52671 + Interview Questions in Chemistry in Class 11

1.	K_a values for Ha,HB and HD are 10 ^(-5),10 ^(-7)and 10 ^(-9)respectively. Whch of the following will be correct for decimolar aqueous solution of NaA, NaB and NaD at 25^(@)C ?
Answer» `(pH)_(N_BA) lt (pH) _(NaB) ` `(pH) _(NAD) lt(pH) _(NaB) ` ` (pH)_(NAA) lt (pH) _(NaD) ` `(pH) _(NaB) =7` SOLUTION :` pH= 7+ ( (pKa+log c)/( 2)) , pH uparrow if pKa uparrow `

Discussion

2.	K_a of NH_4^+ acid is 1.77xx10^(-5). Then give the equation and ionization constant of its conjugate base.
Answer» Solution :The CONJUGATE base of `NH_4^+`is `NH_3` and its IONIZATION CONSTANT is `K_b` So, `K_b=K_w/K_a=(1.0xx10^(-14))/(1.77xx10^(-5))=5.64xx10^(-10)`

Discussion

3.	K_a of CH_3COOH is 1.76 xx 10^(-5) at 298 K temperature. Calculate dissociation constant of its conjugate base.
Answer» SOLUTION :`5.9xx10^(-10)`

Discussion

4.	K_a of acetic acid is 1.8 xx 10^(-5). What volume of one litre 1 M acetic acid be diluted, so that the pH of resulting solution will be twice of the original value ?
Answer» SOLUTION :`2.77 XX 10^4 `

Discussion

5.	K_a of acetic acid is 1.8 xx 10^(-5). How much sodium acetate is to be added to llitre 0.1M acetic acid to make a buffer of pH = 4 ?
Answer» SOLUTION :`1.48g`

Discussion

6.	PK_b for NH_3= 4.7. When the total concentra-tion of buffering agents is 0.6 mol L^(-1), the pH of buffer with NH_3 and NH_4CI is 9. What is the weight of NH_3 in the buffer solution ? log 2=0.4
Answer» SOLUTION :`3.4g L^(-1)`

Discussion

7.	K_(a) for HCN is 5 xx 10^(-10) at 25^(@)C . For maintaining a constant pH of 9 , the volume of 5M KCN solution required to be added to 10 ml of2M HCN solution is
Answer» 4 ML 7.95 ml 2 ml 9.3 ml ANSWER :C

Discussion

8.	K_c values respectively for the reactions, H_2SO_3 leftrightarrow H^+ +HSO_3^(-) and HSO_3^(-) leftrightarrow H^+ +SO_3^(2-) and 2 times 10^-2 and 6 times 10^-8 mol L^-1 Calculate the K_c for the reaction H_2SO_3 leftrightarrow 2H^+ + SO_3^(2-)
Answer» SOLUTION :`1.09 XX 10^(-9)`

Discussion

9.	K_a for CH_3COOH is 1.8xx10^(-5) and K_b for NH_4OH is 1.8xx10^(-5). The pH of ammonium acetate will be
Answer» 7.005 4.75 `7.0` Between 6 and 7 Solution :GIVEN that , `K_a` for `CH_3COOH=1.8xx10^(-5)` `K_b` for `NH_4OH = 1.8xx10^(-5)` AMMONIUM acetate is a salt of WEAK acid and weak base. `therefore pH=7+(pK_a+pK_b)/2` `=7+([-log 1.8xx10^(-5)]-[-log 1.8xx10^(-5)])/2` `=7+(4.74-4.74)/2`=7.00

Discussion

10.	K_(a)for CH_(3)CO OH is 1.8xx10^(-5) and K_(b) for NH_(4)OHis 1.8xx10^(-5). The pH of ammonium acetate will be
Answer» 7.005 4.75 `7.0` between 6 and 7. Solution :`CH_(3)CO ONH_(4)` is a salt of weak acid and weak base. Hence, `PH = 7 + (1)/(2) (pK_(a) - pK_(b))=7+0=7`

