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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.
| 551. |
Calculate normality of a salt solution [of a metal sulphate] having concetration 21.6% w/v if its superoxide has 16% by mass of oxygen. |
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Answer» Correct Answer - 1 |
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| 552. |
The equivalent weight of phosphoric acid `(H_(3)PO_(4))` in the reaction `NaOH+H_(3)PO_(4) rarr NaH_(2)PO_(4)+H_(2)O` isA. `25`B. `98`C. `59`D. `49` |
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Answer» Correct Answer - B Molecular weight of `H_(3)PO_(4)` is `98` and change in its valency `=1` equivalent wieght of `H_(3)PO_(4)` `=("Molecular weight")/("change in valency")=(98)/(1)=98` |
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| 553. |
Which of the following is not intramolecular redox raction?A. `(NH_(4))_(2)Cr_(2)O_(7)rarrN_(2)+Cr_(2)O_(3)+4H_(2)O`B. `2KClO_(3)rarr2KCl+3O_(2)`C. `2Mn_(2)O_(7)rarr4MnO_(2)+3O_(2)`D. `2ClO_(2)rarr5H_(2)O_(2)overset(2OH^(-))rarr2Cl^(-)+5O_(2)+6H_(2)` |
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Answer» Correct Answer - D A disproprtionation redox change involes oxidation and reduction of same atom. |
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| 554. |
Which fo the following reactions is an example of intramolecular redox reaction ?A. `CaCO_3 (s) rarr CaO(s) + CO_2(g)`B. `2 H_2O(l) rarr 2H_2(g) +O_2(g)`C. `2 NaH(s) rarr 2Na(s) +H_(2)(g)`D. `2 KClO_3 (s) rarr 2KCl(s) +3 O_2(g)` |
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Answer» Correct Answer - D `2 overset(+5)(Cl)overset(-2)(O_(3))(s) rarr 2K overset(-1)(Cl)(s)+3overset(0)(O_(2))(g)` It is an example of intramolecular redox reaction because oxygen of ` KClO_3` is oxidized (-2 to 0) while chlorine of ` KClO_3` is reduced (+5 to -1). It is different from disproportionation reaction as different elements (of the same species) are oxidized and reduced. |
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| 555. |
Which is the intramolecular oxidation-reduction reaction?A. `2KClO_(3) rarr 2KCl+3O_(2)`B. `(NH_(4))_(2)Cr_(2)O_(7) rarr N_(2)+CrO_(3)+4H_(2)O`C. `PCl_(5) rarr PCl_(3)+Cl_(2)`D. All of the above |
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Answer» Correct Answer - D (`a`) `underset(+1)underset(darr)(K)underset(+5)underset(darr)(Cl)underset(-2)underset(darr)(O_(3)), rarr underset(+1)underset(darr)(K)underset(-1)underset(darr)(Cl)+underset(0)(3O_(2))` `Cl` is reduced and `O` is oxidised. (`b`) `underset(-3)underset(uarr)((NH_(4))_(2))underset(+6)underset(uarr)(Cr_(2)O_(7)) rarr underset(0)underset(uarr)(N_(2))+underset(+3)underset(uarr)(Cr_(2)O_(3))` `N` is oxidised and `Cr` is reduced (`c`) `underset(+5)underset(uarr)(P)underset(-1)underset(uarr)(Cl_(5)) rarr underset(+3)underset(uarr)(P Cl_(3))+underset(0)underset(uarr)(Cl_(2))` `P` is reduced and `Cl` is oxidised. |
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| 556. |
The pair of compounds that can exist together isA. `FeCI_(3),SnCI_(2)`B. `H_(g)CI_(2),SnCI_(2)`C. `FeCI_(2),SnCI_(2)`D. `FeCI_(3),KI` |
| Answer» `FeCI_(2) and SnSOI_(2)` can exist together since `Sn^(2+)` ions connot reduce `Fe^(2+)` ions | |
| 557. |
Assertion: In the redox reaction `8H^(+)(aq)+4NO_(3)^(-)+6Cl^(-)+Sn(s) rarr SnCl_(6)^(2-)+4NO_(2)+4H_(2)O`. the reducing agent is `Sn(s)`. Reason In balacing half-reaction, `S_(2)O_(3)^(2-) rarr S(s)`, the number of electrons added on the left is `4`.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-4B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-4C. Statement-1 Is True, Statement-2 is False .D. Statement-1 is True False, Statement-2 is True. |
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Answer» Correct Answer - b |
