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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.
| 101. |
Assuming that hydrogen begaves as an ideal gas, calculate EMF of the cell at `25^(@)C` when `P(1)=640mm and P_(2)=425mm` `Pt|H_(2)(P_(1))|HCl|H_(2)(P_(2))|Pt`A. `0.005V`B. `0.004V`C. `0.003V`D. `0.002V` |
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Answer» Correct Answer - A As the cell is a concentration cell, `E_(cell)^(@)-0` Now, the cell reaction is Left electrode: `H_(2)(P_(1)=640mm)hArr2H^(+)+2e^(-)` Right electrode: `:2H^(+)+2e^(-)hArrH_(2)(P_(2)=425mm)` Net cell reaction: `H_(2)(P_(1)mm)hArrH_(2)(P_(2)mm)` Now, from Nernst equation `=0-(0.059)/(2)log.(425)/(640)=0.005V` |
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| 102. |
The emf of the cell : `H_(2)(g)|"Buffer||Normal"` canlomel elctrolde is `0.6885V` at `25^(@)C`, when barometric pressure is `=760 mm`. What is the pH of the buffer solution? `E_("calomel")=0.28V`A. `9.92`B. `7.92`C. `8.92`D. `6.92` |
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Answer» Correct Answer - D As the right electrode is normal (standard) electrode, the emf of cell may be written as `E_(cell)=E_("calomel")-E_(H^(+)("buffer")//H_(2))E_(cell)=E_("calomel")-E_(H^(+)("buffer")//H_(2))` or, `0.6885=0.28-E_(H^(+)//H_(2))` `thereforeE_(H^(+)//H_(2))=-0.4085V` Now, from Nernst equation, `E_(H^(+)//H_(2))=E_(H^(+)//H_(2))^(0)-(0.059)/(n)log.(P_(H_(2)))/([H^(+)]^(2))` `-0.4085=0-(0.059)/(2)` `or,(logP_(H_(2))-log[H^(+)]^(2))` or, `0.4085=(0.059)/(2)(log1+2P^(H))` `therefore`pH of buffer solution = `6.92` |
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| 103. |
The passage of current through a solution of certain electroylte results in the formation of hydrogen at anode the soltuion isA. Aqueous HClB. Fused `CaH_(2)`C. sulphuric acid in waterD. Aqueous `K_(2)SO_(4)` |
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Answer» Correct Answer - B In a CaH2, Hydrogen has the Oxidation state of -1. Hence hydrogen gas is liberated at anode. |
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| 104. |
Resistance of a conductvity cell filled with a solution of an electrolyte of concentration 0.1 M is 100 `Omega`. The conductivity of this solution is 1.29 `Sm^(-1)`. Resistance of the same cell when filled with 0.02M of the same solution is `520 Omega`. the molar conductivity of 0.02M solution of the electrolyte will be:A. `1.24xx10^(4)"S m"^(2)mol^(-1)`B. `12.4xx10^(4)"S m"^(2)mol^(-1)`C. `124xx10^(4)"S m"^(2)mol^(-1)`D. `1240xx10^(4)"S m"^(2)mol^(-1)` |
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Answer» Correct Answer - A `K=(l)/(R.a),"cell constt."((l)/(a))=1.29xx100=126` Again conductivity of `0.02` M solution `x=(1)/(520)xx129` `^^_(m)=(x xx1000)/(M)=(129)/(520)xx(1000)/(0.02)=1.24xx10^(4)Sm^(2)mol^(-1)` |
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| 105. |
Equivalent conductance of 1 M propanoic acid is 10 `ohm^(-1)" cm^(2)eq^(-1)`. pH of the propanpoic acid solution isA. 7B. `3.3`C. `1.3`D. `6.8` |
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Answer» Correct Answer - C `alpha=(^^_(eq)^(c))/(^^_(eq)^(0))=(10)/(200)=0.05` `[H^(+)]=C alpha =1xx0.05=0.05` `pH=-log[H^(+)]=-log(5xx10^(-2))` `=2 log 10 - log 5 =`2-0.6990=1.3` |
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| 106. |
Molar conductane of KCl increases slowly with decrease in concentration because ofA. increase in degree of ionisationB. increase in total number of current carrying speciesC. weakning of interionic attractions and increase in ionic mobilitiesD. increase in hydration of ions. |
| Answer» Correct Answer - C | |
| 107. |