Discussion

11.	K_a for CH_3 COOH and HCN are respectively 1.8 xx 10^(-5)and 5 xx 10^(-10) Calculate the equilibrium constant for the reaction CH_3 COOH + NaCN hArrCH_3 COONa + HCN.
Answer» SOLUTION :`3.6 XX 10^4`

Discussion

12.	K_(a) for ascorbic acid ("HA sc" )is 5xx10^(-5). Calculate the hydrogen ion concentration and percentage of hydrolysis in an aqueous solution in which the concentration of Asc^(-) ions is 0.02 M.
Answer» Solution :As ascorbic ACID is a weak acid, aqueous solution of `Asc^(-)`means the solution of a salt of weak acid with strong base. For such a salt , degree of hydrolysis is GIVEN by `h=sqrt((K_(w))/(K_(a)xxc))=sqrt((10^(-14))/(5xx10^(-5)xx2xx10^(-2))=10^(-4) = 10^(-4) xx 100 % = 0.01 % ` `{:(Asc^(-),+,H_(2)O,hArr,HAsc,+,H^(-),),(c " mol" L^(-1),,,,,,,),(c -ch ,,,,ch,,ch,):}` `[OH^(-)]=ch=0.02xx10^(-4)=2XX10^(-6) "mol " L^(-1) :. [H^(+)]=(K_(w))/([OH^(-)])=(10^(-14))/(2xx10^(-6))=5xx10^(-9) "mol" L^(-1)`

Discussion

13.	K_a for a weak monobasic acid HA is 1 xx 10^(-4). Calculate (a) hydrolysis constant of KA and (b) neutralisation constant for the reaction of HA with KOH.
Answer» Solution :HYDROLYSIS constant `(K_b)` is given as, `K_h=(k_w)/(k_a) = (1 xx 10^(-14))/( 1 xx 10^(-4)) = 1 xx 10 ^(-10)`NEUTRALISATION constant `(K_a)` for the reaction of HA with KOH is given as, `K_n=K_(H)^(-1) =1 xx 10^(10)`

Discussion

14.	K_(4)[Fe(CN)_(6)] is converted into CO_(3)^(-2), Fe^(+3) ions and NO_(3)^(-) ions. Here_____________
Answer» N. is reduced C. is oxidised Iron is oxidised Eq.wt `("Formula.wt")/(61)` Solution :`{:(FE^(++)rarrFe^(+++),"(OX)",1xx1=1),(OVERSET(+2)(CN^(-))rarroverset(+4)(CO_(3)^(-2)),"(OX)",6xx2=12),(overset((-3))(CN^(-))rarroverset((+5))(NO_(3)^(-)),"(OX)",6xx8=48),(,,61):}` n factor = 61

Discussion

15.	K_(2)Cr_(2)O_(7)+H_(2)SO_(4)+4H_(2)O_(2)overset("ether")to 2X+K_(2)SO_(4)+5H_(2)O, X+6H_(2)SO_(4)to2Y +6H_(2)O+7Z(g) the correct statement (s) regarding the above eq:
Answer» The oxidaion state of CENTRAL STOM in X I s+10 and has butter fly stereucture The OXIDATION state of central atom In Y is +3 and has green coloure Z is colourless paramagnetic gas The oxidation state of central stom in X is +^ and has 2 peroxy linkages with butterfly LIKE stucture. Answer :BCD

Discussion

16.	K_(2)Cr_(2)O_(7)+FeSO_(4)+H_(2)SO_(4)rarr K_(2)SO_(4)+Cr_(2)(SO_(4))_(3)+Fe_(2)(SO_(4))_(3)+H_(2)O a) How many electrons are transferred b) What is the mole coefficient of H,30, in the above reaction?
Answer» Solution :Change in oxidation STATE of Cr of `K_(2)Cr_(2)O_(7)` =(2x +6) -(2x+3)=6 Change in oxidation state of Fe of `FeSO_(4)=+3-(+2)=1` Cris -CROSSING the change `K_(2)Cr_(2)O_(7)-6FeSO_(4)` (a) NUMBER of electrons transferred =6 (B) `K_(2)Cr_(2)O_(7)+6FeSO_(4)+7H_(2)SO_(4) to K_(2)SO_(4)+Cr_(2)(SO_(4))_(3)+3Fe_(2)(SO_(4))_(3)+7H_(2)O` Coefficient of `H_(2)SO_(4)=7`