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| 558. |
`aK_(2)Cr_(2)O_(7)+bKCl+cH_(2)SO_(4)rarrxCrO_(2)Cl_(2)+yKHSO_(4)+zH_(2)O` The above equation balances whenA. ` a=4, b=2, c=6` and `x=6, y=2, z=3`B. ` a=6, b=4, c=2` and `x=6, y=3, z=2`C. ` a=1, b=4, c=6` and `x=2, y=6, z=3`D. `a=2,b=4, c=6` and `x=2, y=6 , z=2` |
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Answer» Correct Answer - C Note that one formula unit of ` K_2Cr_2O_7` contains two Cr atoms. Thus, To balance the number fo Cr atoms on both sides, we sholud have ` x=2` and `a=1`. Thus , option (3) is correct. Also note that this is not a redox reaction as there is no change in the oxidation number of any given element. The formation fo ` CrO_2 Cl_2` cna be described as follows : `{:(K_(2)Cr_(2)O_(7)+2H_(2)SO_(4) rarr2KHSO_(4)+2CrO_(3)+H_(2)O),(2KCl+H_(2)SO_(4) rarr 2KHSO_(4)+2HICl"]"xx2),(CrO_(3)+2HCl rarr CrO_(2)Cl_(2)+H_(2)O"]"xx2),(bar(K_(2)Cr_(2)O_(7)+4HCl+6H_(2)SO_(4) rarr 2CrO_(2)Cl_(2)+6KHSO_(4)+3H_(2)O)):}` |
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| 559. |
Assertion: In the redox reaction `8H^(+)(aq)+4NO_(3)^(-)+6Cl^(-)+Sn(s) rarr SnCl_(6)^(2-)+4NO_(2)+4H_(2)O`. the reducing agent is `Sn(s)`. Reason In balacing half-reaction, `S_(2)O_(3)^(2-) rarr S(s)`, the number of electrons added on the left is `4`.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explantion of the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - B `Sn(0 rarr +4)` It undergoes oxidation, so it is the reducing agent. `S_(2)O_(3)^(2-)+6H^(+)+4e^(-) rarr 2S+3H_(2)O` |
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| 560. |
The ratio of coefficient of `HNO_(3), Fe(NO_(3))_(2)` and `NH_(4)NO_(3)` in the following redox reaction `Fe + HNO_(3) rarr Fe (NO_(3))_(2) + NH_(4)NO_(3) + H_(2)O` are respectivelyA. `1:10:4`B. `10:4:1`C. `4:10:1`D. `10:1:4` |
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Answer» Correct Answer - B `Fe+HNO_(3)rarrFe(NO_(3))_(2)+NH_(4)NO_(3)+H_(2)O` Half reaction method : `Fe rarr Fe^(2+)+2e^(-)..(1)xx4` `NO_(3)^(-)+10H^(o+)+8e^(Theta)rarrNH_(4)^(o+)+3H_(2)O.....(2)` Equation `(1)xx4+(2)` `4Fe+NO_(3)^(-)+10H^(o+)rarr4fe^(2+)+NH_(4)^(o+)+3H_(2)O` `"Add"^(n)` of spectator ion `4Fe+10HNO_(3)rarr4Fe(NO_(3))_(2)+NH_(4)NO_(3)+3H_(2)O` |
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| 561. |
Consider the following reaction: `Zn+Cu^(2+)rarrZn^(2+)+Cu` with reference to the above which one of the following is the correct statement ?A. Zn is reduced to `Zn^(2+)`ionsB. Zn is reduced to `Zn^(2+)ions`C. `Zn^(2+)`ions are oxidised to ZnD. `Cu^(2+)` ions are oxidised to Cu |
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Answer» Correct Answer - B Zn lies above Cu in the activity series and is oxidised to `Zn^(2+)` ions by releasing electrons. `ZnrarrZn^(2+)+32e^(-)` |
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| 562. |
When the reaction: `CL^(-)+ClO_(3)^(-)rarrCl_(2)+H_(2)O` is balanced in acid solution what is the ratio of `Cl^(-)` to `ClO_(3)^(-)` ?A. `(1)/(1)`B. `(2)/(1)`C. `(3)/(1)`D. `(5)/(1)` |
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Answer» Correct Answer - d |
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| 563. |
In the half reaction : `2ClO_(3)^(-)rarrCl_(2)`A. 5 electrons are gainedB. 5 electrons are liberatedC. 10 electrons are gainedD. 10 electrons are liberated |
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Answer» Correct Answer - c |
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| 564. |
`3KClO_(3)+3H_(2)SO_(4)rarr3KHSO_(4)+HClO_(4)+2ClO_(2)+H_(2)O` Equivalent weight of `KClO_(3)` isA. `(M)/(4)`B. `(M)/(2)`C. `(M+(M)/(2))`D. `((M)/(4)+(M)/(2))` |
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Answer» Correct Answer - C Disproportionation reaction `ClO_(3)^(ө)rarrClO_(4)^(ө)+2e^(-)(x=2)` (oxidation) `x-6= -1, x-8= -1` `x=5 , x=7` `e^(-)+ClO_(3)^(ө)rarrClO_(2), (x=1)` (reduction) `x=5, x= -4= 0` `x=4` `Ew=M+(M)/(2)` |
