InterviewSolution
Saved Bookmarks
InterviewSolution
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.
| 2151. |
Define order of a reaction. Illustrate your answer with an example. |
|
Answer» Order of a reaction : The sum of the powers of the concentration terms of the reactants present in the rate equation is called order of a reaction. `to ` Order of a reaction can be `0,1,2,3,` and even a fraction `{:(Eg: 1 N_(2)O_(5) to N_(2) O_(4) +1/2 O_(2)), (rate prop [N_(2) O_(5)]),(therefore " It is a first order reaction."):}|{:(2 2N_(2) O to 2N_(2)+ O_(2)),(rate prop [N_(2) O]^(2)),(therefore It is 2^(nd) "order reaction"):}` |
|
| 2152. |
Mention a reaction for which the exponnts of concentration terms are not the same as their stoichiometric coefficients in the rate equation. |
|
Answer» The following are the reactions for which the exponents of concentration terms are not the same as their stoichiometric coefficients in the rate equation. `CHCl_(3)+ Cl_(2)to C Cl_(4) +HCl " "rate =k[CHCl_(3)][Cl _(2)]^(1//2)` `CH_(3) COOC_(2) H_(5)+ H_(2)O to CH_(3) COOH+ C_(2)H_(5)OH rate =k [CHCOOC_(2)H_(5) ][H_(2)O]^(1//2)` |
|
| 2153. |
Rate of corrosion is maximum when (1) An electrolyte is present in water (2) Metal has low S.R.P. (3) Metal has high standard oxidation potential (4) All of these |
|
Answer» Correct option (4) All of these Explanation: When metal has high standard oxidation potential, it has more tendency to undergo oxidation. In presence of electrolyte, rate of Corrosion is maximum. |
|
| 2154. |
Which is correct increasing order of deposition? (1) Na+ < Mg+2 < Zn+2 < Ag+ (2) Ag+ < Zn+2 < Mg+2 < Na+ (3) Mg+2 < Na+ < Zn+2 < Ag+ (4) Mg+2 < Zn+2 < Na+ < Ag+ |
|
Answer» Correct option (1) Na+ < Mg+2 < Zn+2 < Ag+ Explanation: Increasing order of deposition is related to the order of reduction and oxidation potential (in accordance with preferential discharge theory) Na+ < Mg+2 < Zn+2 < Ag+ |
|
| 2155. |
For galvanic cell, `Ag|AgCl(s),CsCl(0.1M)||CsBr(10^(-3)M),AgBr(s)|Ag.` Calculate the `EMF` generated and assign correct polarity to each electrode for spontaneous or exergonic process at `25^(@)C`. Given `:. K_(sp)` of `AgCl=3.0xx10^(-10),K_(sp)` of `AgBr=4.0xx10^(-13)`. |
|
Answer» `LHS` electrode `K_(sp)=[Ag^(o+)][Cl^(c-)]=[Ag^(o+)]_(a)(0.1)` `3.0 xx 10^(-10)=[Ag^(o+)]_(a)(0.1)` `:. [Ag^(o+)]_(a)=(3.0xx10^(-10))/(0.1M)=3.0xx10^(-9)M` `RHS` electrode`:` `K_(sp)=[Ag^(o+)]_(c)[Br^(c-)]` `4.0xx10^(-13)=[Ag^(o+)]_(c)(10^(-3)M)` It is a concentration cell, `:.E^(c-)._(cell)=0` Half cell reactions are Anode reaction `:` `Ag(s) hArr Ag^(O+)Ag^(o+)(aq)+e^(-)` Cathode reaction `: Ag^(o+)(` cathode `)+e^(-) hArr Ag(s)` Cell reaction `:` `ulbar(Ag^(o+)(cathode)hArrAg^(o+)(anode))` Thus, `E_(cell)=E^(c-)._(cell)-(0.059)/(1)log .([Ag^(o+)]_(a))/([Ag^(o+)]_(c))` `=0-(0.059)/(1)log.(3.0xx10^(-9))/(4xx10^(-10))(` Take`0.059~~~0.06)` `=-0.06[log 30-2 log 2]` `=-0.06[1.48-2xx03]` `=-0.0528V` `E_(cell)=-ve,` suggest that cell is non`-` spontaneous or endergonic `(DeltaG=+ve)`. For spontaneous process, the polarity of electrodes has to be reversed, `i.e.,` change anode to cathode and cathode to anode. |
|
| 2156. |
The ratio of volumes of `H_(2) and O_(2)` liberated on electrolysis of water is ____A. `1 : 2`B. `1 : 3`C. `2 : 1`D. `3 : 1` |
|
