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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.
| 51. |
The number of electrons delivered at the cathode during electrolysis by a current of 1 ampere in 60 seconds is (charge on the electron=`1.60xx10^(-19)C`)A. `6xx10^(23)`B. `6xx10^(20)`C. `3.75xx10^(20)`D. `7.48xx10^(23)` |
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Answer» Correct Answer - C Charge flowing during electrolysis`=Ixxt` `=1xx60=60` coulombs But charge flowing=number of electrons flowing`xx`charge on each electron `therefore60=nxx1.60xx10^(-19)` or `n=(60)/(1.60xx10^(-19))=37.5xx10^(19)=3.75xx10^(20)` |
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| 52. |
Under what conditions is `E_(cell)=0` or `Delta_(r)G=0`? |
| Answer» `E_(cell)=0` or `Delta_(r)G=0` when the cell reaction reaches equilibrium. | |
| 53. |
Can `E_(cell)^(@)` and `Delta_(r)G^(@)` for cell reaction ever be equal to zero? |
| Answer» No (`E_(cell)` an be zero but not `E_(cell)^(@)`). Further, `DeltaG^(@)=-nFE_(cell)^(@)`. As `E_(cell)^(@)ne0`, therefore, `Delta_(r)G^(@)` also cannot be zero. | |
| 54. |
The realtionship between standard `emf(E_("cell")^(@))` of a galvanic cell and standard Gibbs energy change `(Delta_(r)G^(@))` for the chemi-cal reaction of the cell isA. `Delta_(r)G^(@) = nFE_("cell")^(@)`B. `Delta_(r)G^(@) = nF//E_("cell")^(@)`C. `Delta_(r)G^(@) = -nFE_("cell")^(@)`D. `Delta_(r)G^(@) = -nF//E_("cell")^(@)` |
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Answer» Correct Answer - C Some of the most important results of elecrochemicsty are the realatlionship amonge cell `emf`, Gibbs-energy change and equilibrium constant. The Gibbs enerfy change `DeltaG` (decreases) for a reaction equals the maximum useful work of the reaction `Delta_(r)G = w_max` for a voltaic cell, this work is the electrical work, `-nFE_("cell")` (where n is the number of moles of electrons transferred in a reaction), so when the reactants and products are in their standard states, we have `Delta_(r)G = -nFE_("cell")^(@)` where `Delta_(r)G^(@)` is measured in joules, `E_("cell")^(@)` is measured in volts, `F` is the faraday, the charge on `1` mole of electrons, `9.65 c 10^(4)C`. With this equation, emf measurement becomes an important source of thermodynamic information. Alternatively, thermodynamic data can be used to calculate cell emfs. Due to negative sign, the direction of spontaneous change gives a negative value for `Delta_(r)G^(@)` but a positive value for `E^(@)` The two quantitative measures of the driving force of a chemical reaction are: the Gibbs-energy change `DeltaG` (a ther-mochemical quantity) and the cell potential (a thermochemical quantity). |
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| 55. |
Can `E_("cell")^(@)` or `Delta_(r)G^(@)` for cell reaction ever be equal to zero? |
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Answer» No, otheriwse the reaction become non-feasible The reaction is feasible only at `E_("cell")^(@)=` positive or `Delta_(r)G^(@)`=negative. when `E^(@)=Delta_(r)G^(@)`=0 the reaction reaches at equilibrium. |
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| 56. |
The standard electrode potential for the half cell reactions are `Zn^(2+)+2e^(-)rarrZn ,E^(@)=-0.76V` `Fe^(2+)+2e^(-)rarrFe,E^(@)=-0.44V` The EMF of the cell reaction `Fe^(2+)+ZnrarrZn^(++)+Fe` isA. `-0.32V`B. `-1.20V`C. `+1.20V`D. `+0.32V` |
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Answer» Correct Answer - D `Fe^(2+)+ZnrarrZn^(2+)+Fe` `ZnrarrZn^(2+)+2e^(-)` `ZnrarrZn^(2+)+2e^(-)` `Fe^(2+)+2e^(-)rarrFe` `Zn|Zn^(2+)||Fe^(2+)|Fe` `EMF^(@)=E_(Fe^(2+)//Fe)^(@)-E_(Zn^(2+)//Zn)^(@)` `=-0.44-(-0.76)` `=0.76-0.44=+0.32V` |
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| 57. |
Depict the galvanic cell in which the cell reaction is `Cu+2Ag^(+)rarr2Ag+Cu^(2+)` |
