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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. |
`Cu^(2+)+2e^(-)toCu,E^(@)=+0.34V,Ag^(+)+1e^(-)toAg,E^(@)=+0.80V` (i) Construct a galvanic cell using the above data. (ii) For what concentration of `Ag^(+)` ions will the emf of the cell be zero at `25^(@)C`, if the concentration of `Cu^(2+)` is 0.01 M? (log 3.919=0.593). |
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Answer» Correct Answer - `5xx10^(-9)M` (i) For `E_(cell)^(@)` to be +ve, oxidation will occur at copper electrode and reduction at silver electrode. Hence, the cell will be represented as : `Cu|Cu^(2+)||Ag^(+)|Ag` (ii) `Cu+2Ag^(+)toCu^(2+)+2Ag,n=2` `E_(cell)=E_(cell)^(@)-(0.0591)/(2)"log"([Cu^(2+)])/([Ag^(+)]^(2))`. hence, `0.46V-(0.0591)/(2)"log"(0.1)/([Ag^(+)]^(2))`. calculate `[Ag^(+)]` |
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| 102. |
Calculate the reduction potential for the following half cell reaction at 298 K. `Ag^(+)(aq)+e^(-)toAg(s)` `"Given that" [Ag^(+)]=0.1 M and E^(@)=+0.80 V`A. 0.741 VB. 0.80 VC. `-0.80`VD. `-0.741`V |
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Answer» Correct Answer - A `Ag^(+) + e^(-) to Ag` `E_"cell"=E_"cell"^@-0.0591/1 "log" 1/([Ag^+])` `E_"cell"=0.80-0.0591/1 "log" 1/0.1` =0.80-0.591=0.741 V |
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| 103. |
The standard reduction potentials for two reactions are given below: `AgCl(s)+e^(-)toAg(s)+Cl^(-)(aq),E^(@)=0.22V` ltBrgt `Ag^(+)(aq)+e^(-)toAg(s),E^(@)=0.80V` Calculate the solubility product of AgCl under standard conditions of temperature (298K). |
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Answer» Subtracting second eqn. from the first eqn., we get `AgCl(s)hArrAg^(+)(aq)+Cl^(-)(aq),E^(@)=-0.58V` Applying nernst equation, `E=E^(@)-(0.0591)/(1)"log"([Ag^(+)][Cl^(-)])/([AgCl(s)])` Putting `AgCl(s)=1` and at equilibrium `E=0` `E^(@)=0.0591log[Ag^(+)][Cl^(-)]=0.0591logK_(sp)` `-0.58=0.0591" log "K_(sp)" or log "K_(sp)=-9.8139=overline(10).1861` or `K_(sp)="Antilog "overline(10).1861=1.535xx10^(-10)`. |
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| 104. |
The standard reduction potential for two reactions are given below `AgCl(s)+e^(-)rarrAg(s)+Cl^(-)(aq) ,E^(@)=0.22V` `Ag^(+)(aq)+e^(-)rarrAg(s),E^(@)=0.80V` The solubility product of AgCl under standard conditions of temperature is given byA. `1.6xx10^(-5)`B. `1.5xx10^(-8)`C. `3.2xx10^(-10)`D. ` 1.5xx10^(-10)` |
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Answer» Correct Answer - D For the equilibrium `AgCl(s)overset(H_(2)O)hArrAg^(+)(aq)+Cl^(-)(aq)` `k_(sp)=[Ag^(+)][Cl^(-)]` `AgCl(s)+e^(-)rarrAg(s)+Cl^(-)(aq)E^(@)=0.22V` `Ag(s)rarrAg^(+)(aq)+e^(-),E^(@)=-0.80V` `AgCl(s)rarrAg^(+)(aq)+Cl^(-)(aq)` `E^(@)=(0.22-0.80)V` `=-0.58V` |
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| 105. |
The standard rectuion potentials for two half-cell reactions are given below `Cd^(2+)(aq)+2e^(-)rarrCd(s),E^(@)=-0.40V` `Ag^(+)(aq)+e^(-)rarrAg(s),E^(@)=0.80V` The standard free energy change for the reaction `2Ag^(+)(aq)+Cd(s)rarr2Ag(s)+Cd^(2+)(aq)` is given byA. 115.8kJB. `-115.8kJ`C. `-231.6kJ`D. `231.6kJ` |
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Answer» Correct Answer - C ` 2Ag^(+)(aq)+2e^(-)rarr2Ag(s) :E^(@)= 0.80V` `Cd(s)rarrCd^(2+)(aq)+2e^(-):E^(@)=0.400` `2Ag^(+)(aq)+Cd(s)rarrCd^(2+)(aq)+2Ag(s)` `E^(@)=1.20V` `DeltaG=-nE^(@)F` `=-2xx1 .20xx96500` `=-231.6kJ` |
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| 106. |
The standard reduction potential for two reactions are given below `AgCl(s)+e^(-)rarrAg(s)+Cl^(-)(aq) ,E^(@)=0.22V` `Ag^(+)(aq)+e^(-)rarrAg(s),E^(@)=0.80V` The solubility product of AgCl under standard conditions of temperature is given byA. `1.613 xx 10^(-5) M^(2)`B. `1.535 xx 10^(-8) M^(2)`C. `3.213 xx 10^(-10) M^(2)`D. `1.535 xx 10^(-10) M^(2)` |