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| 565. |
If `K_(2)Cr_(2)O_(7)` is source of `Cr_(2)O_(7)^(2-)`, what is the normality of solution containing `4.9 g` of `K_(2)Cr_(2)O_(7)` in `0.1` litre of solution? |
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Answer» Correct Answer - 1 Normality of `K_(2)Cr_(2)O_(7) = (w)/(E xx V("in litre"))` `= (4.9)/((294//6) xx 0.1) = 1N (because E_(K_(2)Cr_(2)O_(7)) = (M)/(6))` |
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| 566. |
Which of the following solutions will exactly oxidize `25mL` of an acid solution of `0.1 M Fe` (`II`) oxalate?A. `25 mL` of `0.1 M KMnO_(4)`B. `25 mL` of `0.2 M KMnO_(4)`C. `25 mL` of `0.6 M KMnO_(4)`D. `15 mL` of `0.1 M KMnO_(4)` |
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Answer» Correct Answer - D `{:("Oxidation",,"Reduction"),(Fe^(3+)toFe^(3+)+e^(-),,MnO_(4)^(-)toMn^(2+)+5e^(-)),(C_(2)O_(4)^(2-)to2CO_(2)+2e^(-),,):}` Balance equation: `5FeC_(2)O_(4)+3MnO_(4)^(-) rarr` Product `(M_(1)V_(1))/(5)=(M_(2)V_(2))/(3)=(0.1xx25)/(5)=0.5` For exact neutralisation `(M_(2)V_(2))/(3)` should by equal to `0.5` (`a`) `(M_(2)V_(2))/(3)=(0.1xx25)/(3)=0.83` (`b`) `(M_(2)V_(2))/(3)=(0.2xx25)/(3)=1.66` (`c`) `(M_(2)V_(2))/(3)=0.6xx25)/(3)=5` (`d`) `(M_(2)V_(2))/(3)=(0.1xx15)/(3)=0.5` |
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| 567. |
A `1.10 g` sample of copper ore is dissolved and the `Cu^(2+)` of is treated with excess `KI`. The liberated `I_(2)` requires `12.12 mL` of `0.10M Na_(2)S_(2)O_(3)` solution for titration. Find the `%` copper by mass in ore. |
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Answer» Correct Answer - 7 `{:(Cu^(2+),+e,rarr,Cu^(+),),(,2I^(-),rarr,I_(2)+2e,),(2S_(2)O_(3)^(2-),,rarr,S_(4)O_(6)^(2-),+2e):}` `"Meq. of" Cu^(2+) = "Meq. Of liberated" I_(2)` `= "Meq. of" Na_(2)S_(2)O_(3)`. `= 12.12 xx 0.1 xx 1 = 1.212` `:. (w_(Cu^(2+)))/(63.6//1) xx 1000 = 1.212` `:. w_(Cu) = w_(Cu^(2+)) = 0.077 (because Cu overset(H_(2)SO_(4))rarr CuSO_(4))` `:. %Cu = (0.077)/(1.10) xx 100 = 7%` |
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| 568. |
`0.804 g` sample of iron ore was dissolved in acid. Iron was oxidised to `+2` state and it requires `47.2 mL` of `0.112N KMnO_(4)` solution for titration, Calculate `%` of `Fe` and `Fe_(3)O_(4)` in ore. |
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Answer» `"Meq.of" Fe^(2+)="Meq.of" KMnO_(4)` `"Meq.of" Fe^(2+)=47.2xx0.112=5.2864` `:. (w)/(56//2)xx1000=5.2864` `because Fe^(2+)rarrFe^(3+)+1e` `therefore w_(Fe^(2+))=0.296 g` `therefore %` Purity of `Fe=(0.296xx100)/(0.804)=36.82%` Now `Fe_(3)O_(4)rarr3Fe` `3xx56gFe` is obtained from `232 g Fe_(3)O_(4)`. `therefore % "of" Fe_(3)O_(4)=(0.409)/(0.804)xx100=50.87%` |
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| 569. |
`4I^(-)+Hg^(2+)rarrHgI_(4)^(2-)` , `1` mole each of `Hg^(2+)` and `I^(-)` will form….. Mole `HgI_(4)^(2-)`:A. `1 "mole"`B. `0.5 "mole"`C. `0.25 "mole"`D. `2 "mole"` |
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Answer» Correct Answer - C `4` mole of `I^(-)` will give `1` mole of `HgI_(4)^(2-)` with `1` mole of `Hg^(2+)`. The mole ratio order is `4:1::1` |
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| 570. |
In `C+H_(2)O rarr CO+H_(2)`, `H_(2)O` acts asA. Oxidising agentB. Reducing agentC. (`a`) and (`b`) bothD. None of these |
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Answer» Correct Answer - A In this reaction `H_(2)O` acts as oxidising agent. |
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| 571. |
What is the oxidation number of the metal atom in the following ions? `(i) [Fe(CN)_(6)]^(3) ,(ii) MnO_(4), (iii) [Cr(H_(2)O)_(6)]^(3^(+)` |