An aqeous solution of X is added slowly to an aqueous solution of Y as shown in List-I. The variation in conductivity of these reaction in List-II. Match List-I with List-II and select the correct answer using the code given below the lists: A. `{:(P,Q,R,S),(3,4,2,1):}`B. `{:(P,Q,R,S),(4,3,2,1):}`C. `{:(P,Q,R,S),(2,3,4,1):}`D. `{:(P,Q,R,S),(1,4,3,2):}` |
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Answer» Correct Answer - A `(P) underset(x)(C_(2)H_(5))_(3) N + underset(Y)(CH_(3)COOH)rarr (C_(2)H_(5))_(3) NHCH_(3) COO^(-)` Initially conductivity increases due to ion formation after that it becomes practially constant because X alone can not form ions. Hence (3) is the correct match. `(Q) underset(x)(Ki(0.1M))+underset(Y)(AgNO_(3)(0.01M))rarr AgI darr +KNO_(3)` Number of ions in the solution remains constant until all the `AgNO_(3)` precipitated as `AgI`. therefore conductance increases due to increase in number of ions. Hence (4) is correct match. (R) Initially conductance decreases due to the decrease in the number of `bar(O)H` ions tthereafter it slowly increases due to the increases in number of `H^(+)` ions. Hence (2) is the correct match. (S) Initially it decreases due to decrease in `H^(+)` ions and then increases due to the increases in `bar(O)H` ions. Hence (1) is the correct match. |
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| 108. |
The conductivity of 0.1 n NaOH solution is 0.022 S`cm^(-1)`.When equal volume of 0.1 N HCl solution is added, the conductivity of resultant solution is decreased to 0.0055 S `cm^(-1)` . The equivalent conductivity in `Scm^(2)equi^(-1)` of NaCl solution is :A. `0.0055`B. `0.11`C. 110D. 55 |
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Answer» Correct Answer - C Normality of resulatant solution of `=(0.1xxV)/(VxxV)=0.05N` `Lambda_(eq)=1xx(0.0055)/(0.05)="110 S cm"^(2)eq^(-1)` |
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| 109. |
The conductivity of `"0.1N NaOH"` solution is `0.022 S cm^(-1)`. To this solution equal volume of `" 0.1 N HCl` solution is added which results into decrease of conductivity of solution to `0.0055 S cm^(-1)`. The equivalent conductivity of `NaCl` solution in `S cm^(2)" equiv"^(-1)` is `:`A. `0.011`B. `110`C. `0.0055`D. `55.0` |
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Answer» Correct Answer - 2 Normality of resulting solution `=(0.1V)/(2V)=0.05N` `wedge_(eq)=(Kxx1000)/(N)=(0.0055xx1000)/(0.05)=110` |
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| 110. |
The standard oxidation potential for `Mn^(3+)` ion acid solution are `Mn^(2+)overset(-1.5V)rarrMn^(3+) overset(-1.0V)rarr MnO_(2)`. Is the reaction `2Mn^(3+)+2H_(2)O rarr Mn^(2+)+MnO_(2)+4H^(+)` spontaneous under conditions of unit activity ? What is the change in free energy ?A. spontaneous ,`-48250J`B. nonspontaneous, `+48250J`C. no change in free energyD. spontaneous , `-96500J` |
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Answer» Correct Answer - 1 `DeltaG=DeltaG^(@)+2.303RTlogQ` `Q=1("activity"=1)` `DeltaG=DeltaG^(@)` `Mn^(+3)+e^(-)rarrMn^(+2)" "DeltaG_(1)^(@)=-1xxFxx1.5` `Mn^(+3)+2H_(2)OrarrMnO_(2)+4H^(+)+e^(-)" "DeltaG_(2)^(@)=-1xxF(-1)` `bar(2Mn^(3+)+2H_(2)Orarr Mn^(2+)+MnO_(2))+4H^(+)" "DeltaG^(@)=DeltaG_(2)^(@)=-48250J(DeltaGlt0)` |
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| 111. |
Given that `E_(Fe^(3+)|Fe)^(0)` and `E_(Fe^(2+)|Fe)^(0)` are `-36V` and `-0.439V`, respectively . The value of `E_(Fe^(3+),Fe^(2+)|Pt)^(0)` would beA. `(-0.36-0.439)V`B. `(-0.36+0.439)V`C. `[3(-0.36)+2(-0.439)]V`D. `[3(-0.36)-2(-0.439)]V` |