| Answer» In a galvanic cell, oxidation half reac tion is written on left hand side and reduction half reaction. Is on right hand side. Salt bridge is represented by parallel lines `Cu|Cu^(2+)||Ag^(+)|Ag`. | |
| 58. |
Depict the galvanic cell in which the cell reaction is Cu + 2Ag+ → 2Ag + Cu2+ |
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Answer» Cu|Cu2+|| Ag+|Ag |
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| 59. |
Under what condition is `E_("cell")^(@)=0` or `Delta_(r)G=0`? |
| Answer» Both can be equal to zero when the reaction is is in a state of equilibrium. | |
| 60. |
Reduction potential for the following h alf-cell reaction are `Zn rarr Zn rarr Zn^(2+)+2e^(-),E_(Zn^(2+)//Zn)^(@) = -0.76V` `Fe rarr Fe^(2+)+2e^(-),E_(Fe^(2+)//Fe)^(@) = -0.44V` The emf for the cell reaction `Fe^(2+)Zn rarr Zn^(2+)+Fe` will beA. `+0.32V`B. `-0.32V`C. `+1.20V`D. `-1.20V` |
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Answer» Correct Answer - A (a) `E_(Zn^(2+)//Zn)^(@)=-0.76V` `E_(Fe^(2+)//Fe)^(@)=-0.44V` Cell reaction, `Fe^(2+)+Zn to Zn^(2+) +Fe` `E_(cell)^(@)=E_("(cathode)")^(@)-E_("(anode)")^(@)` `=-0.44 -(-0.76)` `=+0.32V` |
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| 61. |
What does the negative sign in the expression `E_(Zn^(2+)//Zn)^(@)=-0.76V` mean? |
| Answer» It implies that Zn is more reactive than hydrogen or it is a stronger reducing agent.In a cell containing zinc electrode and standard hydrogen electrode present in two half cells, zinc will be oxidised to `Zn^(2+)` ions while `H^(+)` ions will get reduced to hydrogen. | |
| 62. |
Aqueous copper sulphate solution and aqueous silver nitrate solution are electrolysed by 1 ampere current for 10 minutes in separate electrolytic cells. Will the mass of copper and silver deposited on the cathode be same of different? Explain your answer. |
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Answer» The mass of copper and silver deposited on the cathode in the two electrolytic cells will not be the same. This is according to second law of Electrolysis. The law states that When the same quantity of charge is passed through the solutions of different electrolytes connected in series, the mass of the substances deposited at the respective electrodes are directly proportional to their equivalent masses. In the present case. `("Mass of Cu deposited")/("Mass of Ag deposited")=("Gram equivalent mass of " Cu)/("Grass equivalent mass of " Ag)=(31.75)/(108)` This means that the masses deposited are different and not same. |
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| 63. |
Which of the following statement is correct?A. `E_("cell")` and `Delta_(r)G` of cell reaction both are extensive propertiesB. `E_("cell")` and `Delta_(r)G` of cell reaction both are intensive propertiesC. `E_("cell")` is an intensive property while `Delta_(f)G` of cell reaction is an extensive propertyD. `E_("cell")` is an extensive property while `Delta_(r)G` of cell reaction is an intensive property |
| Answer» `E_("cell")` is an intensive property as it does not depend upon mass of species ( number of pa rticles) but `Delta_(r)G` of the cell reaction is an extensive property because this depends upon mass of species (number of particles) | |
| 64. |
In `H_(2) - O_(2)` fuel cell the reaction occuring at cathode isA. `2 H_(2) + O_(2) to 2 H_(2) (l)`B. `H^(+) + OH^(-) to H_(2)O`C. `O_(2) + 2 H_(2)O + 4e^(-) to 4OH^(-)`D. `H^(+) + e^(-) to (1//2) H_(2)` |
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Answer» Correct Answer - C In `H_(2) - O_(2)` fuel cell , the reaction at cathode is `O_(2) + 2 H_(2)O + 4 e^(-) to 4 OH^(-)` . |
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| 65. |
Give the electrode reactions occuring at the anode and at the cathode in `H_(2),O_(2),` fuel cell. |