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Answer» Correct Answer - B From eqs. (i) and (ii) `AgCl(s) hArr Ag^(+)+Cl^(-), E^(@)=0.22-0.80=-0.58 V` `E_("cell")=E_("cell")^(@)-0.0591/2 log [Ag^(+)][Cl^(-)]` `0=-0.58-0.0591/2 log [Ag^(+)][Cl^(-)]` `K_(sp)=[Ag^(+)][Cl^(-)]=1.535xx10^(-10) M^(2)` |
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| 107. |
The following facts are availabel `:` `2A^(c-)+B_(2) rarr 2B^(-)+A_(2),` `2C^(c-)+B_(2) rarr No reaction, ` `2D^(c-)+A_(2) rarr 2A^(c-)+D_(2)` Which of the following statement is correct ?A. `E_(C^(-) | C_(2))^(@) gt E_(B^(-) | B_(2))^(@) gt E_(A^(-) | A_(2))^(@) gt E_(D ^(-) D_(2))^(@)`B. `E_(C^(-) | C_(2))^(@) lt E_(B^(-) | B_(2))^(@) lt E_(A^(-) | A_(2))^(@) lt E_(D^(-) | D_(2))^(@)`C. `E_(C^(-) | C_(2))^(@) lt E_(B^(-) | B_(2)) ^(@) gt E_(A^(-) | A_(2))^(@) gt E_(D^(-) | D_(2))^(@)`D. `E_(C^(-) | C_(2))^(@) gt E_(B^(-) | B_(2))^(@) lt E_(A^(-) | A_(2))^(@) lt E_(D^(-) | D_(2))^(@)` |
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Answer» Correct Answer - B Given examples are of non-metals . A non-metal having higher reduction potential replaces the other from its solution Thus `E_(RP_(D))^(@) gt E_(RP_(A))^(@) gt E_(RP_(B))^(@)` Hence only choice is (b) because `E_(OP_(B))^(@) gt E_(OP_(C))^(@)` as B replaces C from solution . |
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| 108. |
Aluminium is more reactive than Fe. But Al is less easily corroded than iron becauseA. Al is noble metalB. Iron forms both mono and divalent ionsC. Oxygen forms a protective oxide layerD. Fe undergoes reaction easily with `H_(2)O` |
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Answer» Correct Answer - C Aluminium forms a protective oxide layer but iro does not. |
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| 109. |
The `e.m.f.` of a cell corresponding to the reaction : `Zn_((s))+2H_((aq.))^(+) rarr underset((0.1 M))(Zn^(2+))+underset((1 atm))(H_(2(g))` is `0.28 V` at `25^(@)C` and `E_(Zn//Zn^(2+))^(@) = 0.76V`. (i) Write half cell reactions. (ii) Calculate `pH` of the solution at `H` electrode. |
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Answer» (i) Anode: `Zn rarr Zn^(2+)+2e` Cathode: `2H^(+)+2e rarr H_(2)` (ii) `E_(cell) = E_(OP_(Zn))^(@) - (0.059)/(2)"log"[Zn^(2+)]+E_(RP)^(@)` `-(0.059)/(2)"log"([H^(+)]^(2))/(P_(H_(2)))` or `E_(cell) = E_(OP_(Zn))^(@) + 0 +(0.059)/(2)"log"([H^(+)]^(2))/(P_(H_(2)) xx [Zn^(2+)])` `= 0.28 = 0.76 + (0.059)/(2)"log"([H^(+)]^(2))/(0.1 xx 1)` or `([H^(+)]^(2))/(1) = - (0.48 xx 2)/(0.059) = -16.271` `([H^(+)]^(2))/(1) = 5.3556 xx 10^(-17)` or `[H^(+)] = 2.3142 xx 10^(-9)` `:. pH = -log[H^(+)] = 8.635` |
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| 110. |
Ionisation constant of a weak acid (HA) in terms of `A_(m)^(oo)` and `A_(m)` is:A. `K_a=(CLambda_m^oo)/(Lambda_m-Lambda^oo)`B. `K_a=(CLambda_m^2)/(Lambda_m^oo(Lambda_m^oo-Lambda_m)`C. `K_a=(CLambda(Lambda_m^oo)^2)/(Lambda_m^oo(Lambda_m^oo-Lambda_m)`D. None of these |
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Answer» Correct Answer - B |
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| 111. |
When a concentrated solution of an electrolyte is dilutedA. its specific conductance increasesB. its equivalent conductivity decreasesC. its specific conductivity decreases and equivalent conductivity increasesD. Both specific and equivalent conductivity increases |
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Answer» Correct Answer - C |
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| 112. |
Given `E_(Cl_(2)//Cl^(-))^(@)=1.36V,E_(Cr^(3+)//Cr)^(@)=-0.74V` `E_(Cr_(2)O_(7)^(2-)//Cr^(3+))^(@)=1.33V,E_(MnO_(4)^(-)//Mn^(2+))^(@)=1.51V` Among the following, the strongest reducing agent isA. `Cr^(3+)`B. `Cl^(-)`C. `Cr`D. `Mn^(2+)` |
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Answer» Correct Answer - C Greater the standard oxidation potential (or lesser is the standard reduction potential), more easily the substance is oxidized and hence stronger is the reducing is oxidized and hence stronger is the reducing agent. As `E_(Cr^(+3)//Cr)^(@)` (i.e., standard reduction potential) is lowest, hence, Cr will be strongest reducing agent. |