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Answer» `(i) Fe in [Fe(CN)_(6)]^(3^(+))ion` In this ion, the oxidation number of CN ions is -1 Let the oxidation number of Fe ber x. write the oxidation number of Fe above its symbol and that of `CN^(-)` ion above its formula. `overset(x)Feoverset(-1)(CN)_(6)` Caclulate the oxidation num,bners of all the species `x+6(-1)=x-6` Since `[Fe(CN)_(6)]^(3)` is an anion the sum of the oxidation j number must be -3 `x-6=-3 or x=+3.` `(ii) Mn in MnO_(4)` ion Let the oxidation number of atoms above their symbols `overset(x)Mn overset(-2)O_(4)` Calculate the oxidation number of all the atoms since `[MnO_(4))]` is an anion , the sum of the oxidation numbers must bve-1 `therefore x-8=-1 orx=-21+8=+7` `Cr in [Cr(H_(2)O)_(6)]^(3+) ion` As `H_(2)O` is a neurtral molecule its oxidation number is taken as 0 Write the oxidation number of `Cr` above its symbol and that of `H_(2)O` above its formula. `overset(x)Croverset(O)(H_(2))O_(6)` Since `[Cr(H_(2)O)_(6)]^(3+)` is an ion , the sum of the oxidation numbers must be+3 `x=+3` |
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| 572. |
Identify the correct statement about `H_(2)O_(2)`A. It acts as reducing agent onlyB. It acts as both ocidising and reducing agentC. It is neither an oxidiser nor reducesD. It acts as oxidising agent only |
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Answer» Correct Answer - B It acts as both oxidizing and reducing agent. |
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| 573. |
The elememt which forms oxides in all oxidation states ` +I ` to +V is.A. ` N`B. `P`C. ` As`D. ` Sb` |
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Answer» Correct Answer - A N shows a wide spectrum of oxidation states from ` -3 ` to ` +5`. It forms oxides in all oxidation states from ` +1` to `+5` : ` N_2 O (+I), NO (+II) , N_2O_3( +III), NO_2 (+IV)` , and `N_2O_5 (+V)`. |
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| 574. |
Arrange the oxides of nitrogen in increasing order of oxidation state of N from +1 to +5.A. `N_(2)O lt N_(2)O_(3) lt NO_(2) lt N_(2)O_(5) lt NO`B. `N_(2)O lt NO lt N_(2)O_(3) lt NO_(2) lt N_(2)O_(5)`C. `N_(2)O_(5) lt NO_(2) lt N_(2)O_(3) lt NO lt N_(2)O`D. `NO lt N_(2)O lt NO_(2) lt N_(2)O_(3) lt N_(2)O_(5)` |
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Answer» Correct Answer - B Oxides: `N_(2)O rarr + 1, NO rarr +2` `N_(2)O_(3) rarr + 3, NO_(2) rarr +4, N_(2)O_(5) rarr +5` |
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| 575. |
The oxidation number of `S` in `Na_(2)S_(4)O_(6)` isA. `(5)/(2)`B. `(3)/(2)`C. `(3)/(5)`D. `(2)/(3)` |
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Answer» Correct Answer - A `Na_(2)overset(*)(S_(4))O_(6)` `2+4x-12=0` `4x=10 x=(10)/(4) x=(5)/(2)`. |
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| 576. |
The oxides which cannot act as reductant: (I) `CO_(2)` (II) `SO_(3)` (III) `P_(4)O_(10)` (IV) `NO_(2)`A. (I),(II),(III)B. (II),(III),(IV)C. (I),(II),(IV)D. (III),(IV) |
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Answer» Correct Answer - A `C` in `CO_(2), S` in `SO_(3)` and `P` in `P_(4)O_(10)` have `+4, +6, +5` oxidation states which represents their maximum value and thus cannot act as reductant. |
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| 577. |
In the following equation, `MnO_(2)` acts as `MnO_(4)^(2-) + 2H_(2)O + 2E^(-) rarr MnO_(2) + 4overset(Ө)(OH)`A. Oxidising agentB. Reducing agentC. Both oxidising and reducing agent.D. Neither oxidising nor reducing agent. |
| Answer» Correct Answer - B | |
| 578. |
In the following conversion of sulphide of phosphorous `P_(4)S_(3)rarrP_(2)O_(5)+SO_(2)` Equivalent weight of `P_(4)S_(3)` (molecular weight=M) is :A. `(M)/(14)`B. `(M)/(18)`C. `(M)/(32)`D. `(M)/(38)` |