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Answer» At cathode `:O _(2(g))+2H_(2) O_((l))+4e^(-)to 4OH _((aq))^(-)` At anode `: 2H_(2(g))+4HO_((aq))^(-)to 4H_(2) O_((l))+4e^(-)` Overall reaction `: 2H_(2(g))+O_(2(g))to 2H_(2)O_((l))` |
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| 66. |
In `H_(2) - O_(2)` fuel cell the reaction occuring at cathode is:A. `2H_(2)+O_(2)rarr2H_(2)O(l)`B. `H^(+)+OH^(-)rarrH_(2)O`C. `O_(2)+2H_(2)O+4e^(-)rarr4OH^(-)`D. `H^(+)+e^(-)rarr1//2H_(2)`. |
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Answer» Correct Answer - C For details see review (already discussed there) |
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| 67. |
An electrochemical cell stops working after some time becauseA. Electode potential of both the electrodes become equalB. Electrode potential of both the electrodes become equal.C. One of the electrdoe is eaten awayD. The reaction starts proceeding in opposite direction. |
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Answer» Correct Answer - B EmF=`E_(c)-E_(a)` `Emf="zero" [because E_(c)=E_(a)]` |
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| 68. |
Which of the following reactions occurs at the anode during the recharging of lead storage battery ?A. `Pb+SO_(4)^(2-)rarrPbSO_(4)+2e^(-)`B. `Pb+PbO_(2)+H_(2)SO_(4)rarr2PbSO_(4)+2H_(2)O`C. `PbSO_(4)+2e^(-)rarrPb+SO_(4)^(2-)`D. `2PbSO_(4)+2H_(2)OrarrPb+PbO_(2)+2H_(2)SO_(4)` |
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Answer» Correct Answer - C During recharging, cathode is -ve electrode to which electrons are supplied from outer source. Therefore, the cell reaction involved at cathode is `PbSO_(4)+2e^(-)rarrPb+SO_(4)^(2-)` |
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| 69. |
The standard e.m.f. of a cell, involving one electron change is found to be 0.591V at `25^(@)C`. The equilibrium constant of the reaction is `(F=96,500C" "mol^(-1),R=8.314KJ^(-1)" "mol^(-1))`A. `1.0xx10^(10)`B. `1.0xx10^(5)`C. `1.0xx10^(1)`D. `1.0xx10^(30)` |
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Answer» Correct Answer - A `E_(cell)=E_(cell)^(o)-(0.0591)/(n)logK_(C)` At 298K `E_(cell)=0,0=0.591-(0.591)/(n)logK_(C)` `logK_(C)=(0.591xx1)/(0.0591)=10,K_(c)="anti log 10"=1xx10^(10)` |
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| 70. |
96500C electricity is passed through `CuSO_(4)` the amount of copper precipitated isA. 0.25 moleB. 0.5moleC. 1.0moleD. 2.00mole |
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Answer» Correct Answer - B `underset(1"mole")(Cu^(2+))+underset(2"Faraday"=2xx96500C)(2e^(-))rarrCu` 2xx96500C deposits 1 mole of copper 96500C deposits 0.5 mole of copper. |
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| 71. |
The standard e.m.f. of a cell involving one electron charge is found to be 0.591V at `25^(@)C` the equilibrium constant of the reaction is (F=96500C `"mol"^(-1)`:R=8.314 `JK^(-1)"mol"^(-1)`)A. `1.0xx10^(1)`B. `1.0xx10^(30)`C. `1.0xx10^(10)`D. `1.0xx10^(5)` |
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Answer» Correct Answer - C At 298 `E^(@)=(0.0591)/(n)"log" k_(c)` `0.591=(0.0591)/(1)"log"K_(c)` `"log" k_(c)=(0.5910)/(0.0591)` ` "log"k_(c)=10` `K_(c)=10^(10)` |
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| 72. |
The limmiting molar conductivites `^^ ^(@)` for `NaCl_(2) ` Kbr and KCl are 126,152 and 150 S `cm^(2)` respectively the `^^ ^(@)` for NaBr isA. `128Scm^(2)"mol"^(-1)`B. `302Scm^(2)"mol"^(-1)`C. `278S cm^(2)"mol"^(-1)`D. `176Scm^(2)"mol"^(-1)` |
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Answer» Correct Answer - A `^^_(NaBr)^(@)=^^_(NaCl)^(@)+^^_(KBr)^(@)-^^_(KCl)^(@)` `=126+152-150=128 S cm^(2)"mol"^(-1)` |
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| 73. |