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| 113. |
Given `E_(Cl_(2)//Cl^(-))^(@)=1.36V,E_(Cr^(3+)//Cr)^(@)=-0.74V` `E_("Cr_(2)O_(7)^(2-)//Cr^(3+))^(@)=1.33V,E_(MnO_(4)^(-)//Mn^(2+))^(@)=1.51V` Among the following, the strongest reducing agent isA. `Cr^(3+)`B. `Cl^(-)`C. `Cr`D. `Mn^(2+)` |
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Answer» Correct Answer - C `{:("Species","Oxidation Potential"),(Cr^(3+),E_(Cr^(3+)//Cr_(2)O_(7)^(2-))^(@)=-1.33 V),(Cl^(-),E_(Cl^(-)//Cl_(2))^(@)=-1.36 V),(Cr,E_(Cr//Cr^(3+))^(@)=+0.74 V),(Mn^(2+),E_(Mn^(2+)//MnO_(4)^(-))^(@)=-1.51 V):}` Greater is the positive value of standard oxidation potential more is reducing behavious. Thus, Cr will be strongest reducing agent. |
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| 114. |
In the following process of disproportionation `2CIO_(3)^(-)hArrCiO_(2)^(-)+CIO_(4)^(-)` `E_(CIO_(4)^(-)//CIO_(3)^(-))^(@)=+0.36V,E_(CIO_(3)^(-)//CIO_(2)^(-))^(@)=+0.33V` If initial concentration of chloride ion was 0.1M, calculate the equilibrium concentration of perchlorate ion. |
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Answer» The two reactions of the given reaction will be `CIO_(3)^(-)(aq)+H_(2)O(l)toCIO_(4)^(-)(aq)+2H^(+)(aq)+2e^(-)` `underline(" "CI_(3)^(-)(aq)+2H^(+)(aq)+2e^(-)toCIO_(2)^(-)(aq)+H_(2)O(l)" ")` Adding `2CIO_(3)^(-)(aq)hArrCIO_(2)^(-)(aq)+CIO_(4)(aq)` `E_(cell)^(@)=0.33-0.36=-0.03V` `E=E^(@)-(0.059)/(n)logQ` At equilibrium, `E=0,Q=K_(eq)`. also n=2, hence `0=-0.03-(0.059)/(2)logK_(eq)` or `logK_(eq)=-(0.03xx2)/(0.059)=-1=overline(1)` or `K_(eq)=0.1` `{:(,2CIO_(3)^(-)(aq),hArr,CIO_(3)^(-)(aq),+,CIO_(4)^(-)(aq)),("Initial conc.",0.1M,,0,,0),("At eqm.",0.1-2x,,x,,x):}` `K=(x xx x)/((0.1-2x)^(2))=0.1` or `(x)/(0.1-2x)=sqrt(0.1)=0.316` or `x=0.0316-0.632` or `1.632x=0.0316` or `x=0.0194M` |
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| 115. |
For the reaction `NO_(3)^(-)toNO_(2)` (acidic medium), `E^(@)=0.790V` `NO_(3)^(-)toNH_(3)OH` (acidic medium), `E^(@)=0.731V` Calculate the pH at which at the above two half reactions will have same E values (Assume the concentrations of all the species to be unity). |
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Answer» `NO_(3)^(-)+2H^(+)+e^(-)toNO_(2)+H_(2)O," "E^(@)=0.790V` `NO_(3)^(-)+7H^(+)+6e^(-)toNH_(2)OH+2H_(2)O,E^(@)=0.731V` Since E values for both the reactions are same `E_(NO_(3)^(-)//NO_(2))=E_(NO_(3)^(-)//NH_(3)OH)` `thereforeE_(NO_(3)^(@)//NO_(2))+(0.059)/(1)"log"([H^(+)][NO_(3)^(-)])/([NO_(2)])=E_(NO_(3)^(-)//NH_(2)OH)^(@)+(0.059)/(6)"log"([H^(+)]^(7)[NO_(3)^(-)])/([NH_(2)OH])` or `0.790+0.059log[H^(+)]^(2)=0.731+(0.059)/(6)log[H^(+)]` (as concentrations of all species =1, given) or `0.790+0.118log[H^(+)]=0.731+0.0688log[H^(+)]` or `(0.118-0.0688)log[H^(+)]=0.731-0.790` or `0.0492log[H^(+)]=0.059` or `-log[H^(+)]=(0.059)/(0.0492)=1.1992` or `pH=1.1992` |
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| 116. |
Which one is wrong if electrolysis of `CH_(3)COONa(aq)` is made using `Pt` electrodes ?A. `pH` of solution increases.B. Molar ratio of gases at anode and cathode is `3:1`C. `[CH_(3)COO^(c-)]` in solution decreases.D. The molar ration of gases at anode and cathode is `2:1` |
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Answer» Correct Answer - d Anode`:2CH_(3)COO^(c-)rarr C_(2)H_(6)+2CH_(2)+2e^(-)` Cathode `:2H^(o+)+2e^(-) rarr H_(2)` `(CO_(2):H_(2)=2:1)` |
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| 117. |
For a given reaction `:M^((x+n))+n e^(-)rarr M^(c+),E^(c-).(red)` is known along with `M^(x+n)` and `M^(x+)` ion concentrations. ThenA. `n` can be evaluatedB. `x` can be evaluatedC. `(x+n)` can be evaluatedD. `n,x,(x+n)` can be evaluated |
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Answer» Correct Answer - a `E_(red)=E^(c-)._(red)+(0.059)/(n)log.([M^(x+n)])/([M^(x+)])` |
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| 118. |