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Answer» Correct Answer - 4 `nFE_(Fe^(3+)//Fe)^(@)=nEF_(Fe^(3+)//Fe^(2+))^(@)+nEF_(Fe^(2+)//Fe)^(@)` `3xx(-0.36)=1xxE_(Fe^(3+)//Fe^(2+))^(@)+2xx(-0.439)` `E_(Fe^(3+)//Fe^(2+))^(@)=3(-3.6)-2xx(-0.439)` |
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| 112. |
`E^(@)` in the given diagram isA. `0.51 V`B. `0.61 V`C. `0.71 V`D. `0.81 V` |
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Answer» Correct Answer - B `ClO_(3)^(-)+2H_(2)O+4e^(-)rarr ClO^(-)+4OH^(-) Delta G_(1)^(@)` `ClO^(-)+H_(2)O+e^(-)rarr(1)/(2)Cl_(2)+2OH^(-) Delta G_(2)^(@)` `(1)/(2)Cl_(2)+e^(-)rarr Cl^(-) Delta G_(3)^(@)` `ClO_(3)^(-)+3H_(2)O+6e^(-)rarr Cl+6OH^(-)` `Delta G^(@)=Delta G_(1)^(@)+Delta G_(2)^(@)+Delta G_(3)^(@) , Delta G = -nF E^(@)` `-6FE^(@)=-4Fxx0.54-1Fxx0.45-1Fxx1.07` `E^(@)=+(3.68)/(6)=0.61` |
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| 113. |
The equilibrium constant (K) for the reaction `Cu(s)+2Ag^(+) (aq) rarr Cu^(2+) (aq)+2Ag(s)`, will be [Given, `E_(cell)^(@)=0.46 V`]A. `3.9xx10^(15)`B. `3.9xx10^(10)`C. `1.9xx10^(15)`D. `1.9xx10^(9)` |
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Answer» Correct Answer - A `E_(cell)=0rArrE_(cell)^(@)=(0.059)/(2)logK` `rArr logK=15.6rArrK=10^(15.6)=3.9xx10^(15)` |
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| 114. |
The standard oxidation potential of `Ni//Ni^(2+)` electrode is `0.236 V`. If this is combined with a hydrogen electrode in acid solution, at what `pH` of this solution will be measured e.m.f. be zero at `25^(@)C`? Assume that `[Ni^(2+)] = 1 M` and `P_(H_(2)) = 1 atm`.A. 4B. 1C. 2D. 3 |
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Answer» Correct Answer - A `E^(@)Ni^(+2)//Ni=-0.236V` `E_(cell)^(@)=+0.236` `0=0.236-(0.06)/(2)log.([Ni^(+2)])/([H^(+)]^(2))` |
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| 115. |
The standard electrode potential for the following reaction is `+1.33V`. What is the potential at `pH =2.0 ?` `Cr_(2)O_(7)^(2-)(aq,1M)+14H^(+)(aq)+6e^(-) rarr 2Cr^(3+)(aq. 1M)+7H_(2)O(l)`A. `+1.820V`B. `+1.990V`C. `+1.608V`D. `+1.0542V` |
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Answer» Correct Answer - 4 `E_(Cr_(2)O_(7)^(2-)|Cr^(3+))=E_(Cr_(2)O_(7)^(2-)|Cr^(3+))^(@)-(0.0591)/(6)xx(log.([Cr^(3+)]^(2))/([Cr_(2)O_(7)^(2-)])xx(1)/([H^(+)]^(14)))` `E_(Cr_(2)O_(7)^(2+)|Cr^(3+))=1.33-(0.0591)/(6)log.(1)/((0.01)^(14))" "rArr" "1.0542V` |
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| 116. |
The oxidation potential of a hydrogen electrode is related to the pH of the solution by the equation at `25^(@)` CA. `-0.059 xx pH`B. `0.059 xx pH`C. `(0.059)/(pH)`D. `0.059+pH` |
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Answer» Correct Answer - B O.P of hydrogen electrode `=0.059 xx p^(H)` |
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| 117. |
The standard oxidation potential of `Ni//Ni^(2+)` electrode is `0.236V`. If this is combined with a hydrogen electrode in acid solution. At what pH of the solution will the measured emf be zero at `25^(@)C` ? Assume `[Ni^(2+)]=1M` and `p_(H_(2))=1` atm |
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Answer» Correct Answer - 4 `{:(Ni rarr Ni^(2+) + 2e ,E_(OP)^(@) = 0.236V),(2H^(+) +2e rarr H_(2),E_(RP)^(@) = 0):}` `:. E_(cell)^(@) = 0` `:. E_(cell)^(@) + (0.059)/(2) log_(10).([H^(+2)]^(2))/([Ni^(2+)])` or `-log H^(+) = 4 :. pH = 4` |
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| 118. |
Calculate the potential of hydrogen electrode in contact with a solution whose `pH=10`. |