Answer» `H_(2)O rarr H^(+) + OH^(-)` Cathode : `4OH^(-) rarr 2H_(2)O + O_(2) + 4e^(-)` Anode : `4H^(+) + 4e^(-) rarr 2H_(2)` `:.` The ratio of volume of `H_(2) and O_(2) "is " 2 : 1` |
|
| 2157. |
STATEMENT-1: Molar conductivity indreases with decrease in concentration for weak electrolysis.STATEMENT-2: No. o fions increases and o. of ions per unit volume decreases due to dilution.A. If both the statements are TRUE and STATEMENTS-2 is the correct explantion of STATEMENTS-12B. If both the statements are TRUE but STATEMENTS-2 is NOT the correct explanation of STATEMENTS-14C. If STATEMENTS-1 is TRUE and STATEMENTS-2 is FALSED. If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE |
|
Answer» Correct Answer - A |
|
| 2158. |
During electrolysis of `H_(2)O` , the molar ratio of `H_(2)` and `O_(2)` formed isA. ` 2 : 1`B. `1 : 2`C. `1 : 3`D. `1 : 1` |
|
Answer» Correct Answer - A `H_(2)O to H_(2) + 1//2 O_(2)` `therefore` The ratio of `H_(2)` to `O_(2)` `1 : 0.5 therefore 2 : 1` |
|
| 2159. |
Some Indian scientists tried to use a metal x for electroplating iron pillar in Mehrauli but they ended up with Ecell of the reaction to be negative. They concluded that (1) Reaction is spontaneous (2) Reaction is non-spontaneous (3) Reaction is reversible (4) Reaction is non-reversible |
|
Answer» Correct option (2) Reaction is non-spontaneous Explanation: For electroplating Iron a metal ‘x’ is used. Ecell is negative, it means that no reaction takes place and the reaction is non-spontaneous. |
|
| 2160. |
Assertion (A) : Passage of 48,250C of electricity through cupric sulphate and ferrous sulphate solutions results in the deposition 0.5 moles each of iron and copper at the respective cathodes. Reason (R): Number of equivalents of a metal deposited at cathode is equal to the number of faradays of electricity passed through the electrolyte.A. Both A and R are true and R is the correct explanation of AB. Both A and R are true and R is not the correct explanation of AC. A is correct and R is wrongD. A is wrong and R is correct |
| Answer» 48,250 C is equal to 0.5 Faraday. Number of equal alents of metal deposited is equal to the number of faradays of electricity passed through the electrolyte and that means 0.5 equilvalents of copper and iron are deposited. Since they are bivalent ions, 0.5 equialents correspond to 0.25 moles | |
| 2161. |
1 coulomb of charge passes through solution of `AgNO_(3)` and `CuSO_(4)` connected in series and the concentration of two solution being in the ratio `1 : 2` . The ratio of amount of Ag and Cu deposited on Pt electrode isA. `107.9 : 63.54`B. `54 : 31.77`C. `107.9 : 31.77`D. `54 : 63 .54` |
|
Answer» Correct Answer - C The ratio of amount deposited during the same charge take place in the ratio of their equivalent weights and is independent of solution concentration . `therefore` 1g Eq. of Ag = 1 g Eq. of Cu `therefore 107.9 -= 31.77` |
|
| 2162. |
A cell is to be constructed to show a redox change `:` `Cr+2Cr^(3+)hArr3Cr^(2+)`. The number of cells with different `E^(c-)` an `n` but same value of `DeltaG^(c-)` can be made `(` Given `E^(c-)._(Cr^(3+)|Cr^(2+))=-0.40V,E^(c-)._(Cr^(3+)|Cr)=-0.74V,` adn `E^(c-)._(Cr^(2+)|Cr)=-0.91V)`A. `1`B. `2`C. `3`D. `4` |
|
