq)+3SO_(3)^(2-)(aq)+8H^(+) rarr 2Cr^(3+)(aq)+3SO_(4)^(2-)(aq)+4H_(2)O`A. `Cr_(2)O_(7(aq))^(2-)+3SO_(2(g))+2H_((aq))^(+)rarr 2Cr_((aq))^(3+) rarr 2Cr_((aq))^(3+)+3SO_(4(aq))^(2-)+H_(2)O_((l))`B. `2Cr_(2)O_(7(aq))^(2-)+3SO_(2(g))+4H_((aq))^(+)rarr 4Cr_((aq))^(3+)+3SO_(4(aq))^(2-)+2H_(2)O_((l))`C. `Cr_(2)O_(7(aq))^(2-)+3SO_(2(g))+14H_((aq))^(+)rarr 2Cr_((aq))^(3+)+3SO_(4(aq))^(2-)+7H_(2)O_((l))`D. `Cr_(2)O_(7(aq))^(2-)+6SO_(2(g))+7H_((aq))^(+)rarr 2Cr_((aq))^(3+)+6SO_(4(aq))^(2-)+7H_(2)O_((l))` |
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Answer» Correct Answer - A `Cr_(2)O_(7)^(2-)+SO_(2)rarrCr^(3+)+SO_(4)^(2-)` ( in acidic solution) Oxidation half equation : `SO_(2)+2H_(2)O rarr SO_(4)^(2-)+4H^(+)+2e^(-)` ...(i) Reduction half equation : `Cr_(2)O_(7)^(2-)+14H^(+)+6e^(-)rarr 2 Cr^(3+)+7H_(2)O` ...(ii) Multiplying eqn. (i) by 3 and adding to eqn. (ii) we get `Cr_(2)O_(7)^(2-)+3SO_(2)+2H^(+) rarr 2 Cr^(3+)+3SO_(4)^(2-)+H_(2)O` |
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| 743. |
In the neutralization of `Na_(2)S_(2)O_(3)` using `K_(2)Cr_(2)O_(7)` by idometry, the equivalent weight of `K_(2)Cr_(2)O_(7)` isA. `M//2`B. `M//6`C. `M//3`D. `M` |
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Answer» Correct Answer - B `(Cr^(6+))_(2)+6erarr2Cr^(3+)` `:. E_(K_(2)Cr_(2)O_(7))=(M)/(6)` |
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| 744. |
The equivalent weight of `HNO_(3)` (molecular weight `=63`) in the following reaction is `3Cu+8HNO_(3)rarr3Cu(NO_(3))_(2)+2NO+4H_(2)O`A. `(4xx63)/(3)`B. `(63)/(5)`C. `(63)/(3)`D. `(63)/(8)` |
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Answer» Correct Answer - D In this case only `2 mol` of `NO_(3)^(ө)` undergo reduction. `3e^(-)+NO_(3)^(ө)rarrNO(x=3)` (reduction) `x-6= -1` `x-2=0` `x=5, x=2` `6 mol` of `HNO_(3)` are not changing so `6NO_(3)^(ө)` are added in the reaction to get `3 mol` of `Cu(NO_(3))_(2)`. `:. Ew=M+(M)/(3)=(4M)/(3)=(4xx63)/(3)` |
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| 745. |
The equivalent weight of `H_(2)SO_(4)` in the following reaction is `Na_(2)Cr_(2)O_(7)+3SO_(2)+H_(2)SO_(4)rarr3Na_(2)SO_(4)+Cr_(2)(SO_(4))_(3)+H_(2)O`A. `98`B. `(98)/(6)`C. `(98)/(2)`D. `(98)/(8)` |
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Answer» Correct Answer - B In this reaction, `6mol` of `e^(-)` are involved with `1mol` if `H_(2)SO_(4)` in this redox reaction. `H_(2)SO_(4)` acts here as acidic medium. `6e^(-)+Cr_(2)O_(7)^(-2)rarr2Cr^(3+)(x=6)`, (oxidation) `2x-14= -2, 2x=6` `2x=12` `SO_(2)rarrSO_(4)^(2-)+2e^(-)]xx3(x=6)` `x-4=0, x-8 = -2` `x=4, x=6` So, `Ew of H_(2)SO_(4)=(M)/(6)=(98)/(6)` |
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| 746. |
A reducing agent in a redox reaction undergoesA. a decrease in oxidation numberB. an increase in oxidation numberC. loss of electronsD. gain of electrons |
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Answer» Correct Answer - B::C A reducing agent is the one which loses electrons. |
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| 747. |
Which of the following reactions is (are) redox reactions ?A. `6CO_(2)+6H_(2)O overset("Sun light")(rarr) C_(6)H_(12)O_(6)+6O_(2)`B. `KCN +AgCN rarr K[Ag(CN)_(2)]`C. `Zn +2HCl rarr ZnCl_(2)+H_(2)`D. `BaCl_(2)+H_(2)SO_(4) rarr BaSO_(4)+2HCl` |
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Answer» Correct Answer - A::C Reaction which involves change in O.N of atoms is a redox reaction. |
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| 748. |
Which of the following act both as an oxidising as well as reducing agent ?A. `H_(2)O_(2)`B. `H_(2)S`C. `SO_(2)`D. `HNO_(2)` |
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Answer» Correct Answer - A::C::D Element neither having highest nor lowest O.N can act both as O.A and a R.A |
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| 749. |
Which of the following reactions is (are) not redox reaction (s) ?A. `NH_(3)Cl + KOH rarr NH_(3)+H_(2)O +KCl`B. `4KCN +Fe(CN)_(2) rarr K_(4)[Fe(CN)_(6)]`C. `2Rb + 2H_(2)O rarr 2RbOH + H_(2)`D. `2CuI_(2) rarr 2CuI + I_(2)` |
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Answer» Correct Answer - A::B Reactions which do not involve a change in O.N of atoms is not a redox reaction. |
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| 750. |
What is the oxidation state of `Co[Co(H_(2)O)_(5)Cl]^(2+)` ?A. `+2`B. `+3`C. `+1`D. `+4` |
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Answer» Correct Answer - C Since, the O.N of `SO_(4)^(2-)` is -2, therefore the total charge on the complex ion, `[Fe(H_(2)O)_(5)NO]` is +2. Since No is paramagnetic molecule, its unpaired electron is transferred to iron. As a result, NO carries +1 charge. Further, since `H_(2)O` is a neutral molecule, its O.N is zero. Putting the values of all the oxidation numbers, the O.N of Fe in the complex ion `= x+0 xx 5 + 1=2` `:. n = +1` Thu the O.N of Fe is +1 and the correct structure of the `|Fe(H_(2)O)_(5)NO|SO_(4)` is `|Fe^(+1)(H_(2)O)_(5)NO|SO_(4)`. |
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