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Answer» If pH of solutions is 10 then its `[H^(+)]` ion concentration will be `10^(-10)M`. Let us consider a reduction half cell `H^(+)(10^(-10)M)|H_(2)(1 "atm")|Pt` Electrode process : `2H^(+)(10^(-10)M)+2e^(-)hArrH_(2)(1"atm")(n=2)` `Q=(PH_(2))/([H^(+)]^(2))=(1)/([10^(-10)])=10^(20)` According to to Nernst equation `E_(H^(+)//H_(2))=E_(H^(+)//H_(2))^(0)-(0.059)/(n)"log Q"` `=0-(0.059)/(2)"log10"^(20)=-0.59V` |
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| 119. |
Calculate the value of `Lambda_(m) ^prop` for `SrCl_(2)` in water at `25^(@)C` from the following data `:` `{:(Conc. (mol//l t),,0.25,,1),(Lambda_(m)(Omega^(-1)cm^(2)mol^(-1)),,260,,250):}`A. 270B. 260C. 250D. 255 |
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Answer» Correct Answer - 1 `Lambda_(m)=Lambda_(m)=-bsqrt(c)` `260=Lambda_(n)^(oo)-0.5b" "...(1)` `250=Lambda_(m)^(oo)-b" "...(2)` On solving `(1) & (2)`, we get `Lambda_(m)^(oo)=270Omega^(-1)cm^(2)mol^(-1)` |
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| 120. |
What will occur if a block of copper metal is dropped into a beaker containing a solution of `1M ZnSO_(4)`?A. The copper metal will bdissolve and zinc metal will be depositedB. The copper metal will dissolve with evolation of oxygen gas.C. The copper metal will dissolve with evolution of hydrogen gasD. No reaction will occur |
| Answer» Correct Answer - D | |
| 121. |
The Kohlrausch law is related toA. Conductance of ions at infinite dilutionB. Independent migration of ionsC. Both 1 and 2D. Neither 1 and 2 |
| Answer» Correct Answer - C | |
| 122. |
`Ag|AgCl|Cl^(-)(C_(2))||Cl^(-)(C_(1))||AgCl|Ag` for this cell `DeltaG` is negative if `:`A. `C_(1)=C_(2)`B. `C_(1) gt C_(2)`C. `C_(2) gt C_(1)`D. Both `(1)` and `(3)` |
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Answer» Correct Answer - 2 Given cell of electrolyte concentration cell `E_(cell)=(RT)/(F)ln.([Ag^(+)]_(RHS))/([Ag^(+)]_(LHS))` where `[Ag^(+)]=(K_(sp))/([Cl^(-)])` . For spontaneous chemical change conc. Of `Ag^(+)` of cathodic compartment must be greater than anodic compartment. |
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| 123. |
A Daniel cell constructed in the laboratory. The voltage observed was `0.9V` instead of `1.10V` of the standard cell. A possible explanation isA. Molar ratio of `Zn^(+2),Cu^(2+)`is `2:1`B. The Zn electrode has thrice the surface of Cu electrodeC. `[Zn^(2+)]lt[Cu^(2+)]`D. `[Zn^(2+)]gt[Cu^(2+)]` |
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Answer» Correct Answer - D `E_(cell)=E_(cell)^(@)-(0.059)/(n)log.([Zn^(2+)])/([Cu^(2+)])` `E_(cell) " decrease indicates" [Zn^(2+)]gt[Cu^(2+)]` |
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| 124. |
Electrolysis of selt solutions is due to to the formation ofA. ElectronB. IonsC. OxidesD. Acids |
| Answer» Correct Answer - B | |
| 125. |
The reactions taking place at anode and cathode of a cell respectively areA. Reduction, oxidationB. Oxidation, reductionC. Hydrolysis, oxidationD. Reduction hydrolysis |
| Answer» Correct Answer - B | |
| 126. |
Chemical passivity is possible withA. Conc `HNO_(3)`B. AirC. Both 1 and 2D. Metal oxides |
| Answer» Correct Answer - C | |
| 127. |
Which of the following is correct regarding mechanical passivityA. Visible metal oxide film is formedB. Visible metal oxide film prevents dissociation of metalC. Fe, Mn and Pb exhibit mechanical passivityD. All the above |
| Answer» Correct Answer - D | |
| 128. |
Match the Column-I with Column-II NAR_CHM_V02_XII_C01_E01_084_Q01.png" width="80%">A. (a-r)(b-q)(c-q,r)(d-p,s)B. (a-q)(b-r)(c-q,s)(d-p,r)C. (a-r)(b-q)(c-q,s)(d-p,r)D. (a-r)(b-q)(c-p,s)(d-s,r) |
| Answer» Correct Answer - B | |
| 129. |
Dilute nitric acid on electrolysis using platinum electrodes yieldsA. both oxygen & hydrogen at cathodeB. both oxygen & hydrogen at anodeC. `H_(2)` at cathode and `O_(2)` at anodeD. |