The `EMF` of `Ni-Cad` battery is dependent of `:`A. `Cd(OH)_(2)`B. `Ni(OH)_(2)`C. `overset(c-)(O)H`D. None of these |
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Answer» Correct Answer - d The net redox change `:` `NiO_(2)(s)+Cd+2H_(2)O rarr Ni(OH)_(2)(s)+Cd(OH)_(2)(s)` |
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| 119. |
For `Cr_(2)O_(7)^(2-)+14H^(+)+6e^(-)to2Cr^(3+)+7H_(2)O,E^(@)=1.33V` At `[Cr_(2)O_(7)^(2-)]=4.5` millimole, `[Cr^(3+)]=15` millimole, E is 1.067V. Calculate the pH of the solution. |
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Answer» For the given reaction, `Cr_(2)O_(7)^(2-)+14H^(+)+6e^(-)to2Cr^(3+)+7H_(2)O` `E=E^(@)=-(0.0591)/(6)"log"([Cr^(3+)]^(2)[H_(2)O]^(7))/([Cr_(2)O_(7)^(2-)][H^(+)]^(14))` `1.067=1.33-9.85xx10^(-3)"log"([15xx10^(-3)]^(2)[1]^(7))/([4.5xx10^(-3)][H^(+)]^(14))` `(-0.263)/(-9.85xx10^(-3))=2log(15xx10^(-3))-log(4.5xx10^(-3))-14log(H^(+))` `26.7=2(-1.82)+2.34+14pH` or `pH=(28)/(14)=2`. |
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| 120. |
Write the chemistry of recharging the elad storage battery, highlighting all the materials that are involved during reacharging. |
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Answer» A lead storage battery consists of anode of lead, cathode of a grid of lead packed with lead dioxide `(PbO_(2))` and 38% solution of sulphuric acid as electrolytic. When the battery is in use, the following reactions take place: ltBrgt At anode: `Pb(s)+SO_(4)^(2-)(aq)toPbSO_(4)(s)+2e^(-)` . . . (i) At cathode: `PbO_(2)(s)+SO_(4)^(2-)(aq)+4H^(+)(aq)+2e^(-)toPbSO_(4)(s)+2H_(2)O(l)` . . . (ii) Overall reaction: `overline(Pb(s)+PbO_(2)(s)+2H_(2)SO_(4)(aq)to2PbSO_(4)(s)+2H_(2)O(l)" ")` On charging the battery, the reverse reactio takes place, i.e., `PbSO_(4)` deposited on the electrodes is oververted back into Pb and `PbO_(2)` and `H_(2)SO_(4)` is regenerated. Reverse of reaction (i) will be reduction and hence will take place at cathode. reverse of reaction (ii) will be oxidation and hence will take place at anode. |
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| 121. |
Consider the reaction `Cr_(2)O_(7)^(2-)+14H^(+)+6e^(-)to2Cr^(3+)+7H_(2)O` What is the quantity of electricity in couolombs needed to reduce 1 mole of `Cr_(2)O_(7)^(2-)`? |
| Answer» From the given reaction, 1 mol fo `Cr_(2)O_(7)^(2-)` ions require `6F=6xx96500C=579000C` of electricity for reduction to `Cr^(3+)`. | |
| 122. |
The electrode with reaction `:Cr_(2)O_(7)^(2-)(aq)+14H^(o+)(aq)+6e^(-) rarr 2Cr^(3+)(aq)+7H_(2)O` can be represented asA. `Pt|H^(o+)(aq),Cr_(2)O_(7)^(2-)(aq)`B. `Pt|H^(o+)(aq),Cr_(2)O_(7)^(2-)(aq),Cr^(3+)(aq)`C. `Pt_(H2)|H^(o+)(aq),Cr_(2)O_(7)^(2-)`D. `Pt_(H2)|H^(o+)(aq),Cr_(2)O_(7)^(2-)(aq),Cr^(3+)(aq)` |
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Answer» Correct Answer - d `Cr_(2)O_(7)^(2-)` reduces to `2Cr^(3+)`, so much be represented at cathode. `H_(2)` is oxidized to `2H^(o+)`, so must be representedd at anode. So the cell representation is as shown in `(d)`. |
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| 123. |
Consider the reaction `:` `Cr_(2)O_(7)^(2-)+14H^(o+)+6e^(-) rarr 2Cr^(+3)+8H_(2)O` What is the quantity of electricity in coulombs needed to reduce `1 mol e` of `Cr_(2)O_(7)^(2-)` ? |
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Answer» From the reaction,1 mol of `Cr_(2)O_(7)^(2-)` require 6 F `=6xx96500=579000C` `therefore579000C` of electrcity are required for reduction of `Cr_(2)O_(7)^(2-)` to `Cr^(3+)`. |
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| 124. |
Consider the reaction, `Cr_(2)O_(7)^(2-) +14H^(+) +6e^(-) to 2Ce ^(3+) +7H_(2)O` What is the quantity of electricity in coulombs needed to reduce 1 mole `Cr_(2) O_(7)^(2-)` ? |
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Answer» 1 mole `Cr_(2)O_(7)^(2-)` required 6 Faraay electricity (6F) `=6xx96500C=5.79xx10^(5)C` Quantity of electricity `=5.79 xx10^(5)` Coulomb |
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| 125. |