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Answer» Correct Answer - c |
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| 579. |
Balance the following redox reactions by ion electron method: a. `MnO_(4)^(Θ)(aq)+I^(Θ)(aq) rarr MnO_(2)(s)+I_(2)(s)` (in basic medium) b. `MnO_(4)^(Θ)(aq)+SO_(2)(g) rarr Mn^(2+)(aq)+HSO_(4)^(Θ)(aq)` (in acidic solution) c. `H_(2)O_(2)(aq)+Fe^(2+)(aq) rarr Fe^(3+)(aq)+H_(2)O(l)` (in acidic solution) d. `Cr_(2)O_(7)^(2-)+SO_(2)(g) rarr Cr^(3+)(aq)+SO_(4)^(2-)(aq)` (in acidic solution) |
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Answer» (a) Step 1 : The two half reactions involved in the given reaction are : Oxidation half reaction `overset(-1)(I)_((aq)) to overset(0)(I)_(2 (s))` Reduction half `overset(+7)(M)n O_(4 (aq))^(-) to overset(+4)(M) n O_(2 (aq))` Step 2 : Balancing l in the oxidation half reaction , we have: `2I_((aq))^(-) to I_(2 (s))` Now , to balance the charge , we add `2 e^(-)` to the RHS of the reaction . `2 I_((aq))^(-) to I_(2 (s)) + 2 e^(-)` Step 3 : In the reduction half reaction , the oxidation state of Mn has reduced from `+7` to +4 . Thus, 3 electrons are added to the LHS of the reaction. `MnO_(4 (aq))^(-) + 3 e^(-) to MnO_(2 (aq))` Now , to balance the charge , we add 4 `OH^(-)` ions to the RHS of the reaction as the reaction is taking place in a basic medium. `MnO_(4(aq))^(-) + 3e^(-) to MnO_(2 (aq)) + 4 OH^(-)` Step 4 : In this equation , there are 6 O atoms on the RHS and 4 O atoms on the LHS . Therefore , two water molecules are added to the LHS . `MnO_(4 (aq))^(-) + 2 H_(2)O + 3 e^(-) to Mn O_(2 (aq)) + 4 OH^(-)` Step 5 : Equalising the number of electrons by multiplying the oxidation half reaction by 3 and the reduction half reaction by 2, we have: `6I_(aq)^(-) to 3I_(2 (s)) + 6 e^(-)` `2MnO_(4 (aq))^(-) + 4H_(2)O + 6 e^(-) to MnO_(2 (s)) + 8 OH_((aq))^(-)` Step 6 : Adding the two half reactions, we have the net balanced redox reaction as: `6 I_((aq))^(-) + 2 MnO_(4 (aq))^(-) + 4 H_(2)O_((l)) to 3 I_(2 (s)) + 2 Mn O_(2 (s)) + 8 OH_((aq))^(-)` (b)Following the steps as in part (a), we have the oxidation half reaction as: `SO_(2 (g)) + 2 H_(2)O_((l)) to HSO_(4 (aq))^(-) + 3 H_((aq))^(+) + 2 e_((aq))^(-)` And the reduction half reaction as : `MnO_(4 (aq))^(-) + 8 H_((aq))^(+) + 5 e^(-) to Mn_((aq))^(2+) + 4 H_(2)O_((l))` Multiplying the oxidation half reaction by 5 and the reduction half reaction by 2, and then by adding them, we have the net balanced redox reaction as: `2MnO_(4 (aq))^(-) + 5 SO_(2 (g)) + 2H_(2)O_((l)) + H_((aq))^(+) to 2 Mn_((aq))^(2+) + 5 HSO_(4 (aq))^(-)` (c) Following the steps as in part (a) , we have the oxidation half reaction as : `Fe_((aq))^(2+) to Fe_((aq))^(3+) + e^(-)` And the reduction half reaction as : `H_(2)O_(2 (aq)) + 2 H_((aq))^(+) + 2 e^(-) to 2H_(2)O_((l))` Multiplying the oxidation half reaction by 2 and then adding it to the reduction half reaction, we have the net balanced redox reaction as: `H_(2)O_(2 (aq)) + 2 Fe_((aq))^(2+) + 2 H_((aq))^(+) to 2 Fe_((aq))^(3+) + 2 H_(2)O_((l))` (d) Following the steps as in part (a) , we have the oxidation half reaction as : `SO_(2 (g)) + 2H_(2)O_((l)) to SO_(4 (aq))^(2-) + 4 H_((aq))^(+) + 2e^(-)` And the reduction half reaction as : `Cr_(2)O_(7 (aq))^(2-) + 14 H_((aq))^(+) + 6 e^(-) to 2 Cr_((aq))^(3+) + 7 H_(2)O_((l))` Multiplying the oxidation half reaction by 3 and then adding it to the reduction half reaction, we have the net balanced redox reaction as: `Cr_(2)O_(7 (aq))^(2-) + 3 SO_(2 (g)) + H_((aq))^(+) to 2 Cr_((aq))^(3+) + 3 SO_(4 (aq))^(2-) + H_(2)O_((l))` |
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| 580. |
Which ordering of compounds is according to the decreasing order of the oxidation state of nitrogen ?A. `HNO_(3),NO,NH_(4)Cl,N_(2)`B. `HNO_(3),NO,N_(2),NH_(4)Cl`C. `Hno_(3),NH_(4)Cl,NO,N_(2)`D. `NO,HNO_(3),NH_(4)Cl,N_(2)` |