Answer» Correct Answer - C For `I` : `{:(Cr|Cr^(3+)||Cr^(3+),Cr^(2+)|Pt),(CrrarrCr^(3+)+3e,E^(@)=+0.74V),(3Cr^(3+)+3erarr3Cr^(2+),E^(@)=-0.40V):}/(Cr+2Cr^(3+)rarr3Cr^(2+)(n=3))` `E^(@) = 0.74 - 0.40 = 0.34 V` `DeltaG^(@) = 3 xx 0.34 xx F = 1.02 F` For (II) `{:(Cr|Cr^(2+)||Cr^(3+)|Cr),(3Crrarr3Cr^(3+)+6e,E^(@)=0.91V),(2Cr^(3+)+6erarr2Cr,E^(@)=-0.74V):}/(Cr+2Cr^(3+)rarr3Cr^(2+)(n=6))` For `E^(@) = 0.91 - 0.74 = 0.17 V` `DeltaG^(@) = 0.17 xx 6 xx F = 1.02 F` For (III) `{:(Cr|Cr^(2+)||Cr^(3+)|Cr^(2+)|Pt),(CrrarrCr^(2+)+2e,E^(@)=0.91V),(2Cr^(3+)+2erarr2Cr^(2+),E^(@)=-0.40V):}/(Cr+2Cr^(3+)rarr3Cr^(2+)(n=2))` `E^(@) = 0.91 - 0.40 = 0.51 V` `:. DeltaG^(@) = 2 xx 0.51 xx F = 1.02F` |
|
| 2163. |
The molecular conductivity of strong electrolyteA. increases lineraly with concentrationB. increases linearly with concentration in a linear fashionC. decreases lineraly with concentrationD. decrases with square root of concentration in a linear fashion. |
|
Answer» Correct Answer - D `^^_(m)^(@)= ^^_(m)^(@)-Asqrt(C)` |
|
| 2164. |
Salts of `A` (atomic weight `7`), `B` (atomic weight `27`) and `C` (atomic weight `48`) were electolysed under idential condition using the same quanity of electricity. It was found that when `2.1 g` of `A` was deposited, the weights of `B` and `C` deposited were `2.7` and `7.2 g`. The valencies `A, B` and `C`respectively:A. `3, 1` and `2`B. `1, 3` and `2`C. `3, 1` and `3`D. `2, 3` and `2` |
|
Answer» Correct Answer - B Eq. of `A =` Eq. of `B =` Eq. of `C` `(2.1)/(7//n_(1)) = (2.7)/(27//N_(2)) = (7.2)/(48//n_(3))` `0.3n_(1) = 0.1n_(2) = 0.15n_(3)` `:. N_(1) = (n_(2))/(3) = (n_(3))/(2) = (n_(1), n_(2), n_(3))` are intergers. |
|
| 2165. |
A current of 5 amp is passed through a solution of NaCl for 3.25 hrs . The weight of NaOH formed isA. 6.50 gB. 17.25 gC. 24.25 gD. `13.0` g |
|
Answer» Correct Answer - C 1gM.mass NaCl`-=`1 g M. mass of NaOH 58.5 NaCl `-=` 40 g NaOH `:.1 F= 96500` Celectricity passed `-= 1g` eq. mass NaOH formed `-=40`g of NaOH formed `:.Q=I xxt=5xx3.25xx60xx 60C=Wt. of NaOH` `:. 58500 C =Wt.of NaOH` Wt.of `NaOH=(ixxtxxR)/F=(58500xx 40)/96500=24.25 g` |
|
| 2166. |
Same quantity of electricity was passed through solutions of salts of elements A,B and C with atomic weights 7, 27 and 48 respectively. The masses of A,B and C deposited were 2.1 g,27.7g and 7.2 respectively. The valencies of A,B and C respectively areA. 3,2 and 1B. 1,2 and 3C. 1,3 and 2D. 2,3 and 2 |
|
Answer» Correct Answer - C Suppose valencies of A,B and C are a,b, and c respectively. Then their equivalent weights will be (At. Wt/valance) 7/a,27/b and 48/c. when same quantity of electricity is passed, masses deposited are in the ratio of their equivalent weights, i.e., `("Mass of A")/("Eq. wt. of A")=("Mass of B")/("Eq. wt of B")=("Mass of C")/("Eq. wt of C")` `therefore(2.1)/(7//a)=(2.7)/(27//b)=(7.2)/(48//c)` or 0.3 a=0.1, b=0.15c, or 3a=b=1.5c. thus, if b=1, then `a=(1)/(3)`, `c=(1)/(1.5)=(2)/(3)` `therefore`Ratio `a:b:c=(1)/(3):1:(2)/(3)=1:3:2` |
|
| 2167. |
The electrochemical equivalent of silver is `0.001114` g . When an electric current of 0.5 ampere is passed through an aqueous silver nitrate solution for 200 seconds , the amount of silver deposited isA. 1.118 gB. `0.1118 g`C. `5.590` gD. `0.5590 g` |
|
Answer» Correct Answer - B `W = Z xx C xx t = 0.00118 xx 0.5 xx 200 = 0.1118` g |
|
| 2168. |
Which one of the following ions has highest limiting molar conductivity?A. `Na^(+)`B. `Mg^(2+)`C. `K^(+)`D. `Ca^(2+)` |