Discussion

17.	K_1 and K_2 for carbonic acid are 4.45 xx 10^(-11) and 4.09 xx 10^(-11). Calculate the concentrations of HCO_(3)^(-) . CO_(3)^(2-) and pH of centimolar H_2 CO_3.
Answer» SOLUTION :`6.7 XX 10^(-5) M ,4.7 xx 10^(-11) M and4.2`

Discussion

18.	K_(1) and K_(2) are the equilibrium constants for the recation respectively. N_(2(g))+O_(2(g))overset(K_(1))(hArr)2NO_((g)) 2NO_((g))+O_(2(g))overset(K_(2))(hArr)2NO_(2(g)) what is the equilibrium constant for the reaction NO_(2(g))hArr1//2N_(2(g))+O_(2(g))
Answer» `(1)/(SQRT(K_(1)K_(2)))` `(K_(1)=K_(2))^(1//2)` `(1)/(2K_(1)K_(2))` `((1)/(K_(1)K_(2)))^(3//2)` Answer :A

Discussion

19.	K_(1) and K_(2) are velocity constant of forward and backward reaction. The equilibrium constant K_c of the reaction is........
Answer» SOLUTION :`(K_1)/(K_2)`

Discussion

20.	K_(1) & K_(2) for oxalic acid are 6.5 xx 10^(-2) and 6.1 xx 10^(-5) respectively . What will be the [OH^(-)] in a 0.01 M solution of sodium oxalate
Answer» `9.6 xx 10^(-6)` `1.4 xx 10^(-1)` `1.2 xx 10^(-6)` `1.3 xx 10^(-8)` Solution :The hydrolysis of `C_(2)O_(4^(2-))` is as follows `C_(2)O_(4^(2-)) + H_(2)O hArr HC_(2)O_(4)^(-) + OH^(-)` `[OH^(-)] = sqrt((K_(W) xx C)/(K_(2))) = sqrt((10^(-14) xx 10^(-2))/(6.1 xx 10^(-5)))` `= 1.2 xx 10^(-6)`

Discussion

21.	K_1 and K_2 are the equilibrium constants for the reactions respectively. N_2(g)+ O_2(g) overset(K_1)hArr 2NO(g) 2NO(g)+ O_2(g) overset(K_2)hArr 2NO_2(g) What is the equilibriumconstant for the required reaction NO_2(g) hArr 1/2N_2(g)+ O_2(g)
Answer» `1/(sqrtK_1K_2)` `(K_1= K_2)^(1/2)` `1/(2K_1K_2)` `(1/(K_1K_2))^(3/2)` Solution :LET equilibrium constant for the required reaction be K. Then, `K_1 = ([NO]^2)/([N_2][O_2]),K_2=([NO_2]^2)/([NO]^2[O_2]),K=([N_2]^(1/2)[O_2])/([NO_2])` `sqrtK_1 = [NO]/([N_2]^(1/2)[O_2]^(1/2)),sqrtK_2 = ([NO_2])/([NO][O_2]^(1/2))` `SQRT(K_1.K_2) = ([NO_2])/([N_2]^(1/2)[O_2])` `:. K =1/(sqrt(K_1K_2))`

Discussion

22.	K^(+)ions are essential for
Answer» METABOLISM of GLUCOSE inside the CELL The SYNTHESIS of proteins Activation of CERTAIN enzymes All Answer :D

Discussion

23.	K+ClrarrKCl. This is an example of
Answer» oxidation reduction a REDOX reaction none of these Answer :C

Discussion

24.	K = 6.2 xx 10^(-8) for the reaction Ag(NH_3)_(2)^(+) hArrAg^(+)+ 2NH_3. If solubility product of AgCl is 1.8 xx 10^(-10), calculate the concentration of complex in 1 M aqueous NH_3
Answer» SOLUTION :`0.054 M`