Suggest a list to metals that are extracted electrolytically. |
| Answer» Those metals which are highly reactive and have large negative `E^(0)` values can be exerted electrolytically. This type of metals are powerful reducing agents. Eg: Sodium, Postassium, Calcium, Magnesium etc. | |
| 126. |
Assertion: Conductivity of all electrolytes decreases on dilution. Reason: on dilution, number of ions per unit volume decreases.A. Both assertion and reason are true and the reason is the correct explanation of assertion.B. Both assertion and reason are true and the reason is not the correct explanation of assertion.C. Assertion is true but the reason is false.D. Both assertion and reason are false. |
| Answer» Correct Answer - A | |
| 127. |
Assertion : Conductivity of all electrolytes decreases on dilution. Reason : On dilution number of ions per unit volume decreases.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. |
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Answer» Correct Answer - A Reason is the correct explanation for assertion. |
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| 128. |
Why does the conductivity of a solution decreases with dilution ? |
| Answer» The conductivity of a solution is linked with the the number of ions present per unit volume. With dilution, these decrease and the corresponding conductivity or specific conductance of the solution decreases. | |
| 129. |
Statements: (i) Unit of specific conductivity is `ohm^(-1) cm^(-1)`. (ii) Specific conductivity of strong electrolytes decreases on dilution. (iii) The amount of an ion discharged during electrolysis does not depend upon resistance. (iv) The unit of electrochemical equivalence is g/coulomb.A. All are correctB. All are wrongC. Only (i), (ii) and (iv) are correctD. Only (ii), (iii) and (iv) are correct |
| Answer» Correct Answer - A | |
| 130. |
Suggest a way to determine the `wedge_(m)^(@)` value of water. |
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Answer» `wedge_(m)^(@)(H_(2)O)=lamda_(H^(+))^(@)+lamda_(OH^(-))^(@)` we find out `wedge_(m)^(@)(HCl),wedge_(m)^(@)(NaOH) and wedge_(m)^(@)(NaCl)`. Then `wedge_(m)^(@)(H_(2)O)=wedge_(m)^(@)(HCl)+wedge_(m)^(@)(NaOH)-wedge_(m)^(@)(NaCl)` |
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| 131. |
Assertion: `wedge_(m)` for weak electrolytes shows a sharp incrase when the electrolytic solution in diluted. Reason: For weak electrolytes, degree of dissociation increases with dilutio of solution.A. Both assertion and reason are true and the reason is the correct explanation of assertion.B. Both assertion and reason are true and the reason is not the correct explanation of assertion.C. Assertion is true but the reason is false.D. Both assertion and reason are false. |
| Answer» Correct Answer - A | |
| 132. |
Calculate the equilibrium constant for the reaction `Cu(s)+2Ag^(+)(aq)hArrCu^(2+)(aq)+2Ag(s)` Given that `E_(Ag^(+)//Ag)^(@)=0.80V and E_(Cu^(2+)//Cu)^(@)=0.34V` |
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Answer» The cell may be represented as: `Cu|Cu^(2+)(aq)||Ag^(+)(aq)|Ag` `E_(cell)^(@)=E_(RHS)^(@)-E_(LHS)^(@)=0.80V-(0.34V)=046V` `E_(cell)^(@)=(0.0591)/(n)" log "K_(c)` For the given reaction, `n=2,E_(cell)^(@)=0.46V` `therefore0.46=(0.0591)/(2)logK_(c)" or "logK_(c)=(0.46xx2)/(0.0591)=15.5668" "thereforeK_(c)"Antilog "15.5668=3.6xx10^(15)`. |
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| 133. |
Calculate `wedge_(m)^(@)` for `CaCl_(2)` and `MgSO_(4)` from the following data: `lamda^(@)` (S `cm^(2)mol^(-1)):Ca^(2+)=119.0,Mg^(2+)=106.0,Cl^(-)=76.3,SO_(4)^(2-)=160.0`. |
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Answer» By Kohlrausch law, `wedge_(m)^(@)(CaCl_(2))=lamda_(Ca^(2+))^(@)+2lamda_(Cl^(-))^(@)=119.0" S "cm^(2)mol^(-1)+2xx76.3" S "cm^(2)mol^(-1)` `=(119.0+152.6)" S "cm^(2)mol^(-1)=271.6" S "cm^(2)mol^(-1)` `wedge_(m)^(@)(MgSO_(4))=lamda_(Mg^(2+))^(@)+lamda_(SO_(4)^(2-))^(@)=106.0" S "cm^(2)mol^(-1)+160.0" S "cm^(2)mol^(-1)=266" S "cm^(2)mol^(-1)`. |