|
Answer» Correct Answer - D Ion `Na^(+)K^(+)Mg^(2+)Ca^(2+)` 50-70 73.50 106.0 119.0 `^^_(m)^(@)` `(S cm^(2)"mol"^(-1))` The size of hydrated `Na^(+)` ion is largest and that of `Ca^(2+)` is the smallest. Hence they follow the above order. |
|
| 2169. |
What current is to be passed for `0.25` sec for deposition of certain weight of metal which is equal to its electrochemical equivalent ?A. 4AB. 100AC. 200AD. 2A |
|
Answer» Correct Answer - A Electrochemical equivalent is the weight deposited by 1 coulomb `Q=1xxt "or" 1=1xx0.25` or 1=4A. |
|
| 2170. |
Salts of `A` (atomic weight `7`), `B` (atomic weight `27`) and `C` (atomic weight `48`) were electolysed under idential condition using the same quanity of electricity. It was found that when `2.1 g` of `A` was deposited, the weights of `B` and `C` deposited were `2.7` and `7.2 g`. The valencies `A, B` and `C`respectively:A. 1 , 2 and 3B. 1 , 3 and 2C. 3 , 1 and 2D. 2 , 3 and 1 |
|
Answer» Correct Answer - B When small quantity of current is passed through solution of different salts , the amount of substanced deposited are in the ratio of their equivalent masses . Valency = `("At. Mass")/("Eq. mass") = ("At.mass")/("Mass of the substance")` `therefore` valency of A `:` valency of B `:` valency of C `= (7)/(2.1) : (27)/(2.7) : (64)/(9.6)` = `(1)/(0.3) : (1)/(0.1) : (2)/(0.3)` `= (1)/(0.3) : (3)/(0.1) : (2)/(0.3)` |
|
| 2171. |
What current is to be passed for `0.25` sec for deposition of certain weight of metal which is equal to its electrochemical equivalent ?A. `4 A`B. `100 A`C. `200 A`D. `2 A` |
|
Answer» Correct Answer - A `W=Zit` `w/Z=It` `1=It=Ixx0.25` `I=4` amp. |
|
| 2172. |
`EMF` of the cell `|Ag|AgNO_(3)(0.1M)||K Br(1 N),AgBr(s)|Ag is -0.6V` at `298K` `AgNO_(3) ` is `80%` and `KBr` is `60%` dissociated. Calculate `a.` Solubility and `b. K_(sp)` of `AgBr` at `298 K`. |
|
Answer» The above cell is a concentration cell. `:. E^(c-)._(cell)=0.0V` The half cell reactions are Anode reaction `: Ag(anode) hArrAg^(o+)(0.1M)+e^(-)` and Cathode reaction `:` `Ag^(o+)(cathode) +e^(-)hArrAgs(cathode)` Cell reaction `ulbar( :Ag^(o+)(cathode)hArrAg^(o+)(0.1 M) _(anode))` Since `AgNO_(3)` is `80%` ionized. `:. [Ag^(o+)]_(a)=0.1xx(80)/(100)=0.08M` `:.E_(cell)=E^(c-)._(cell)-(0.06)/(1)log.([Ag^(o+)]_(a))/([Ag^(o+)]_(c))(` Take `0.059~~0.06)` `-0.6 V=0-0.06log.(0.08)/([Ag^(o+)]_(c))` `log[Ag^(o+)]_(c)=log0.08-10=(log_(2)^(3)-log 100)-10` `=0.9-2-10=-11.1` `:. [Ag^(o+)]_(c)=Antil o g(-11.1)=Antil og(bar(12).9)=7.9xx10^(-12)` `~~8xx10^(-12)` Since `K Br` is `60%` dissociated, `:.[Br^(c-)]=1xx(60)/(100)=0.6M` `:. K_(sp)=[Ag^(o+)][Br^(c-)]` `=8 xx 10^(-12)xx0.6xx10^(-12)M^(2)` Solubility `(S) = sqrt(K_(sp))=sqrt(4.8)xx10^(-6)M^(2)` `=2.19xx10^(-6)M` |
|
| 2173. |
If the atomic mass of M is x , the electrochemical equivalent of M in the solution of `M_(2)(SO_(4))_(3)` will beA. `(3x)/(F)`B. `(x)/(3F)`C. `(2x)/(F)`D. `(x)/(F)` |
|
Answer» Correct Answer - B Valency of M in `M_(2) (SO_(4))_(3) ` is 3 Eq. mass of M = `("Atomic mass")/("valency") = (x)/(3)` `therefore` Electrochemical equivalent = `("Equivalent mass")/(96500) = (E)/(F) = (x)/(3F)` |
|
| 2174. |
Given `:` `Fe^(2+)(aq)+2e^(-) rarr Fe(s), " "E^(c-)=-0.44V` `Al^(3)+3e^(-) rarr Al(s)," "E^(c-)=-1.66V` `Br^(2+)+2e^(-)rarr 2Br^(c-)(aq)" "E^(c-)=-1.08V` The decreasing order of reducing power is `………………………………..` . |