| Answer» Correct Answer - C | |
| 130. |
The standard emf of the cell. `Cd(s) | CdCl_(2) (aq) rightarrow 2Ag(s) + Cd^(2+) (aq) + 2Cl^(-)(aq)` is 0.6915 V at `0^(@)` and 0.6753 V at `25^(@)C`. The `DeltaH^(@)` of the reaction at `25^(@)C` is:A. `167.26 kJ//mol`B. `-167.26 kJ//mol`C. `40 K//mol`D. `-40 K cal//mol` |
| Answer» Correct Answer - D | |
| 131. |
In which of the following `(E_("cell")-E_("cell")^(@))=0`A. `Zn|Zn^(2+)(0.01 M)||Ag^(+)(0.1M)|Ag(s)`B. `Pt(H_(2)(1 atm)|pH=1||Zn^(2+)(0.01 M)|Zn`C. `Pt|H_(2)(1 atm) pH = 1||Zn^(2+)(1 M)|Zn`D. `Pt|H_(2)(1 atm)H^(+)(0.01 M)||Zn^(2+)(0.01 M)|Zn` |
| Answer» Correct Answer - A | |
| 132. |
For an electrode reaction written as `M^(n+)+n e^(-)rarr M` , `E_("red")=E_("red")^(@)-(RT)/(nf)ln(1)/([M^(n+)])` `=E_("red")^(@)-(0.059)/(n)"llog"(1)/([M^(n+)])` at 298 k For the cell reaction `aA+bB rarr xX+yY` `rArr E_("cell")=E_("cell")^(@)-(RT)/(nf)"log"([X]^(x)[Y]^(y))/([A]^(a)[B]^(b))` For pure solids, liquids or gases at lampt. molarconc = 1 Std. free energy change `Delta G^(@)=-nfE^(@)` where f - for faraday = 96,500 c, n = no. of `e^(-)` `Delta G=-2.303 RT log K_(c )` where `K_(c )` is equilibrium constant `K_(c )` can be calculated by using `E^(@)` of cell. The emf of the cell `Zn//zn^(2+)(0.01M)//Fe^(2+)(0.001M)//Fe` at 298 K is `0.2905`. The value of equilibrium constant for cell reaction siA. `e^((0.32)/(0.0295))`B. `1-^((0.32)/(0.0295))`C. `10^(0.26//0.0295)`D. `10^((0.32)/(0.059))` |
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Answer» Correct Answer - B `Zn+Fe^(2+)rarr Fe+Zn^(2+)` `E=E^(@)-(0.059)/(n)log Ke` Given E = 0.2905 i.e., `0.2905 = E^(@)-(0.059)/(2)"log"(0.01)/(0.001)` `E^(@)=0.2905+0.0295 log 10=0.32` `E^(@)=(0.059)/(n) log K_(eq) =0.32 =(0.059)/(2)log K_(eq)` `K_(eq)=10^(0.32//0.0295)` |
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| 133. |
A half cell is prepared by `K_(2)Cr_(2)O_(7)` in a buffer solutoin of pH = 1. Concentration of `K_(2)Cr_(2)O_(7)` is 1 M. To 3 litre of this solution 570 gm of `SnCl_(2)` is added which is oxidised completely to `SnCl_(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 |
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Answer» Correct Answer - A `40.K_(2)Cr_(2)O_(7)+3SnCl_(2)+H^(+)rarr SnCl_(4)+2Cr+3` mole of `SnCl_(2)=3 " " therefore` mole of `Cr^(+3)=2` |
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| 134. |
For an electrode reaction written as `M^(n+)+n e^(-)rarr M` , `E_("red")=E_("red")^(@)-(RT)/(nf)ln(1)/([M^(n+)])` `=E_("red")^(@)-(0.059)/(n)"llog"(1)/([M^(n+)])` at 298 k For the cell reaction `aA+bB rarr xX+yY` `rArr E_("cell")=E_("cell")^(@)-(RT)/(nf)"log"([X]^(x)[Y]^(y))/([A]^(a)[B]^(b))` For pure solids, liquids or gases at lampt. molarconc = 1 Std. free energy change `Delta G^(@)=-nfE^(@)` where f - for faraday = 96,500 c, n = no. of `e^(-)` `Delta G=-2.303 RT log K_(c )` where `K_(c )` is equilibrium constant `K_(c )` can be calculated by using `E^(@)` of cell. On the basis if information avaliable from the reaction `(4)/(3)Al+O_(2)rarr(2)/(3)Al_(2)O_(3) Delta G=-827 kJ//mol^(-1)` The minimum emf required to carry out an electrolysis of `Al_(2)O_(3)` isA. `2.14 v`B. `4.28 v`C. `6.42 v`D. `8.56 v` |
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Answer» Correct Answer - A For 1 mole of `O_(2)` `O_(2)rarr20^(-) i.e., 2//3xx30^(-)` or 4/3 mole of Al to change into `Al^(3+)` this n = 4 Thus `Delta G=-nFE` `E=-(Delta G)/(nF)=(-827000)/(4xx96500)=2.14v` |
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| 135. |