The concentration of potassium ions inside a biological cell is least twenty time higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is `M_((s))|M^(+)(aq,0.05" molar")||M^(+)(aq,1" molar")|M_((s))` For the above electrolytic cell the magnitude of the cell potential `|E_(cell)|=70mV`. Q. For the above cellA. `E_(cell) lt 0, DeltaG gt 0`B. `E_(cell) gt 0,DeltaG lt 0`C. `E_(cell) lt 0, DeltaG^(@) gt 0`D. `E_(cell) gt 0, DeltaG^(@) lt 0` |
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Answer» Correct Answer - B `E_(cell)=0-(2.303RT)/(F)"log"(0.05)/(1)` =a positive value =70mV Hence `DeltaG lt 0`. [given] |
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| 74. |
The concentration of potassium ions inside a biological cell is least twenty time higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is `M_((s))|M^(+)(aq,0.05" molar")||M^(+)(aq,1" molar")|M_((s))` For the above electrolytic cell the magnitude of the cell potential `|E_(cell)|=70mV`. Q. if the 0.05 molar solution of `M^(+)` is replaced by a 0.0025 molar `M^(+)` solution,then the magnitude of the cell potential would beA. 35 mVB. 70 mVC. 140 mVD. 700 mV |
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Answer» Correct Answer - C `E_(cell)=0-(2.303RT)/(F)"log"(0.0025)/(1)` `=(2.303RT)/(F)log(0.05)^(2)=2xx70mV=140mV`. |
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| 75. |
The concentration of potassium ions inside a bilological cell is at least twenty times higher than the outisde. The resulting potential difference across the cell is important in several process such as transmission of nerve impuses and maintaning the ion balance. A simple model for such a concentration cell involving a metal M is `M(s)|M^(+)(aq,0.05" molar")||M^(+)(aq,1" molar")|M(s)` For the above electrolytic cell, the magnitude of the cell potential `|E_(cell)|=70mV`. Q. For the above cellA. `E_(cell)lt0,DeltaGgt0`B. `E_(cell)gt0,DeltaGlt0`C. `E_(cell)lt0,DeltaG^(@)gt0`D. `E_(cell)gt0,DeltaG^(@)lt0` |
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Answer» Correct Answer - B For the given concentration cell, `E_(cell)=(2.303RT)/(nF)"log"(C_(2))/(C_(1))` `=(2.303RT)/(F)"log"(1)/(0.05)=+ve,` i.e., `E_(cell)gt0`. For the reaction to be spontaneous, `DeltaG lt 0`. |
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| 76. |
A lead storage battery consists of a lead anode and a grid of lead packed with lead dioxide as the cathode. The electrolyte taken is 39% `H_(2)SO_(4)` by mass having a density of 1.294 g `mL^(-1)`. The battery holds 3.5 L of the acid. During the discharge of the battery, the density `H_(2)SO_(4)` falls from 1.294 g `mL^(-1)` to 1.139 g `mL^(-1)` which is 20% `H_(2)SO_(4)` by mass Q. The reacton occurring at the andoe during charging isA. `Pb^(2+)+2e^(-)toPb`B. `Pb^(2+)+SO_(4)^(2-)toPbSO_(4)`C. `PbtoPb^(2+)+2e^(-)`D. `PbSO_(4)+2H_(2)Oto2PbO_(2)+4H^(+)+SO_(4)^(2-)+2e^(-)` |
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Answer» Correct Answer - D At anode, oxidation occurs. `PbSO_(4)` is oxidized to `PbO_(2)` according tor eaction (d). |
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| 77. |
A lead storage battery consists of a lead anode and a grid of lead packed with lead dioxide as the cathode. The electrolyte taken is 39% `H_(2)SO_(4)` by mass having a density of 1.294 g `mL^(-1)`. The battery holds 3.5 L of the acid. During the discharge of the battery, the density `H_(2)SO_(4)` falls from 1.294 g `mL^(-1)` to 1.139 g `mL^(-1)` which is 20% `H_(2)SO_(4)` by mass Q. Moles of sulphuric acid lost during discharge isA. 9.88B. 8.88C. 2.32D. 1.16 |
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Answer» Correct Answer - A Mass of the solution before discharge `=3500mLxx1.294" g "mL^(-1)=4529g` mass of `H_(2)SO_(4)` present in this solution `=(39)/(100)xx4529g=1766.31g` mass of the solution of after discharge `=3500mLxx1.139" g "mL^(-1)=3986.5g` mass of `H_(2)SO_(4)` present in this soluton, `=(20)/(100)xx3986.5=797.3g` Loss in mass of `H_(2)SO_(4)` during discharge `=1766.31-797.3=969.01g` Moles of `H_(2)SO_(4)` lost during discharge `=(969.01)/(98)=9.88786` |