|
Answer» Correct Answer - `(Al>Fe>Br^(c-))` `(AlgtFegtBr^(c-))` More the oxidation potential, more the reducing power. `:. E^(c-)._(o x i d (Al|Al^(3+))(+1.66C))gtE^(c-)._( o x i d(Fe|Fe^(2+)))(0.044V)gt` `E^(c-)._(o x i d(2Br^(c-)|Br_(2)))(-1.08V)` |
|
| 2175. |
What is wrongly stated about electrochemical seriesA. It is the representation of element in order of increasing or decreasing standard electrode reduction potentialB. It does not compare the relative reactivity of metalsC. It compares relative strengths of oxidising agentsD. `H_(2)` is centrally placed element |
|
Answer» Correct Answer - B Electrochemical series compare the relative reactivity of metals. |
|
| 2176. |
The e.f.m. of the cell `Ag|Ag^(+)(0.1M)||Ag^(+)(1M)|Ag` at 298K isA. 0.0059VB. 0.059VC. 5.9VD. 0.59V |
|
Answer» Correct Answer - B `Ag|Ag^(+)(.1m)||Ag^(+)1M|Ag|` `E_(cell)=(2.303RT)/(nF)"log"(c_(1))/(c_(2))=(0.059)/(1)"log"(1)/(0.1)` `=0.059log10=0.059`volt. |
|
| 2177. |
The electrochemical equivalent for zinc `(` atomic weight `=6.4)` is `……………………………` . |
|
Answer» Correct Answer - `(3.4xx10^(-4))` `(3.4xx10^(-4)` Electrochemical `Eq=(Ew)/(1F)=(65.4//2)/(96500C)` `[Ew of Zn=(Aw)/(2)]` `=3.4xx10^(-4)` |
|
| 2178. |
The `EMF` of the cell `:` `Pt, H_(2)(1atm)|H^(o+)(aq)||AgCl|Ag` is `0.27` and `0.26V` at `25^(@)C`, respectively. The heat of the reaction occuring inside the cell at `25^(@)C` is `………………….. kJ K^(-1)` |
|
Answer» Correct Answer - `(-54.8kJ K^(-1))` `(-54.9k J K^(-1))` `((delE)/(delT))_(P)=((0.26-0.27)/(35-25))=(-0.01)/(10)V K^(-1)[n_(cell)=1]` `DeltaH=-nF[E_(cell)(at 25^(@)C)-T((delE)/(delT))_(P)]` `DeltaH=-1xx96500[0.27-298(-(0.01)/(10))]` `DeltaH=-1xx96500Cxx0.568V K^(-1)=-54812 J K^(-1)` `=-54.8 k J K^(-1)` |
|
| 2179. |
Electrode potentials `(E_("red")^(@))` of 4 element A,B, C,D are -1.36,-0.32,0,-1.26V respectively. The decreasing reactivity order of these elements isA. A,D,B and CB. C,B,D and AC. B,D,C and AD. C,A,D and B |
|
Answer» Correct Answer - B Lower is the reduction potential more easily a metal is oxidised and more is its reactivity. |
|
| 2180. |
When `E_(Ag^+//Ag)^(o)=0.8` volt and `E_(Zn^(2+)//Zn)^(o)=-0.76` volt, which of the following is correctA. `Ag^(+)` can be reduced by `H_(2)`B. Ag can oxidise `H_(2)` into `H^(+)`C. `Zn^(2+)` can be reduced by `H_(2)`D. Ag and reduce `Zn^(2+)` ion |
|
Answer» Correct Answer - A Because `H_(2)` has greater reduction potential so it reduced at `Ag^(+)`. |
|
| 2181. |
Electrode potentials of five elements A,B,C,D and E are respectively -1.36,, -0.32,0,-1.26 and -0.42. The reactivity order of these elements are in the order ofA. A,D,E,B and CB. C,B,E,D and AC. B,D,E,A and CD. C,A,E,D and B |
|
Answer» Correct Answer - A Greater the oxidation potential, greater is the reactivity. |
|
| 2182. |
Amongst the following electrodes the one with zero electrode potential isA. Calomel electrodeB. Standard hydrogen electrodeC. Glass electrodeD. Gas electrode |
|
Answer» Correct Answer - B Standard hydrogen electrode have zero electrode potential. |
|
| 2183. |
In which cell the free energy of a chemical reaction is directly convertd into electricityA. Leclanche cellB. Concentration cellC. Fuel cellD. Lead storage battery |
|
Answer» Correct Answer - C Fuel cell converts the chemical energy into electrical energy. |