Discussion

25.	Justify the position of hydrogen in the periodic table on the basis of its electronic configuration.
Answer» Solution :Hydrogen is the first element in the periodic table. However, its placement in the periodic table has been a subject of DISCUSSION in the past. Hydrogen has electronic configuration `1s^1` on one hand, its electron configuration is similar to the outer electronic configuration II `(ns^1)` of alkali metals, which belong to the first group of the periodic table. On the other hand, like halogens, it is short by, one electron to the corresponding NOBLE gas configuration. Hydrogen, therefore has resemblance to alkali metals, which lose one electron to form unipositive ions. Like alkali metals hydrogen forms oxides, halides and sulphides. However, unlike alkali metals, it has a very HIGH IONIZATION enthalpy and does not possess metallic characteristics under normal conditions. `Delta_i H` of Li is 520 kJ `"mol"^(-1)`,F is 1680 kJ `"mol"^(-1)` and that of H is 1312 kJ `"mol"^(-1)` Like halogens, hydrogen forms a diatomic molecule, combines with elements to form hydrides and a large number of COVALENT compounds. However, in terms of reactivity, it is very low as compared to halogens. Inspite of the fact that hydrogen, to a certain extent resembles both with alkali metals and Halogens, it differs from them as well. Loss of the electron from Hydrogen atom results in nucleus `(H^+)` of ~ `1.5xx10^(-3)` pm size. This is extremely small as compared to normal atomic and ionic sizes of 50 to 200 pm. As a consequence, `H^+` does not exist freely and is always associated with other atoms or molecules. Thus, it is unique in behaviour and is, therefore, best placed separately in the periodic table

Discussion

26.	Justify the position of hydrogen in the periodic table on the basis of its electronic configuration
Answer» SOLUTION :For ANSWER, CONSULT SECTION 9.1

Discussion

27.	Justify the position of Hydrogen in the periodic table.
Answer» Solution :Hydrogen has been placed at the TOP of the alkali metals family. Hydrogen and alkali metals resemble in the following ASPECTS. 1. Electronic configuratin: Hydrogen has one electron in its VALANCE shell like the alkali metals `H-1s^(1)`. 2. ELECTROPOSITIVE character: As both hydrogen and alkali metals form monovalent cations by losing the electron in the valance shell. `HtoH^(+)+e^(-)`

Discussion

28.	Justify the position of hydrogen in the periodic table?
Answer» Solution :Hydrogen resembles alkali metals in the following aspects (i) Electronic configuration `1s^(1)` as alkali metals have `ns^(-1)` Hydrogen forms unipositive `H^+` ion like alkali METAL `Na^+,K^+` (iii) Hydrogne forms halides (HX), oxides `(H_2O)`, PEROXIDES `(H_2O_2)` like alkali metals `(NaX,Na_2O,NaO_2)` Hydrogen also acts as reducing agent like alkali metals.

Discussion

29.	Justify the position of f-block elements in the periodic table.
Answer» ANSWER :the SYMMETRY of PERIODIC table should not be DISTURBED

Discussion

30.	Justify the given statement with suitable examples- "the Properties of the elements are a periodic function of their atomic numbers".
Answer» Solution : There are so MANY physical properties of elements such as melting points, boiling points, heats of fusion and vaporisation, ENERGY of atomisation, etc., which show periodic variations. The cause of periodicity in properties is the repetition of same OUTER configurations after certain regular intervals. e.g., `1^(st)` group elements have almost same outer electronic configuration, i.e, `ns^(1)` `""_(3)Li=1s^(2),2s^(1)` `""_(11)Na=1s^(2),2s^(2),2p^(2),3s^(1)` `""_(19)K=1s^(2),2s^(2),2p^(6),3s^(2),3p^(6),3s^(1)` `""_(37)Rb=1s^(2),2s^(2),2p^(6),3s^(2),3p^(6),3d^(10),4s^(2),4P^(6),5s^(-1)` `""_(87)Fr=1s^(2),2s^(2),2p^(6),3s^(2),3d^(10),4s^(2),4p^(6),4d^(10),4f^(14),5s^(2),5p^(6),5d^(10),6p^(6),7s^(1)` Due to same configuration all alkali metals have similar properties. e.g., SODIUM and potassium are soft and reactive metals. They forms basic oxides and their basic character increases as we go from top to bottom. They form +ve ion while lossing of one electron. All the elements of `17^(th)` group (halogen family) have same outermost shell electronic configuration, i.e., `ns^(2)np^(5)` and thus have similar properties. `""_(9)F=1s^(2),2s^(2),2p^(5)` `""_(17)Cl=1s^(2),2s^(2),2p^(6),3s^(2),3p^(5)` `""_(35)Br=1s^(2),2s^(2),2p^(6),3s^(2),3p^(6),3d^(10),4s^(2),4p^(5)` `""_(53)I=Is^(2),2s^(2),3s^(2),3d^(10),4s^(2),4p^(6),4d^(10),5s^(2),4p^(5)` `""_(85)At= 1s^(2),2s^(2),2p^(6),3s^(2),3p^(10),4s^(2),4p^(6),4d^(10),4f^(14),5s^(2),5p^(6),5d^(10),6s^(2),6p^(5)`