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| 78. |
A lead storage battery consists of a lead anode and a grid of lead packed with lead dioxide as the cathode. The electrolyte taken is 39% `H_(2)SO_(4)` by mass having a density of 1.294 g `mL^(-1)`. The battery holds 3.5 L of the acid. During the discharge of the battery, the density `H_(2)SO_(4)` falls from 1.294 g `mL^(-1)` to 1.139 g `mL^(-1)` which is 20% `H_(2)SO_(4)` by mass Q. The number of ampere-hour for which the battery must have been used isA. 2650.5B. 265.05C. 26.505D. 2.6505 |
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Answer» Correct Answer - B Ampere-hour`=("Coulombs")/(3600s)=(964178)/(3600)=262.05` |
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| 79. |
A lead storage battery consists of a lead anode and a grid of lead packed with lead dioxide as the cathode. The electrolyte taken is 39% `H_(2)SO_(4)` by mass having a density of 1.294 g `mL^(-1)`. The battery holds 3.5 L of the acid. During the discharge of the battery, the density `H_(2)SO_(4)` falls from 1.294 g `mL^(-1)` to 1.139 g `mL^(-1)` which is 20% `H_(2)SO_(4)` by mass Q. The amount of charge in coulombs used up by the battery is nearlyA. 954180B. 477090C. 95418D. 47709 |
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Answer» Correct Answer - A During the discharge reaction, 4 moles of `H^(+)` ions i.e., 2 moles of `H_(2)SO_(4)` require 2F, i.e., `2xx96500` coulombs `=(2xx96500xx9.88786)/(2)=954178` coulombs |
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| 80. |
Which of the following solutions can be safely stored in a copper vessel ?A. `ZnSO_(4)`B. `AgNO_(3)`C. `AlCl_(3)`D. all of them. |
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Answer» Correct Answer - B Cu react with `Ag NO_(3)` solution . |
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| 81. |
Write Nernst equation for the reaction : `Zn(s)|Zn^(2+)(aq)||Cu^(2+)(aq)Cu(s)` |
| Answer» `E_(cell)=E_(cell)^(@)-(0.0591)/(2)"log"([Zn^(2+)(aq)])/([Cu^(2+)(aq)])` | |
| 82. |
In a fuel cell, hydrogen and oxygen react to produce electricity. In process, hydrogen gas is oxidized at the anode and oxygen at the cathode. If `67.2L` of `H_(2)` at `STP` reacts in `15mi n` , what is the average current produced ? If the entire current is used for electro`-` deposition of copper from copper `(II)` solution, how many grams of copper will be deposited ? Anode reaction `: H_(2)=2overset(c-)(O)Hrarr2H_(2)O+2e^(-)` Cathode reaction `:O_(2)+2H_(2)O+2e^(-) rarr 4overset(c-)(O)H` |
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Answer» Correct Answer - `190.5g` Moles of `H_(2)` reacting `=(67.2)/(22.4)=3` `:. `Equivalent of `H_(2)` used `=3xx2=6` Now, `(W)/(E_(w))=(It)/(96500)` `:. 6=(Ixx15xx60)/(96500)` `:.I=643.33A` Also, equivalent of `H_(2)=` equivalent of `Cu` formed `:. ` Equivalent of `Cu` deposited `=6` `:. W_(Cu)=6xx(63.5)/(2)=190.5g` Thus, weight of `Cu` deposited `=190.5g` |
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| 83. |
Lead storage battery has anode made up of `………………….` and cathode made up of `………………………..`. |
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Answer» Correct Answer - `Pb,PbO_(2)` `Pb, PbO_(2)` |
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| 84. |
Which of the following solutions can be safely stored in a copper vessel ?A. `ZnSO_(4)`B. `AgNO_(3)`C. `AuCl_(3)`D. All of them. |
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Answer» Correct Answer - A `Cu` cannot reduced `Zn^(2+)` but it reduces `Ag^(+)` as well as `Au^(+3)` ions. |
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| 85. |
Write the Nernst equation for the EMF of the cell `Ni _((s))|Ni +_((aq))^(2+)||Ag_((aq))^(+)|Ag` |
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Answer» Given cell is `Ni_((s))|Ni_((aq))^(2+) ||Ag_((aq))^(+)|Ag` Nernst equation for the cell is `E_(cell)=E_(cell)^(0) +(RT)/(nF)ln ""([Ag^(+)])/([Ni]^(+2))` |