|
| 2184. |
Prove that for two half reactions having potentials `E_(1)` and `E_(2)` which are combined to yield a third half reaction, having a potential `E_(3)`, `E_(3)=(n_(1)E_(1)+n_(2)E_(2))/n_(3)` |
|
Answer» `DeltaG_(3)=DeltaG_(1)+DeltaG_(2)` `-n_(3)FE_(3)=-n_(1)FE_(1)-n_(2)FE_(2)` or `n_(3)E_(3)=n_(1)E_(1)+n_(2)E_(2)` or `E_(3)=(n_(1)E_(1)+n_(2)E_(2))/n_(3)` |
|
| 2185. |
Given `E^(c-)._(Fe^(2+)|Fe)` and `E^(c-)._(Fe^(3+)|Fe^(2+))` are `-0.44` and `0.77V` respectively. If `Fe^(2+),Fe^(3+)` and `Fe` blocks are kept together, then `[Fe^(3+)]` `……………………` and `[Fe^(2+)]…………………….` . |
|
Answer» Correct Answer - (Decreases, increases) ( Decrease, increases ) As `E^(c-)._(Fe^(3+)|Fe^(2+))gtE^(c-)._(Fe^(2+)|Fe)` `:. Fe^(3+)` will get reduces to `Fe^(2+)` and `Fe` will get oxidized to `Fe^(2+)` . Hence `[Fe^(3+)]` decreases and `[Fe^(2+)]` increases. |
|
| 2186. |
The dissociation of a wask electrolyte obeys the law of mass action. It was found byA. OstwaldB. ArrheniusC. BerzeliusD. None of these |
|
Answer» Correct Answer - a Factual statements |
|
| 2187. |
Given `E^(c-)._(Fe^(2+)|Fe)` and `E^(c-)._(Fe^(3+)|Fe^(2+))` are `-0.44` and `0.77V` respectively. If `Fe^(2+),Fe^(3+)` and `Fe` blocks are kept together, then `[Fe^(3+)]` `……………………` and `[Fe^(2+)]…………………….` .A. `Fe^(3+)` increasesB. `Fe^(3+)` decreasesC. `Fe^(2+), Fe^(3+)` remain unchangedD. `Fe^(2+)` decreases |
|
Answer» Correct Answer - b Cell reaction `: 2Fe^(3+)+Fe rarr 3Fe^(2+)` `E^(c-)._(cell)=0.77-(-0.44)=1.21V` `implies Fe^(3+)` and `Fe` will reduce. |
|
| 2188. |
A half cell is prepared by `K_(2)Cr_(2)O_(7)` in a buffer solution of `pH =1`. Concentration of `K_(2)Cr_(2)O_(7)` is `1M`. To 3 litre of this solution `570 gm` of `SnCI_(2)` is added which is oxidised completely to `SnCI_(4)`. Given: `E_(Cr_(2)O_(7)^(-2)//Cr^(+3).H^(+))^(@) = 1.33V, (2.303)/(F) RT = 0.06`, Atomic of mass `Sn = 119, E_(Sn^(+4)//Sn^(+2))^(@) = 0.15` Emf of the cell `Pt|underset((0.1M))(Sn^(+2)),underset((0.2M))(Sn^(+4)),underset((1M))(H^(+)):||:underset((0.2M))(Cr_(2)O_(7)^(2-)),underset((1M))(Cr^(+3)),underset((1M))(H^(+)):|:Pt`A. `-1.18V`B. `1.164V`C. `1.18V`D. None of these |
|
Answer» Correct Answer - B `Cr_(2)O_(7)^(2-) +3Sn^(+2) +14H^(+) rarr 3Sn^(+4) +2Cr^(+3) +7H_(2)O` `E = (1.33 -0.15) -(0.06)/(6) log. ((0.2)^(3)(1)^(2))/((0.2)^(1)(0.1)^(3)(1)^(4))` `= 1.164 V`. |
|
| 2189. |
Calculate the solubility product of `Co_(2)[Fe(CN)_(6)]` in water at `25^(@)C`. Given, conductivity of saturated solutions of `Co_(2)[Fe(CN)_(6)]` is `2.06 xx 10^(-6) Omega^(-1)cm^(-1)` and that of water used is `4.1 xx 10^(-7) Omega^(-1)cm^(-1)`. The ionic molar conductivities of `Co^(2+)` and `[Fe(CN)_(6)]^(4)` are `86.0 Omega cm^(2)mol^(-1)` and `444.0 Omega^(-1) cm^(2)mol^(-1)`, respectivly.A. `7.87xx10^(-7)`B. `7.87xx10^(-6)`C. `7.87xx10^(-8)`D. `7.87xx10^(-9)` |
|
Answer» Correct Answer - a `k_(Co_(2)[Fe(CN)_(6)])=2.06xx10^(-6)=0.41xx10^(-6)=1.65xx10^(-6)` `wedge_(mCo_(2)[Fe(CN)_(6)])=2wedge_(m)Co^(2+)+wedge_(m)[Fe(CN)_(6)]^(4-)` `implieswedge_(m)=1000(k)/(c)impliesc=2.7xx10^(-6)M` `implies K_(spCo_(2)[Fe(CN)_(6)])=4c^(3)=7.87xx10^(-17)M^(3)` |
|
| 2190. |