Discussion

31.	Justify the given statement with suitable examples -"the properties of the elements are a periodic function of their atomic numbers"?
Answer» Solution :There are numerous physical properties of elements such as melting points, boiling points, heats of fusion and vaporisation, energy of atomisation, etc, which show periodic variations. The case of periodicity in properties is the repetition of similar outer electronic configuration after certain regualr intervals. e.g., all the elements of 1st group (ALKALI metals) have similar outer electronic configuration. i.e., `ns^(1)`. `._(3)Li = 1s^(2).2s^(1)` `._(11)Na=1s^(2), 2s^(1), 2P^(6), 3s^(1)` `._(19)K=1s^(2), 2s^(2), 2p^(6), 3s^(2), 3p^(6), 4s^(1)` `._(37)Rb=1s^(2), 2s^(2), 2p^(6), 3s^(6), 3d^(10), 4s^(2), 4P^(6), 5s^(1)` `._(56)Cs=1s^(2), 2s^(2), 2p^(6), 3s^(2), 3p^(6), 3d^(1), 4s^(2), 4p^(6), 4d^(10), 5s^(2), 5p^(6), 6s^(1)` `._(87)Fr=1s^(2),2s^(2),2p^(6),3s^(2),3p^(6),3d^(10),4s^(2),4p^(6),4d^(10),4f^(14)` `5s^(2), 5p^(6), 5d^(10)6s^(2), 6p^(6), 7s^(1)` Therefore, due to similar outermost shell electronic configuration all alkali metals have similar properties. e.g., sodium and potassium both are soft and reactive metals. They all form unpositive ion by the LOSE of one electron. Similarly, all the elements of 17th group (halogens) have similar outermost shell electronic configuration, i.e., `ns^(2)np^(5)` and thus possess similar properties. `._(9)F=1s^(2) , 2s^(2), 2p^(5)` `._(17)Cl=1s^(2), 2s^(2), 2p^(6), 3s^(2), 3p^(5)` `._(35)Br=1s^(2), 2s^(2), 2p^(6), 3s^(2), 3p^(6), 3d^(10), 4s^(2), 4p^(6), 4d^(10), 5s^(2), 5p^(5)` `._(53)I= 1s^(2), 2s^(2), 2p^(6), 3s^(2), 3p^(6), 3d^(10), 4s^(2), 4p^(6), 4d^(10), 5s^(2), 5p^(5)` `._(85)At=1s^(2), 2s^(2), 2p^(6), 3s^(2), 3p^(6), 3d^(10), 4s^(2), 4p^(6), 4d^(10), 4f^(14), 5s^(2), 5p^(6), 5d^(10), 6s^(2), 6p^(5)`

Discussion

32.	Justify the given statement with suitable examples the properties of the elements are a periodic function of their atomic numbers''.
Answer» Solution :(i)The cause of periodicity in properties the REPETITION of similar outer electronic CONFIGURATION after certain REGULAR intervals. (ii)Eg : Consider elements of group 1 s `_(3)Li-1s^(2)2s^(1)` `_(11)Na-1s^(2)2s^(2)2p^(6)3s^(1)` Due to similar OUTERMOST electronic configuration they exhibit similar properties.