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Answer» Correct Answer - b |
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| 581. |
Which ordering of compound is according to the decreasing order of the oxidation state of nitrogen?A. `HNO_(3)`, `NO`, `N_(2)`,` NH_(4)Cl`B. `HNO_(3)`, `NO`, `NH_(4)Cl`,` N_(2)`C. `HNO_(3)`, `NH_(4)Cl`, `NO`,` N_(2)`D. `NH_(4Cl)`, `N_(2)`, `NO`, `HNO_(3)` |
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Answer» Correct Answer - A `overset(+5)(HNO_(3))`, `overset(+2)(NO)`, `overset(0)(N_(2))`, `overset(-3)(NH_(4))Cl` |
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| 582. |
Which ordering of compound is according to the decreasing order of th e oxidation state of nitrogen?A. `HNO_(3), NO, NH_(4)Cl, N_(2)`B. `HNO_(3),NO,N_(2),NH_(4)Cl`C. `HNO_(3),NH_(4)Cl, NO,N_(2)`D. `NO,HNO_(3),NH_(4)Cl,N_(2)` |
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Answer» Correct Answer - B Let the oxidation state of `N` is `a`. `{:(HNO_(3)rArr+1+a+3(-2)=0,,:. a= +5,,),(NOrArra+1(-2)=0,,:. a=+2,,),(NH_(4)ClrArra+4(+1)+(-1)=0,,:. a=-3,,),(N_(2)rArr2a=0,,:. a=0,,):}` Decreasing order of the oxidation state of `N` is `{:(HNO_(3)",",NO",",N_(2)",",NH_(4)Cl,),(+5,+2,0,-3,):}` |
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| 583. |
In the reaction ` xAs_2S_3 + y NO_3^(-)+ H^(+) rarr AsO_4^(3-) + S + NO + H_2O`.A. ` x=4, y=9`B. ` x=2,y=10`C. ` x=4, y=11`D. ` x=3, y=10` |
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Answer» Correct Answer - D `overset(+3)(As_(2))overset(-2)(S_(3))+overset(+5)(N)overset(-2)(O_(3))+overset(+1)(H^(+))rarr overset(+5)(As)overset(-2)(O_(4)^(3-))+overset(0)(S)+overset(+2-2)(NO)+overset(+1)(H_(2))overset(-2)(O)` The oxidation number of As increases from ` +3` to ` + 5`, of S increases from ` -2 ` to 0, whilen of N decreases from ` +5`, to `+2` . This implies that both As and S of ` As_2S_3` are oxidized while N of ` NO_3^(-)` is reduced. The change in oxidation numbers of As and S must be considered together as they must maintain their atomic ration fo ` 2: 3`. Oxidation haf-reaction: ` As_2S_3 rarr AsO_4^(3-) + S` Reduction half-reaction ` NO_3^(-) rarr NO` Now we balance the two half -reactions (a) Balance all atoms excpt H and O atoms ` As_2S_3 rarr 2 AsO_4^(3-) + 3S` ` NO_3^(-) rarr NO` (b) Account for the change in oxidation number. Total inccrease in the oxidation numbern of As `( + 6 rarr + 10 )` is of four units, totla increase in the oxidation number fo ` S ( -6 rarr 0)` is of six units, and decrease in the oxidation number of ` N (+ 5 rarr +2)` is fo three units . `As_2S_3 rarr 2AsO_4^(3-) + 3 S + 10e^(-)` `NO_3^(-) + 3 e^(-) rarr NO` (c ) Balance ionic charges by adding ` H^(+)` ions as the reaction is carried out in acidic medium `As_(2)S_(3) rarr 2AsO_(4)^(3-) +3S+16H^(+) +10e^(-)` `NO_(3)^(-)+4H^(+)+3e^(-) rarr NO` (d) Balance H and O atoms by adding `H_(2)O` molecules: `As_(2)S_(3)+8H_(2)O rarr 2AsO_(4)^(3-)+3S+16H^(+)+10e^(-)` `NO_(3)^(-)+4H^(+)+3e^(-) rarr NO +2H_(2)O` To equalize the number of electrons lost amnd gained, multiply oxidation half-reaction by 3 and resuction half-reaction by 10. Finally add them, cancelling electrons on both sides: `3As_(2)S_(3)+10NO_(3)^(-)+4H_(2)O rarr 6AsO_(4)^(3-)+9S+8H^(+)` |
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| 584. |
200 ml of 0.01 M `KMnO_(4)` oxidise 20 ml of `H_(2)O_(2)` sample in acidic medium. The volume strength of `H_(2)O_(2)` isA. 2.8 volumeB. 5.6 volumeC. 0.5 volumeD. 0.25 volume |
| Answer» Correct Answer - A | |
| 585. |
Which of the following refers ton the original description of oxidation ?A. Addition of oxygenB. Addition fo electronegative elementC. Removal fo hydrogenD. Removal of electropositive element |