A half cell is prepared by `K_(2)Cr_(2)O_(7)` in a buffer solution of `pH =1`. Concentration of `K_(2)Cr_(2)O_(7)` is `1M`. To 3 litre of this solution `570 gm` of `SnCI_(2)` is added which is oxidised completely to `SnCI_(4)`. Given: `E_(Cr_(2)O_(7)^(-2)//Cr^(+3).H^(+))^(@) = 1.33V, (2.303)/(F) RT = 0.06`, Atomic of mass `Sn = 119, E_(Sn^(+4)//Sn^(+2))^(@) = 0.15` Number of moles of `Cr^(+3)` formed areA. 2B. 6C. 4D. 3 |
|
Answer» Correct Answer - A Cathode: `Cr_(2)O_(7)^(2-) +6e^(-) +14H^(+) rarr 2Cr^(+3) +7H_(2)O` Anode: `[Sn^(2+)rarr Sn^(+4) +2e^(-)] xx 3` Net cell reaction: `Cr_(2)O_(7)^(2-) +3Sn^(+2) +14H^(+) rarr 3Sn^(+4) +2Cr^(+3) +7H_(2)O` `{:("Initial conc.",1M,1M,10^(-1)M,-,-,-),(At.,(2)/(3)M,-,10^(-1),1M,(2)/(3)M,-):}` completion Moles of `Cr^(+3) =(2)/(3) xx 3 = 2` |
|
| 2191. |
Calculate the solubiltit and solubility product of `Co_(2) [Fe(CN)_(6)]` is water at `25^(@)C` from the following data: conductivity of a saturated solution of `Co_(2)[Fe(CN)_(6)]` is `2.06 xx 10^(-6) Omega^(-1) cm^(-1)` and that of water used `4.1 xx 10^(-7) Omega^(-1) cm^(-1)`. The ionic molar conductivites of `Co^(2+)` and `Fe(CN)_(6)^(4-)` are `86.0 Omega^(-1)cm^(-1) mol^(-1)` and `44.0 Omega^(-1) cm^(-1) mol^(-1)`. |
|
Answer» Correct Answer - `K_(sp) = 7.682 xx 10^(-7)` `lambda_(M) = (k xx 1000)/(M)` `(86 xx 2+444) = ((2.06 xx 10^(-6) -4.1xx10^(-7))xx1000)/(s)` `s = 2.678 xx 10^(-6)` `K_(sp) = 4s^(3) = 7.687 xx 10^(-17)` |
|
| 2192. |
A sample of water from a large swimming pool has a resistance of `10000Omega` at `25^(@)C` when placed in a certain conductace cell. When filled with `0.02M KCI` solution, the cell has a resistance of `100 Omega` at `25^(@)C, 585 gm` of `NaCI` were dissolved in the pool, which was throughly stirred. A sample of this solution gave a resistance of `8000 Omega`. Given: Molar conductance of `NaCI` at that concentration is `125 Omega^(-1) cm^(2) mol^(-1)` and molar conductivity of `KCI` at `0.02 M` is `200W^(-1) cm^(2) mol^(-1)`. Conductivity `(Scm^(-1))` of `H_(2)O` is:A. `4 xx 10^(-2)`B. `4 xx 10^(-3)`C. `4 xx 10^(-5)`D. None of these |
|
Answer» Correct Answer - C `[NaCI] = (585//58.5)/(V(l))` `k_(NaCIsolution) = (1)/(8000) xx 0.4 = 5 xx 10^(-5)` `^^_(m) =(K(NaCI)xx1000)/((585//58.5)/(V)) =125` `K(NaCI) xx V(l) xx 100 = 125, V(l) = (125)/(100 xx k(NaCI))` `k(H_(2)O) = (1)/(10000) xx 0.4 = 4 xx 10^(-4) = 4 xx 10^(-5)` |
|
| 2193. |
A sample of water from a large swimming pool has a resistance of `10000Omega` at `25^(@)C` when placed in a certain conductace cell. When filled with `0.02M KCI` solution, the cell has a resistance of `100 Omega` at `25^(@)C, 585 gm` of `NaCI` were dissolved in the pool, which was throughly stirred. A sample of this solution gave a resistance of `8000 Omega`. Given: Molar conductance of `NaCI` at that concentration is `125 Omega^(-1) cm^(2) mol^(-1)` and molar conductivity of `KCI` at `0.02 M` is `200W^(-1) cm^(2) mol^(-1)`. Volume (in Litres) of water in the pool is:A. `1.25 xx 10^(-5)`B. `1250`C. `12500`D. None of these |
|
Answer» Correct Answer - A `k(NaCI) = 10^(-5)` `V = (125)/(100 xx 10^(-5)) = (125)/(10^(-3)) = 1.25 xx 10^(5)L` |
|
| 2194. |
A sample of water from a large swimming pool has a resistance of `10000Omega` at `25^(@)C` when placed in a certain conductace cell. When filled with `0.02M KCI` solution, the cell has a resistance of `100 Omega` at `25^(@)C, 585 gm` of `NaCI` were dissolved in the pool, which was throughly stirred. A sample of this solution gave a resistance of `8000 Omega`. Given: Molar conductance of `NaCI` at that concentration is `125 Omega^(-1) cm^(2) mol^(-1)` and molar conductivity of `KCI` at `0.02 M` is `200W^(-1) cm^(2) mol^(-1)`. Cell constant (in `cm^(-1))` of conductane cell is:A. 4B. 0.4C. `4 xx 10^(-2)`D. `4 xx 10^(-5)` |