Discussion

33.	Justify the following statements : Many thermodynamically feasible reactions do not occur under ordinary conditions.
Answer» Solution :It is because HEAT ENERGY is REQUIRED to overcome activation energy.

Discussion

34.	Justify the following statements : Reaction with Delta G^() lt 0 always have an equilibrium constant greater than 1.
Answer» SOLUTION :`Delta G^(@) = -2.303 RT log K` If `K GT 1, Delta G^(@)` will be LESS than zero because products formed are more than that of reactants, i.e., process is spontaneous in the forward direction

Discussion

35.	Justify the following statements : The entropy of a substance increases on changing from liquid to vapour state at any temperature.
Answer» Solution :The ENTROPY of VAPOUR is more than that of the LIQUID and therefore entropy increases during VAPORISATION.

Discussion

36.	Justify the following statements : (a)Reactions with DeltaG^(@)lt 0alwayshave an equilibrium constantgreater than 1. (b)Many thermodynamically enthalpyfeasible reactions do not occur under ordinary conditions. (c ) At low temperature, enthalpychange dominates the DeltaGexpression and at high temperatures, it is entropy whch dominatest the value of DeltaG.
Answer» SOLUTION :(a)Aready discussedalready (b) Under ordinaryconditions, the average energy of the reactants MAY be less than THRESHOLD energy.They requiresome activation energy to initiate the reaction. (c )`DeltaG = DELTAH - T DELTAS`. At low temperature, `TDeltaS` is small. Hence,`DeltaH ` dominates. At high temperatures,`T DeltaS`is large, ie., `DeltaS` dominates the value of`DeltaG`.

Discussion

37.	Justify the following statements : At low temperatures, enthalpy change dominates the Delta G expression and at high temperatures it is the entropy which dominates the value of Delta G.
Answer» Solution :`Delta G = Delta H - T Delta S`, At low TEMPERATURES `Delta H GT T Delta S` whereas at HIGH temperatures `T Delta S gt Delta H "":. Delta G` decreases, i.e., becomes NEGATIVE.

Discussion

38.	Justify the following statements : An exothermic reaction is always thermodyamically spontaneous.
Answer» Solution :It is FALSE. Exothermic process are not ALWARS SPONTANEOUS. If `Delta S = -ve` and `T Delta S gt Delta H`, the process will ve non spontaneous EVEN if it is exothermic.

Discussion

39.	Justify the followingstatement : (a)An exothermic reaction isalways thermodynamicallyspontaneous. (b) The entropyof a substance increaseson going from liquidto vapour state at any temperature.
Answer» Solution :(a) EXOTHERMIC REACTIONS are generally thermodynamically spontaneous becauseeven if it is accompaniedby decrease of randomness ( e.g.,in the condensation of a gas or solidification OFA liquid), the heatreleasedis absorbed by the surroundings so that the entropyof the surroundings increases to such an EXTENT that `DeltaS_("total")` is positive. (b) The molecules in the vapour state have greater freedom of MOVEMENT and hence greater randomness than those in the liquid state. Hence, entropy increases in goingfrom liquidto vapour state.

Discussion

40.	Justify the following reaction isredox reaction Fe_2O_3 (s) + 3CO (g) to 2Fe (s) + 3CO_2 (g)
Answer» Solution :`Fe_2O_3` is reduced to Fe(+3 to 0), C is o0XIdised to `CO_2(+2 to +4)`- HENCE a REDOX reaction

Discussion

41.	Justify the given reaction is redox reaction CuO (s) + H_2 (g) toCu (s) +H_2O (g)
Answer» Solution :CUO is reduced to CU(+2 to 0), `H_2` (o to +1)- hence a redox REACTION

Discussion

42.	Justify the following reaction is redox reaction 4NH_3(g) + 5O_2 (g) to 4NO (g) + 6H_2O (g)
Answer» Solution :`NH_3` is o0XIdised to NO (-3 to +2), `O_2` is reduced to `H_2O` (0 to -2) - redox REACTION

Discussion

43.	Justify the following reaction is redox reaction 4BCl_3 (g) + 3LiAlH_4 (s) to 2B_2H_6 (g) + 3LiCl (s) + 3AlCl_3 (s)
Answer» Solution :`BCl_3` is reduced to `B_2H_6` (+3 to -3), `LiAlH_4` is o0XIdised( o0XIdation NUMBER of H in `LiAlH_4` is -1 which INCREASES to +1 in`B_2H_6`)- Redox REACTION.