For an electrode reaction written as `M^(n+)+n e^(-)rarr M` , `E_("red")=E_("red")^(@)-(RT)/(nf)ln(1)/([M^(n+)])` `=E_("red")^(@)-(0.059)/(n)"llog"(1)/([M^(n+)])` at 298 k For the cell reaction `aA+bB rarr xX+yY` `rArr E_("cell")=E_("cell")^(@)-(RT)/(nf)"log"([X]^(x)[Y]^(y))/([A]^(a)[B]^(b))` For pure solids, liquids or gases at lampt. molarconc = 1 Std. free energy change `Delta G^(@)=-nfE^(@)` where f - for faraday = 96,500 c, n = no. of `e^(-)` `Delta G=-2.303 RT log K_(c )` where `K_(c )` is equilibrium constant `K_(c )` can be calculated by using `E^(@)` of cell. `E^(@)` for the cell, `Zn//Zn_((aq))^(2+)//Cu_((aq))^(2+)//Cu` is `1.10 v` at `25^(@)c`. The equilibrium constant for the cell reactoin. `Zn+Cu_((aq))^(2+) hArr Cu+Zn_((aq))^(2+)` is the order ofA. `10^(-37)`B. `10^(37)`C. `10^(-17)`D. `10^(17)` |
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Answer» Correct Answer - B (b) `E_("cell")^(@)=(0.059)/(2)log K : log K =(1.10xx2)/(0.059)` `K=1.9xx10^(37)` |
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| 136. |
Consider the cell potentials `E_(Mg^(2+)|Mg)^(0)=-2.37V` and `E_(Fe^(3+)|Fe)^(0)=-0.04V` The best reducing agent would beA. `Mg^(2+)`B. `Fe^(3+)` decreaseC. `Mg`D. `Fe` |
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Answer» Correct Answer - 3 `Fe^(3+) & Mg^(2+)` are oxidising agent of `Mg` has more negative value of SRP. `E_(anode Mg//Mg^(2+))^(@)=2.37V` `E_(anode Fe//Fe^(3+))^(@)=0.04V` `i.e.` best reducing agent is Mg. |
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| 137. |
The electrochemical cell shown below is a concentration cell. `M|M^(2+)(` saturated solution of sparingly soluble salt, `MX_(2))||M^(2+)(0.001 mol dm^(-3))|M` The `emf` of the cell depends on the difference in the concentration of `M^(2+)` ions at the two electrodes. The `emf` of the cell at `298` is `0.059V`. The solubility product `(K_(sp),mol^(3) dm^(-9))` of `MX_(2)` at 298 based on the information available the given concentration cell is `(` Take `2.303xxRxx298//F=0.059V)`A. `1xx10^(-15)`B. `4xx10^(-15)`C. `1xx10^(-12)`D. `4xx10^(-12)` |
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Answer» Correct Answer - B `M|underset((K_(sp) =?))(M^(+)(sat.))"||"M^(2+) (0.001M)` emf of concentration cell, `E_(cell) = (-0.059)/(n)log. ([M^(+2)]_(a))/([M^(+2)]_(c))` `0.059 = (0.059)/(2)log. ([0.001])/([M^(+2)]_(a))` `[M^(+2)]_(a) = 10^(5) = S` (solubility of salt in saturated solution) `underset((S))(MX_(2))rarr underset((S))(M^(+2)) +underset((2S))(2x^(-)(aq))` `K_(sp) = 4S^(3) = 4 xx (10^(-5))^(3) = 4 xx 10^(-15)` |
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| 138. |
The electrochemical cell shown below is a concentration cell `M//M^(2+)` (saturated solution of a sparingly soluble salt, `MX_(2))||M^(2+)(0.001 mol dm^(-3))|M` The emf of the cell depends on the difference in concentrations of `Mn^(2+)` ions at the two electrodes. The emf of the cell at `298 K` is `0.059 V`. The value of `DeltaG ( kJ "mol"^(-1))` for the given cell is : (take `1 F = 96500 C "mol"^(-1)`)A. `-5.7`B. `5.7`C. `11.4`D. `-11.4` |
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Answer» Correct Answer - D At anode: `M(s) +2X_((aq))^(-) rarr MX_(2(aq)) +2e^(-)` At cathode: `M_((aq))^(+2) +2e^(-) rarr M(S)` n-factor for the cell reaction is 2. `DeltaG =- nFE_(cell) =- 2 xx 96500 xx 0.059` `=- 113873 J//"mole"` `=- 11.387 KJ//"mole" =- 11.4 KJ//"mole"` |
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| 139. |