|
Answer» Correct Answer - B `k = G xx (l)/(A)` `^^_(m) =(k xx 1000)/(C) = (200 xx 0.02)/(1000)` `=(1)/(100) xx (l)/(A) = 0.4 cm^(-10` |
|
| 2195. |
A sample of water from a large swimming pool has a resistance of `9200 Omega` at `25^(@)C` when placed in a certain conductance cell. When filled with `0.02M KCI` solution the cell has a resistance of `85 Omega` at `25^(@)C. 500g` of `NaCI` were dissolved in the pool, which was throughly stirred. A sample of this solution gave a resistance of `7600 Omega`. calculate the volume of water in the pool. Given: Molar conductance of `NaCI` at that concentration is `126.5 Omega^(-1) cm^(2) mol^(-1)` and molar conductivity of `KCI` at `0.02M` is `138 Omega^(-1) cm^(2) mol^(-1)`. |
|
Answer» Correct Answer - `2 xx 10^(5) dm^(3)` For `KCI` `lambda_(M) = (k xx 1000)/(M) rArr 138 (k xx 1000)/(0.02)` `k = 2.76 xx 10^(-3) = (1)/(R) ((l)/(a)) = (1)/(85) ((l)/(a))` `(l//a) = 0.2346` For `H_(2)O: k_(H_(2)O) = (1)/(9200) (l//a)` For NaCI: `k_(NaCI) = (1)/(7600) (l//a)` `lambda_(M) = ((k_(NaCI-k_(H_(2)O)))xx1000)/(M)` `126.5=(((1)/(7600)-(1)/(9200))xx0.2346xx1000)/(M)` ` = 4.2438 xx 10^(-5)` `M = (500)/(58.5xxV) =4.2438 xx 10^(-5)` `V = 201400 L` `V = 2.014 xx 10^(-5)L` |
|
| 2196. |
Expression representing the cell potential (E cell)A. `E_("cathode")+E_("anode")`B. `E_("anode")-E_("cathode")`C. `E_("cathode")-E_("anode")`D. `E_("left")-E_("right")` |
|
Answer» Correct Answer - C `E_(cell)=E_("cathode")-E_("anode")` |
|
| 2197. |
The substance having conductivity at room temperature among the following isA. 0.1N HClB. 0.1 N NaClC. GraphiteD. Glass |
|
Answer» Correct Answer - C Graphite has highest conductivity among the given species because of free delocalised electrons. |
|
| 2198. |
The unit of equivalent conductivity `(Lambda_(eq.))` areA. ohm cmB. `ohm^(-1)cm^(+2) (0.66cm^(-1)("g equivalent")^(-1)`C. `ohm cm^(2)`(g equivalent)D. `S cm^(-2)` |
|
Answer» Correct Answer - B See Comprehensive Review. |
|
| 2199. |
`lambda_(CICH_(2)COONa)=224ohm^(-1)cm^(2)` gm `eq^(-1)` `lambda_(NaCl)=38.2 ohm^(-1)cm^(2)` gm `eq^(-1) lambda_(HCl)`=203 `ohm^(-1)cm^(2)gm eq^(-1)`, what is the value of `lambda_(CICH_(2)COOH)`?A. `288.5ohm^(-1)cm^(2)gm eq^(-1)`B. `289.5ohm^(-1)ohm^(-1)cm^(2)gmeq^(-1)`C. `388.8ohm^(-1)cm^(2)gmeq^(-1)`D. `59.5ohm^(-1)cm^(2)gmeq^(-1)` |
|
Answer» Correct Answer - C `lambda_(ClCH_(2)COOH)=lambda_(ClCH_(2)COONa)+ lambda_(HCl)-lambda_(NaCl)` `=(224+203-38.2)ohm^(-1)cm^(2)(g eq^(-1))` =`388.8ohm^(-1)cm^(2)(g eq^(-1))` |
|
| 2200. |
Assertion : Mercury cell does not give steady potential. Reason : In the cell reaction, ions are not involved in solution.A. Both assertion and reaction are true and the reason is correct explanation for assertionB. Both assertion and reason are true and reason is not correct explanation for assertionC. ) Assertion is true but the reason is False.D. Both assertion and reason are false. |
|
Answer» Correct Answer - D Assertion is fales. Mercury cell actuallt gives a sterady potential (1.35 V) throughout its life. |
|