Discussion

44.	Justify the given reaction is redox reaction 2K (s) + F_2 (g) to 2FK (s)
Answer» SOLUTION :K is o0XIdised (0 to +1), `F-2` is REDUCED (0 to -1) - REDOX REACTION

Discussion

45.	Justify that the reaction : 2Na (s)+H_(2)(g)rarr2NaH(s) is a redo reaction
Answer» SOLUTION :Since NaH is an ionnic compound itmay be represented a `Na^(+)H^(-)`(s) thus `2Na(s)+H_(2)(g)rarr2Na^(+)H^(-)(s)` this reactoincan be SPLIT up into the FOLLOWNG two half reaction `Na(s)rarrNa^(+)+E^(-)]xx2` `H_(2)(g)+2e^(-)rarr2H^(-)` overall redox reaction`2Na(s)+H_(2)(g)rarr2Na^(+)H^(-)(s)` In the first half rection Na is oxiised to `Na^(+)` ion while in the second half reaction `H_(2)` is reduced to `H^(-)` ions therefore the overall reaction is a redox change

Discussion

46.	Justify the following reaction is a redox reaction. CuO_((s)) + H_(2(g)) to Cu_((s)) + H_(2)O_((g))
Answer» Solution :`OVERSET((+2)(-2))(CuO_((s))) + overset((o))(H_(2(g))) + overset((+1)(-2))(H_(2)O_(g))` n the above reaction, oxygen is removed from CuO. So CuO gets REDUCED. Oxygen is added to `H_(2)` to form WATER. So `H_(2)` gets oxidised, i.e. In CuO, OXIDATION number of CU +2 is reduced to 0 whereas in `H_(2)` , oxidation of `H_(2)` 0 is increased to +1. So the above reaction is a redox reaction

Discussion

47.	Justify that the reaction : 2Na_((s))+H_(2(g))to2NaH_((s)) is a redox change.
Answer» SOLUTION :Since in the above reaction the compound formed is an ionic compound, which may also be represented as `Na^(+)N_((s))^(-)`, this suggests that one half reaction in this process is : `2Na_((s))to2Na_((g))^(-)+2E^(-)` and the other half reaction is : `H_(2(g))+2e^(-)to2H_((g))^(-)` This splitting of the reaction under examination into two half reactions automatically reveals that here sodium is oxidised and hydrogen is REDUCED, therefore, the complete reaction is a redox CHANGE.

Discussion

48.	Justify that the reaction : 2Cu_(2)O_((s))+Cu_(2)S_((s))to6Cu_((s))+SO_(2(g)) is a redox reaction. Identify the species oxidised/reduced, which acts as an oxidant and which acts as a reductant.
Answer» Solution :LET us assign oxidation number to each of the species in the REACTION under examination. This RESULTS into : `overset(+1,-2)(2Cu_(2)O_((s)))+overset(+1,-2)(Cu_(2)S_((s)))tooverset(0)(6Cu_((s)))+overset(+4,-2)(SO_(2)(g))` We therefore, conclude that in this reaction copper is reduced from +1 state to zero oxidation state and sulphur is oxidised from -2 state to +4 state. The above reaction is thus a redox reaction. Further, `Cu_(2)O` helps sulphur in `Cu_(2)S` to INCREASE its oxidation number, therefore, Cu(I) is an oxidant, and sulphur of `Cu_(2)S` helps copper both in `Cu_(2)S` itself and `Cu_(2)O` to decrease its oxidation number, therefore, sulphur of `Cu_(2)S` is reductant.