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| 134. |
The conductivity of 0.001028 M acetic acid is 4.95`xx10^(-5)" S "cm^(-1)`. Calculate its dissociation constant if `wedge^(@)` for acetic acid is 390.5 S `cm^(2)mol^(-1)`. |
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Answer» For the given concentration of acetic acid solution, `wedge_(m)=(kappaxx1000)/(c)=(4.95xx10^(-5)S" "cm^(-1)xx1000cm^(3)L^(-1))/(0.001028" mol "L^(-1))=48.15" S "cm^(2)mol^(-1)` `alpha=(wedge_(m))/(wedge_(m)^(@))=(48.15" S "cm^(2)mol^(-1))/(390.5" S "cm^(2)mol^(-1))=0.1233` `{:(,CH_(3)COOH,hArr,CH_(3)COO^(-),+,H^(+)),("Initial conc.",c,,0,,0),("Equilibrium conc.",c-calpha=c(1-alpha),,calpha,,calpha):}` `K=(calpha.calpha)/(c(1-alpha))=(calpha^(2))/(1-alpha)=((0.001028" "molL^(-1))(0.1233)^(2))/(1-0.1233)=1.78xx10^(-5)mol" "L^(-1)` |
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| 135. |
The conductivity of 0.001 mol `L^(-1)` solution of `CH_(3)COOH` is `3.905xx10^(-5)S cm^(-1)`. Calculate its molar conductivity and degree of dissociation `(alpha)`. `("Given":lamda_((H^(+)))^(@)=349.65 S cm^(2)mol^(-1)andlamda^(@)(CH_(3)COO^(-))=40.9 D cm^(2)mol^(-1))` |
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Answer» Correct Answer - `Delta_(m)=39.05 S cm^(2) mol^(-1),a=0.1` Step I. Calcualtion of molar conductance (K) `k=3.905xx10^(5)" S "cm^(-1),C=0.001" mol "L^(-1)=(0.001" mol")/(10^(3) cm^(3))=10^(-6)"mol "cm^(-3)` Molar conductance `(Lambda_(m)) =(k)/(C )=((3.905xx10^(-5)" S "cm^(-1)))/((10^(-6)"mol "cm^(-3)))=39.05" S "cm^(2)mol^(-1)` Step II. Calcualtion of degree of dissociation `(alpha)` `Lambda_(m)^(@)=lamda^(@)(H^(+))+lambda^(@)(CH_(3)COO^(-))` `=(349.6+40.9)=390.5" S "cm^(2)mol^(-1)` `alpha=(Lambda_(m)^(C))/(Lambda_(m)^(o))=((39.05" S "cm^(2)mol^(-1)))/((390.5" S "cm^(2) mol^(-1)))=0.1` |
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| 136. |
`k=4.95xx10^(-5) S cm^(-1)` for a `0.001 M `solution. The reciprocal of the degree of dissociation of acetic acid, if `wedge_(m)^(@)` for acetic acid is `400S cm^(-2) mol^(-1)` is `:`A. 7B. 8C. 9D. 10 |
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Answer» Correct Answer - b `k=4.95xx10^(-5)S cm^(-1), alpha=(wedge_(m))/(wedge^(@)._(m))` `wedge_(m)=1000xx(k)/(C)=1000xx(4.95xx10^(-5))/(0.001)=49.5 S cm^(-1)` `impliesalpha=(49.5)/(400)implies (1)/(alpha)=(400)/(49.5)~~8` |
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| 137. |
The conductivity of `0.001028 M` acetic acid is `4.95xx10^(-5) S cm^(-1)`. Calculate dissociation constant if `wedge_(m)^(@)` for acetic acid is `390.5 S cm^(2) mol ^(-1)`. |
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Answer» `wedge_(m)=(kxx1000)/(M)=(4.95xx10^(-5)S cm^(-1))/(0.001028 mol L^(-1))xx(1000cm^(3))/(L)` `=48.15 S cm^(2)mol ^(-1)` `alpha=(wedge_(m)^(c))/(wedge_(m)^(@))=(48.15S cm^(2) mol^(-1))/(390.5 S cm^(2)mol^(-1))=0.1233` `K_(a)=(calpha^(2))/((1-alpha))=(0.001028mol L^(-1)xx(0.1233)^(2))/((1-0.1233))` `=1.78xx10^(-5)mol L` |
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| 138. |
The conductivity of 0.001 `" mol "L^(-1)` solution of `CH_(3)COOH` is `4.95xx10^(-5)" S "cm^(-1)`. Calculate its molar conductance and degree of dissociation (alpha).`"Given" lambda_((H^(+)))^(@)=349.6" S "cm^(2)mol^(-1), lambda_((CH_(3)COO^(-)))^(@)=40.95 cm^(2)mol^(-1)` |
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Answer» `k=4.95xx10^(-5)" S "cm^(-1)` `C=0.001" mol "L^(-1)=((0.001" mol"))/((10^(3)"cm"^(3)))=10^(-6)" mol "cm^(-3)` Molar conductance at given concentration `Lambda_(m)^(c)=(k)/(C )=((4.95xx10^(-5)" S "cm^(-1)))/((10^(-6)" mol "cm^(-3)))=49.5" S "cm^(2)mol^(-1)`. Molar conductance at infinite dilution. `(Lambda^(@))` `Lambda_(m)^(@)=(lambda_(CH_(3)COO^(-))^(@)+lambda_(H^(+))^(@))=(40.9+349.6)" S "cm^(2)mol^(-1)` `=390.5" S "cm^(2)mol^(-1)` Degree of dissociation `(alpha)=(Lambda_(m)^(c))/(Lambda_(m^(@)))=((49.5" S "cm^(2)mol^(-1)))/((390.5" S "cm^(2)mol^(-1)))=0.127` |