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Answer» Correct Answer - A Orginally, the term "oxidation" was used to describe the addition of oxygen to an element or a compound : `S(s) + O_2(g) rarr SO_2 (g)` `CH_4 (g) + 2O_2 (g) rarr CO_2 (g) + 2H_2 O(1)`. |
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| 586. |
The value of n in ` NO_3^(-) + 4H^(+) + n e^(-) rarr 2 H_2O + NO` is .A. `2`B. `4`C. `5`D. `3` |
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Answer» Correct Answer - D The given equation is balanced atomically . To balance it electrically , we add `3` dlectrons to the LHS. `NO_3^(-) + 4H^(+) + 3e^(-) rarr 2H_2O + NO` . |
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| 587. |
Which one of the following reactions is not an example of redox reaction?A. `Cl_(2)+2H_(2)O+SO_(2) rarr 4H^(+)+SO^(4-)2Cl^(-)`B. `Cu^(++)+Zn rarr Zn^(++)+Cu`C. `2H_(2)+O_(2) rarr 2H_(2)O`D. `HCl+H_(2)O rarr H_(3)O^(-)+Cl^(-)` |
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Answer» Correct Answer - D In reaction `HCl+H_(2)O rarr H_(3)O^(-)+Cl^(-)`, only reduction has taken place, not oxidation. |
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| 588. |
What is the equivalent weight of `C_(12)H_(22)O_(11)` in the following reaction? `C_(12)H_(22)O_(11)+36HNO_(3) rarr 6H_(2)C_(2)O_(4)+36NO_(2)+23H_(2)O`A. `(342)/(36)`B. `(342)/(12)`C. `(342)/(22)`D. `(342)/(3)` |
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Answer» Correct Answer - A `(C_(12))^(@) rarr 6(C_(2))^(6+)+36^(e-)` `:. Eq. wt.=(M)/(36)=(342)/(36)` |
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| 589. |
Which of the following is the best description of the behaviour of bromine in the reaction given below?A. Proton acceptor onlyB. Both oxidised and reducedC. Oxidised onlyD. Reduced only |
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Answer» Correct Answer - B It is a disproportionation reaction. |
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| 590. |
The number of electrons transferred (lost and gained ) during the reactiong `Fe + H_(2)O rarr Fe_3 O_4 + H_2` is .A. `8`B. `6`C. `4`D. `2` |
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Answer» Correct Answer - A `overset(0)(F)e+overset(+1)(H_(2))overset(-2)(O)rarr overset(+8//3)(Fe_(3))overset(-2)(O_(4))+overset(0)(H_(2))` Oxidation half-reaction `3overset(0)(F)e rarroverset(+8//3)(Fe_(3))O_(4)+8e^(-)` Reduction half-reaction `overset(+1)(H_(2))overset(-2)(O)+2e^(-) rarr overset(0)(H_(2))` `4H_(2)O+8e^(-) rarr 4H_(2)` Thus, eight electrons are transferred during the reaction |
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| 591. |
In the reaction, `l_(2) + 2S_(2)O_(3)^(2-) rarr 2l^(-) + S_(4) O_(6)^(2-)`, equivalent weight of iodine will be equal toA. MB. `M//2`C. `M//4`D. `2M` |
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Answer» Correct Answer - B `l_(2)^(0) + 2S_(2)O_(3)^(2-) rarr overset(-1)(2l^(-)) + S_(4)O_(6)^(2-)` `:.` Decrease in ON of iodine per atom `=1` `:.` Decrease in ON of iodine per molecule `= 2 xx 1 = 2` Hence, equivalent weight of iodine `("Molecular weight of iodine")/("Total decrease in ON of iodine per molecule") = (M)/(2)` |
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| 592. |
Which of the following is/are disproportionation reactions ?A. `6NaOH + 3Cl_(2) to NaClO_(3) + 5NaCl + 3H_(2)O`B. `2NaOH + Cl_(2) to NaOCl + NaCl + H_(2)O`C. `H_(2)O_(2) to H_(2)O + (1)/(2)O_(2)`D. `2KClO_(3) to 2KCl + 3O_(2)` |
| Answer» Correct Answer - A::B::C | |
| 593. |
Consider the following reaction `2Fe^(3+) + 2l^(-) rarr 2Fe^(2+) + l_(2)` The half-reactions for the given reaction areA. `2l^(-) rarr l_(2) + 2e^(-)` and `Fe^(3+) + e^(-) rarr Fe^(2+)`B. `l_(2) rarr 2l^(-) + 2e^(-)` and `Fe^(2+) + e^(-) rarr Fe^(3+)`C. `l_(2) rarr 2Theta^(-) + 2l^(-)` and `Fe^(2+) rarr Fe^(3+) + e^(-)`D. None of the above |