Discussion

49.	Justify that the reaction : 2Cu_(2)O(s)+Cu_(2)(s)rarr6Cu(s)+SO_(2)(g) is a redox rection identify the species oxidised / reduced which acts as an oxidant and which acts as a reductant
Answer» Solution :Writing the OXIDATION number of each atom its symbol we have `2Cu_(2)O(s)Cu_(2)rarr6Cu(s)+SO_(2)(g)` here in this reaction the oxidatio number of copper decrease from +1 in `Cu_(2)S` to 0 in copper metal therefore `Cu^(+)` is reduced further the oxidation number of S increase from - 2 in `Cu_(2)S` to +4 in `SO_(2)` therefore `S^(2-)` is oxidised thus the above reaction is a redox reaction further OXYGEN in `Cu_(2)O` helps suphur in `Cu_(2)O` to increase its oxidation number from -2 to+4 therefore `Cu_(2)O` is an oxidant conversely suphur in `Cu_(2)` helps COPER both in `Cu_(2)O` and `Cu_(2)S` to decrease its oxidation number from +1 to ZERO in copper metal therefore `Cu_(2)S` is the reductant

Discussion

50.	Justify that the following reactions are redoxreactions : (a) CuO_((s))+H_(2(g))toCu_((s))+H_(2)O_((g)) (b) Fe_(2)O_(3(s))+3CO_((g))to2Fe_((s))+3CO_(2(g)) ( c) 4BCl_(3(g))+3LiAlH_(4(s))to2B_(2)H_(6(g))+3LiCl_((s))+3AlCl_(3(s)) (d) 2K_((s))+F_(2(g))to2K^(+)F_((s))^(-) (e) 4NH_(3(g))+5O_(2(g))to4NO_((g))+6H_(2)O_((g))
Answer» Solution :(a) `overset(+2)(Cu)overset(-2)(O_((s)))+overset(0)(H_(2(g)))tooverset(0)(Cu_((s)))+overset(+1)(H_(2))overset(-2)(O_((g)))` Here, O is being removed from CuO. Therefore Cu gets reduced. Where O is being added to `H_(2)`, and from `H_(2)O`. Therefore it gets oxidized. Therefore, oxidation number of Cu decreases from +2 to 0. therefore it gets reduced. Where oxidation number of H increases from 0 to +1. Therefore it gets oxidized. Therefore given reaction is a redox reaction. (B) `overset(+3)(Fe_(2))overset(-2)(O_(3(s)))+overset(+2)(3CO_((g)))tooverset(0)(2Fe_((s)))+overset(+4)(3CO_(2(g)))` : Oxidation number of FE gets reduced from +3 to 0. Whereas oxidation number of C increases from +2 to +4. O is being removed from `Fe_(2)O_(3)` and added into CO. `Fe_(2)O_(3)` gets reduced and CO gets oxidized. Therefore given reaction is a redox reaction. ( c) `overset(+3)(4B)overset(-1)(Cl_(3(g)))+overset(+1)(3Li)overset(+3)(Al)overset(-1)(H_(4(s)))tooverset(-3)(2B_(2))overset(+1)(H_(6(g)))+overset(+1)(3Li)overset(-1)(Cl_((s)))+overset(+3)(3Al)overset(-1)(Cl_(3(s)))` Here, oxidation number of B reduces from +3 to -3. therefore their is reduction. Same an oxidation number of hydrogen increases from-1 to +1. Therefore their is oxidation. H is added to `BCl_(3)` and hydrogen is removed from `LiAlH_(4)`. Therefore given reaction is redox reaction. (d) `overset(0)(2K_((s)))+overset(0)(F_(2(g)))to2K^(+)F_((s))^(-)` : Here, oxidation number of K increases from 0 to +1 and oxidation number of F reduces from 0 to -1. Therefore K gets reduced and F gets oxidized. Therefore given reaction is redox reaction. (e) `overset(-3)(4N)overset(+1)(H_(3(g)))+overset(0)(5O_(2(g)))tooverset(+2)(4N)overset(-2)(O_((g)))+overset(+1)(6H_(2))overset(-2)(O_((g)))` : Here, oxidation number of N is INCREASED from -3 to +2. Oxidation number of O is reduced from 0 to -2. Therefore their is reduction. Here, H get removed from `NH_(3)`, and again added therefore `NH_(3)` gets oxidized and `O_(2)`, gets reduced. Therefore given reaction is redox reaction.

Discussion

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