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| 139. |
The equivalent conductances of NACl at concentration c and at infinite dilution are `lambda_(c)` and `lambda_(oo)` respectively. The correct relationship between `lambda_(c)` and `lambda_(oo)` is given as : (where the constant b si positive)A. `lambda_(c)=lambda_(oo)-b sqrt(c)`B. `lambda_(c)=lambda_(oo)+bsqrt(c)`C. `lambda_(c)=lambda_(oo)+bc`D. `lambda_(c)=lambda_(oo)-bc` |
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Answer» Correct Answer - A According to Debye-Huckel relation : `lambda_(c)=lambda_(oo)-b sqrt(c)` |
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| 140. |
Calculate `K_(a)` of acetic acid it its `0.05N` solution has equivalent conductances of `7.36 mho cm^(2)` at `25^(@)C (lambda_(CH_(3)COOH)^(oo) = 390.70)` |
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Answer» Correct Answer - `1.76 xx 10^(-5)` mole/litre Degree of dissociation `(x) =(Lamda_(c))/(Lamda_(0))=(7.36)/(390.7) = 0.0188` For equilibrium `{(0.05,0,0,"Initial conc (moles/litre)"):}` `CH_(3)COOH=CH_(3)COO^(-)+H^(+)` `0.05(1-x) 0.05 x 0.05 x` Equilibrium concentration (for `CH_(3)COOH,0.05 N = 0.05 M`) `K_(a)=(0.05 x xx 0.05 x)/(0.05(1-x))` since `x` is very small `K_(a)=0.05 x^(2)= 0.05 xx(0.0188)^(2)=1.76 xx10^(-5) "mole/L"`. |
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| 141. |
The specific conductance of a `N//10 KCI` solution at `18^(@)C` is `1.12 xx 10^(-2) mho cm^(-1)`. The resistance of the solution contained in the cell is found to be 65 ohms. Calculate the cell constant. |
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Answer» Correct Answer - `0.728 cm^(-1)` Cell constant `= K xx R = 1.12 xx 10^(-2) xx 65` `= 72.8 xx 10^(-2) = 0.728 cm^(-1)`. |
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| 142. |
The sp cond of a saturated solution of `AgCI` at `25^(@)C` after substractin the sp conductances of conductivity of water is `2.28 xx 10^(-6) mho cm^(-1)`. Find the solubility product of `AgCI` at `25^(@)C(lambda_(AgCI)6(oo) = 138.3 mho cm^(2))` |
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Answer» Correct Answer - `2.70 xx 10^(-10)("mole"//"litre")^(2)` For equilibrium, `AgCl = Ag^(+) + Cl^(-) K_(sp)=[Ag^(+)][CL^(-)]` If solubility of `AgCl` in water is, say, x moles/L or x.eq/L `K_(sp) = x.x =x^(2)` `:.` volume containing 1 eq. of `AgCl = (1000)/(x)` `Lamda_(AgCl) = "sp. Cond." x V = 2.28 xx 10^(-6) xx(1000)/(x)` Since `AgCl` is sparingly soluble in water, `Lamda_(AgCl)=Lamda_(AgCl)^(@)=138.3` `:. 2.28 xx 10^(-6) xx (1000)/(x)=138.3` or `x =1.644 xx 10^(-5)` eq./L or mole/L `K_(sp)=x^(2)=(1.644xx 10^(-5))^(-2)=2.70 xx 10^(-10)("mole/L")^(2)`. |
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| 143. |
A cell constructed by coupling a standard copper electrode and a standard magnesium electrode has EMF to 2.7 volts. If the standard reduction potential of copper electrode is +0.34 volt that of magnesium electrode is:-A. `+3.04`voltsB. `-3.04` voltsC. `+2.36` voltsD. `-2.36` volts |
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Answer» Correct Answer - D Mg lies above Cu in electrochemical series and hence Cu electrode acts as cathode. `E_(cell)^(o)=E_(Cu^(++)//Cu)^(o)-E_(Mg^(++)//Mg)^(o)=-2.36V` |
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| 144. |
At 291K, saturated solution of `BaSO_(4)` was found to have a specific conductivity of 3.648`xx10^(-6)ohm^(-1)cm^(-1)`, that of water used being `1.25xx10^(-6)ohm^(-1)cm^(-1)`. Ioinc conductances of `Ba^(2+) and SO_(4)^(2-)` ions are 110 and 136.6 `ohm^(-1)cm^(2)mol^(-1)` respectively. calculate the solubility of `BaSO_(4)` at 291 K. (At masses: `Ba=137,S=32,O=16)` |
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Answer» Correct Answer - `2.266xx10^(-3)gL^(-1)` `kappa(BaSO_(4))=kappa("solution")-kappa("water")." solubility"=(kappaxx1000)/(wedge_(m)^(@))=((3.648-1.250)10^(-6)xx1000)/((110+136.6))xx233" g "L^(-1)`. |
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| 145. |