The electrochemical cell shown below is a concentration cell. `M|M^(2+)(` saturated solution of sparingly soluble salt, `MX_(2))||M^(2+)(0.001 mol dm^(-3))|M` The `emf` of the cell depends on the difference in the concentration of `M^(2+)` ions at the two electrodes. The `emf` of the cell at `298` is `0.059V`. The solubility product `(K_(sp),mol^(3) dm^(-9))` of `MX_(2)` at 298 based on the information available the given concentration cell is `(` Take `2.303xxRxx298//F=0.059V)`A. `1xx10^(-15)`B. `4xx10^(-15)`C. `1xx10^(-15)`D. `4xx10^(12)` |
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Answer» Correct Answer - B For concentration cell, `E_(cell)=(0.0591)/(n)log.(C_(2(RHS)))/(C_(1(LHS)))` `E_(cell)=0.059V,C_(2(RHS))=0.001` `0.059=(0.0591)/(2)log.(0.001)/(C_(1))` `or (2xx0.059)/(0.0591)=log.(0.001)/(C_(1))" or anti log2"=(0.001)/(C_(1))` `thereforeC_(1)=(0.001)/(100)=10^(-5)` `C_(1)"=concentration or solubility of M"^(2+)=10^(-5)` `MX_(2)hArrM_(s)^(2+)+2underset(2S)X^(-)` `K_(sp)=S(2S)^(2)=4S^(3)` `K_(sp)=4xx(10^(-5))^(3)=4xx10^(-15)mol^(3)dm^(-9)` |
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| 140. |
Given the cell reactions `MX_((s))+e^(-)rarrM_((s))+X_((aq))^(-),E^(@)=0.207V` and `M_((aq))^(+)+e^(-)rarrM_((s)),E^(@)=0.799V` The solubility of `MX_((s))` at 298K isA. `1.0xx10L^(-1)`B. `1.0x10^(-9)"mole L"^(-1)`C. `1.0xx10^(-4)"mole L"^(-1)`D. `1.0xx10^(-5)"mole L"^(-1)` |
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Answer» Correct Answer - D `MX_((S))rarrM_((aq))^(+)+X_((aq))^(-)` `E_(cell)^(@)=0.207-0.7999=-0.592V` `log k_(sp)=-10thereforeS=10^(-5)` |
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| 141. |
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 2B. 1,3 and 2C. 3,1 and 3D. 2,3 and 2 |
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Answer» Correct Answer - 2 Weight `prop` equivalent weight `[(w_(1))/(w_(2))=(E_(1))/(E_(2))]` |
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| 142. |
Following techinique is adopted to check suspected drunk drivers. Read the passage and answer the questions at the end of it. When suspected drunk drivers are tested with a Breathalyzer, the alcohol (ethanol) in the exhaled breath is oxidized to ethanoic acid with an acidic solution of potassium dichromate : `3CH_(3)underset("Ethanol")(CH_(2)OH)(aq)+2Cr_(2)O_(7)^(2-)+16H^(+)(aq)rarr underset("Ethanoic acid")(3CH_(3)COOH)(aq)+4Cr^(3+)(aq)+11H_(2)O(l)` The Breathalyzer measures the colour change and produces a meter reading calibrated in terms of blood alcohol content. Colour of the testing solution changes from :A. `Cr_(2)O_(7)^(2-)` is red changing to `Cr^(3+)` (green)B. `Cr_(2)O_(7)^(2-)` is orange changing to `Cr^(3+)` (blue)C. `Cr_(2)O_(7)^(2-)` is orange changing to `Cr^(3+)` (green)D. `Cr_(2)O_(7)^(2-)` is blue changing to `Cr^(3+)` (yellow) |
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Answer» Correct Answer - C `Cr_(2)O_(7)^(2-)` is orange changing to `Cr^(3+)` (green) |
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| 143. |
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. `1.176 V`C. `1.18 V`D. None of these |
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Answer» Correct Answer - B `Cr_(2)O_(6)^(-2)+3Sn^(+2)+14H^(+)rarr 3sn^(+4)+2Cr^(+3)+7H_(2)O` `E^(@)=1.18` `E=1.18-(0.059)/(6)"log"((0.2)^(3)(1)^(2))/((0.2)(0.1)^(3)(1)^(14))` |
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| 144. |
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` Half cell potential `E_(Cr_(2)O_(7)^(-2)//Cr^(+3))^(@)` after teh reaction of `SnCI_(2)` is:A. `1.187`B. `1.191`C. `1.285`D. None of these |
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Answer» Correct Answer - B `Cr_(2)O_(7)^(-2)+14H^(+)+6e^(-)underset("F mole 2")underset("1 mole 3")(rarr.)2Cr^(+3)+7H_(2)O` `E=1.33-(0.059)/(6)"log"(2^(2))/(2(10^(-1))^(14))` |
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| 145. |