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Answer» Correct Answer - A The half-reactions are oxidation reaction as in `2l^(-) rarr l_(2) + 2e^(-)` and reduction reaction as in `Fe^(3+) + e^(-) rarr Fe^(2+)` |
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| 594. |
Which of the following statements regardign `H_(2)SO_(5)` is/are correct ?A. The oxidation number of sulphur is +6B. Two oxygen atoms are present in form of peroxideC. Three oxygen atoms are present in form of oxideD. The oxidation state of sulphur is +8 |
| Answer» Correct Answer - A::B::C | |
| 595. |
Which of the following refers to the original description of reduction?A. Addition of hydrongenB. Addition of electropositive elementC. Removal of oxygenD. Removal of electropositive element |
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Answer» Correct Answer - C In the beginning , reduction was considered as the removal of oxygen from a compound : `2ZnO(s) overset(Delta)(rarr) 2Zn(s) +O_2(g)`. |
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| 596. |
Which of the following reaction does not stick to the cleassical idea of redox reactions ?A. ` 3Fe_(3)O_4(s) +8Al(s) rarr 9 Fe(s) +4Al_2O_2(s)`B. `2Na(s) +H_2(g) rarr 2NaH(s)`C. ` H_2S(g) +Cl_2(g) rarr 2HCl(g) +S(s)`D. None of these |
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Answer» Correct Answer - B In reaction (1), aluminim is oxidizes because oxygen is added to it while ferrous ferric oxide `(Fe_3O_4)` is reduced because oxygen has been removed from it. In reaction (3), chlorine isn reuced due to the addition of hydrogen to it whilen ` H_2S` is oxidized due to the removal of hydrogen from S or we can say due to the addition of chlorine (an electronegative element ) to hydrogen. In reaction (2) , the classical concept of redox reactions fails necause both Na and `H_2` are of similar nature, i.e., both are electropositive elements (possessing the tendency to form positive ions ). Accoring to the classical concept , Na is reduced to `NaH` due to the addition fo hydency to it and `H_2` is also reduced due to the addition fo electropositive element to it, However, with the caraful application of the concept of electronegativity, we conclude that Na is oxidized while hydrogen is reduced. Thus, reaction (3) prompts us to think in terms of another way to define redox reactions. We face similar problem in the following reaction: `Cl_(2)+3F_(2) rarr 2ClF_(3)` as both `Cl_(2)` and `F_(2)` are electronegative elements. |
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| 597. |
The process in which oxidation number increase, isA. reductionB. hydrolysisC. oxidationD. decomposition |
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Answer» Correct Answer - C Oxidation is the process in which oxiation number of effective element in the species increase |
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| 598. |
20 volume `H_(2)O_(2)` solution has a strength of aboutA. 60.86 g/LB. 3.58 NC. 1.79 MD. `3.035%` |
| Answer» Correct Answer - A::B::C | |
| 599. |
The ratio of oxidation states of `Cl` in potassium chloride to that in potassium chlorate isA. `+(1)/(5)`B. `-(1)/(5)`C. `-(2)/(5)`D. `+(3)/(5)` |
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Answer» Correct Answer - B Oxidation state of `Cl` in `KCl = -1` Oxidation state of Cl in `KClO_(3) = +5` `:.` Ratio of oxidation state of `Cl = (-1)/(5)` |
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| 600. |
In the reaction `B_(2)H_(6)+2KOH+2Xrarr 2Y+6H_(2)`, `X` and `Y` are respectivelyA. `H_(2)`, `H_(3)BO_(3)`B. `HCl`, `KBO_(3)`C. `H_(2)O`, `KBO_(3)`D. `H_(2)O`, `KBO_(2)` |
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Answer» Correct Answer - D `B_(2)H_(6)+2KOH+2H_(2)O rarr 2KBO_(2)+6H_(2)` |
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