Equivalent conductance of saturated `BaSO_(4)` is 400 S `cm^(2) eq^(-1)` an d specific conductyance is 8 `xx10^(-5)S cm^(-1)`. Solubility product. `K_(sp)` of `BaSO_(4)` isA. `4xx10^(-8)`B. `1xx10^(-8)`C. `2xx10^(-4)`D. `1xx10^(-4)` |
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Answer» Correct Answer - B `lambda_(m)^(0)(BaSO_(4))=2xxlambda_(eq)^(0)(BaSO_(4))` `=2xx400=800` `lambda_(m)^(0)(BaSO_(4))=(kxx1000)/(M)` `M=(kxx1000)/(lambda_(m)^(0)BaSO_(4))` `=(8xx10^(-5)xx1000)/(800)=10^(-4)` |
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| 146. |
The following results have been obtained during the kinetic studies of the reaction : `2A+B to C+D` Determine the rate law and rate constant for the reaction. |
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Answer» Rate law may be expresswed as Rate `=k[A]^(x)[B]^(y)` `[Rate]_(1)=6.0xx10^(-3) =k(0.1)^(x) (0.1)^(y) -(i)` `[Rate]_(2)=7.2 xx10^(-2) =k(0.3) ^(x) (0.2)^(y)-(ii)` `[Rate]_(3)=2.88xx10^(-1)=k(0.3)^(x)(0.4)^(y)-(iii)` `[Rate]_(4)=2.40xx10^(-2) =k(0.4)^(x)(0.1)^(y)-(iv)` `((Rate)_(1))/((Rate)_(4))=(6.0xx10^(-3))/(2.40xx10^(-2))=(k[0.1]^(x) [0.1]^(y))/(k[0.4]^(x)[0.1]^(y))or 1/4 =((0.1)^(y))/((0.4)^(x))=((1)/(4))^(x) x=1` `((Rate)_(2))/((Rate)_(3))=(7.2xx10^(-2))/(2.88xx10^(-1))=(k[0.3]^(x)[0.2]^(y))/(k[0.3]^(x)[0.4]^(y))or 1/4 =((0.2)^(y))/((0.4)^(y))=[(1)/(2)]^(y)y=2` `therefore` Rate law expression is given by Rate `=k [A][B]^(2)` Rate constant k, can be determined by placing the values of A, B and rate of formation of D. By taking the values for experiment II. Rate `k[A][B]^(2)` `k=(Rate)/([A][B]^(2))=(7.2xx10^(-2)mol L^(-1)"min"^(-1))/((0.3 mol L^(-1) )(0.2 molL^(-1))^(2))=6.0 mol^(-2) L^(2)"min"^(-1)` `therefore k=6.0 mol^(-2)L^(2) min ^(-1).` |
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| 147. |
Calculate aprooximately how much current is necessary to produce oxygen gas at rate of 1 mL per second ? |
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Answer» Correct Answer - 17.23 ampere 1 g-equivalent of oxygen or `22400/4` mL is liberated by 96500 coulomb. |
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| 148. |
At `25^(@)C` molar conductance of `0.1` molar aqueous solution of ammonium hydroxide is `9.54 ohm^(-1) cm^(2) mol^(-1)` and at infinte dilution its molar conductance is `238 ohm^(-1) cm^(2) mol^(-1)` The degree of ionisation of ammonium hydroxide at the same concentration and termperature isA. `20.800%`B. `4.008%`C. `40.800%`D. `2.080%` |
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Answer» Correct Answer - B The degree of ionisation `(alpha)` of ammonium hydrox-ide at the same temperature and concentration is given as follows: `alpha=(Lambda_(m)^(c))/(Lambda_(m)^(oo))` `= (9.54 ohm^(-1)cm^(2)mol^(-1))/(238 ohm^(-1)cm^(2)mol^(-1))` `= 0.04008` Therefore, percentage ionisation will be `= 0.04008 xx 100% = 4.008%` |
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| 149. |
The molar conductivity of a `0.5 mol//dm^(3)` solution of `AgNO_(3)` with electrolytic conductivity of `5.76 xx 10^(-3) S cm^(-1)` at `298 K` isA. `28.8S" "cm^(2)//mol`B. `2.88S" "cm^(2)//mol`C. `11.52S" "cm^(2)//mol`D. `0.086S" "cm^(2)//mol` |
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Answer» Correct Answer - C `lamda_(M)^(o)=(kappaxx1000)/(M)=(5.76xx10^(-3)xx1000)/(0.5)`. `=11.52S" "cm^(2)//mol^(-1)` |
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| 150. |
At `25^(@)C` molar conductance of `0.1` molar aqueous solution of ammonium hydroxide is `9.54 ohm^(-1) cm^(2) mol^(-1)` and at infinte dilution its molar conductance is `238 ohm^(-1) cm^(2) mol^(-1)` The degree of ionisation of ammonium hydroxide at the same concentration and termperature isA. `40.800%`B. `2.08o%`C. `20.800%`D. `4.008%` |
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Answer» Correct Answer - D `%alpha=(lamda^(c))/(lamda^(oo))xx100=(9.54)/(238)xx100=4.008%` |
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