For the half cell At `pH=2`. Electrod potential is `:`A. `1.36V`B. `1.30 V`C. `1.42 V`D. `1.20 V` |
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Answer» Correct Answer - 3 `E=E^(@)-(0.059)/(2)"log of "[H^(+)]^(2)` |
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| 146. |
Using the standard electrode potential values given below, decide which of the statements, `I,II,III` and `IV` are correct. Choose the right answer from `(1) (2)` and `(4)` `Fe^(2+)+2e^(-) hArr " "," "E^(0)=-0.44V` `Cu^(2+)+2e^(-) hArr Cu" "," "E^(0)=+0.34V` `Ag^(+)+e^(-)hArr Fe" "," "E^(0)=+0.80V` `I. ` Copper can displace iron from `FeSO_(4)` solution. `II.` Iron can displace copper from `CuSO_(4)` solution `III.` Silver can displace copper from `CuSO_(4)` solution `IV.` Iron can displace silver from `AgNO_(3)` solution.A. I and IIB. II and IIIC. II and IVD. I and IV |
| Answer» Correct Answer - 3 | |
| 147. |
`0.169` gram of copper is deposited on the cathode by a current of `32` milliamperes passing for 5 hours through a solution of copper sulphate. Determing the current efficiency. [ Atomic weight of `Cu=63.6]`A. `98%`B. `78%`C. `69%`D. `89%` |
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Answer» Correct Answer - 4 Current efficiency`=(0.169)/([(3600xx32xx10^(_3)xx5xx(63.5)/(2))/(96500)])xx100%=89%` |
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| 148. |
A current of 1.70 ampere is passed through 300mL of 0.160M solution of `ZnSO_(4)` for 230 sec with a current efficiency of 90%. Find the molarity of `Zn^(2+)` after the deposition of Zn. Assume the volume of the solution deposition remains constant during elecyrolysis |
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Answer» We know, `i=(1.70xx90)/(100)` ampere `therefore` Eq. of `Zn^(2+)` lost `=(i.t.)/(96500)` `=(1.70xx90xx230)/(100xx96500)=3.646xx10^(-3)` `therefore` Meq.of `Zn^(2+)"lost"=3.646` Initial Meq. `Zn^(2+)=300xx0.160xx2` `[because Mxx2=N" for " Zn^(2+), Meq. = NxxV_(("in.mL."))]` `=48xx2=96` `therefore` Meq. of `Zn^(2+)` left in solution `=96-3.646=92x354` `therefore [ZnSO_(4)]=(92.354)/(2xx300)=0.154M` |
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| 149. |
The e.m.f of the cell `Ni//Ni^(+2)(1M)"//"Cl^(-)(1M)Cl_(2),Pt` `({:(E^(@)Ni^(2+)//Ni=-0.25eV":"),(E^(@)1/2Cl_(2)//Cl^(-)=+1.36eV):})`A. `-1.11V`B. `1.11V`C. `-1.61V`D. `1.61V` |
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Answer» Correct Answer - D `E=E_(C(SRP))-E_(A(SRP))` |
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| 150. |
An aqueous solution of NaCl on electrolysis gives `H_(2(g))Cl_(2(g))` and NaOH according to reaction : `2Cl_((aq.))^(-)+2H_(2)Orarr 2OH^(-)+H_(2(g))+Cl_(g)` A direct current of 25 ampere with a current efficiency of 62% is passes though 20 litre of NaCl solution (20% by weight). Write drown the reactions taking place at teh electrodes. (b) How long will it take to produce 1kg of `Cl_(2)` ? (c ) What will be the molarity od solution with respect to `OH^(-)` ? Assume no loss in volume due to evaporation. |
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Answer» We know, (a) Anode : `2Cl^(-)rarr Cl_(2)+2e` Cathod : `2e+2H_(2)O rarr 2OH^(-)+H_(2)` (b) `w=(E.i.t)/(96500)` `because W_(Cl_(2))=10^(3)g, E_(Cl_(2))=35.5` `10^(3)=(35.5xx25xx62xxt)/(100xx96500)` `therefore` Current efficiency = 62% `t = 175374.83` sec `therefore i = (25xx62)/(100)` ampere `therefore t=48.71` hr (c ) Eq. of `OH^(-)` formed = Eq. of `Cl_(2)` formed `=(10^(3))/(35.5)=28.17` `therefore` Mole `OH^(-)` formed `=28.17` (`because` monovalent) `therefore [OH^(-)]=("mole")/("valume in litre")=(28.17)/(20)=1.408"mol litre"^(-1)` |
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