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| 86. |
The density of `H_(2)SO_(4) …………………………..` in lead storage cell during discharging and calomel electrode is a `………………..` electrode. |
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Answer» Correct Answer - Decrease, reference, hydrogen Decrease, reference, hydrogen |
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| 87. |
A solution is one molar in each of NaCl , `CdCl_(2) , ZnCl_(2) and PbCl_(2)` . To this , tin metal is added . Which of the following is true ? Given : `{:(E_(Pb^(2+) |P)^(@) = -0.126 V "," E_(Sn^(2+)|Sn) = -0.136 V) , (E_(Cd^(2+)|Cd)^(@) = -0.40 V "," E_(Zn^(2+)|Zn)^(@)= -0.763 V), (E_(Na^(+) |Na)^(@) = -2.71 V):}`A. Sn can reduce `Na^(+)` to NaB. Sn can reduce `Zn^(2+)` to ZnC. Sn can reduce `Cd^(2+) ` to CdD. Sn can reduce `Pb^(2+)` to Pb . |
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Answer» Correct Answer - D Check the spontaneity of each reaction by finding the E.M.F. |
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| 88. |
The Nernst equation giving dependence of electrode reduction potential on concentration isA. `E=E^(@)+(2.303RT)/(nF)"log"([M])/([M^(n+)])`B. `E=E^(@)+(2.303RT)/(nF)"log"([M^(n+)])/([M])`C. `E=E^(@)-(2.303RT)/(nF)"log"([M^(n+)])/([M])`D. `E=E^(@)+(2.303RT)/(nF)"log"[M]^(n+)` |
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Answer» Correct Answer - B See important foumulae in Comprehensive Review. |
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| 89. |
What is Nernst equation ? Write the equation for an electrode with electrode reaction `M^(n+)(aq)+ne^(-) hArr M(s).` |
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Answer» The electrode potential at any concentration measured with respect to standard hydrogen electrode is represented by Nernst equation. Nernst equation is `E=E^(0) +(RT)/(nF) ln[M^(x+)]` [Metal Electrodes] Given electrode reaction is `M_((aq))^(n+)+ne^(-) hArrM_((s))` For the above electrode reaction Nernst equation is `E_((M^(n+)//M))=E_((M^(n+)//M))^(0)+(RT)/(nF)ln [Mn^(+)]` Here `E_((M^(n+)//M))=` Electrode potential `E_((M^(n+)//M))=` Standard Electrode poteitial R= gas constant `=8.314` J/K mole F = Faraday = 96487 c/mole T = temperature `[M^(n+)]=` concentration of species `M^(n+)` |
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| 90. |
State and explain Nernst equation with the help of a metallic electrode and a non-metallic electrode. |
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Answer» The electrode potential at any concentration measured with respect to standard hydrogen electrode is represented by Nernst equation. Nernest equation is `E=E^(0) +(RT)/(nF) ln [M^(+)]` [Metal Electrodes] Given electrode reaction is `M_((aq))^(n+)+n e^(-)hArrM_((s))` For the above electrode reaction Nerst equation is `E_((M^(n+)//M))=E_((M^(n+)//M))+(RT)/(nF) ln [M^(n+)]` Here `E_((M^(n+)//M))=` Electrode potential `E_((M^(n+)//M))^(0) =` Standard Electrode potential R = gas constant `=8.314J //K.` mole F = Faraday `=96487c//` mole T = temperature `[M^(n+)]` concentration of species `M^(n+)` For non-metal electrodes : `E=E^(0)-(RT)/(nF)ln C,C=` concentration Example for metal electrode : Given cell is `Ni_((s))|Ni_((aq))^(2+)||Ag _((aq))^(+)|Ag` Nernest equation for the cell is `E_(cell) =E_(cell)^(0) +(RT)/(nF)ln ""([Ag^(+)])/([Ni]^(+2))` Example for non-metal electrode : `pt, Cl_(2) //Cl^(-)` `E=E^(0) -(RT)/(nF)log C` `E=E^(0)-(RT)/(F)log C` |
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| 91. |
`Cl_(20` is stronger `…………………..` agent then `Br_(2)` and `I_(2)` as its reduction potential is `………………………` than that of `Br_(2)` and `I_(2)` |
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Answer» Correct Answer - Oxidizing, higher Oxidizing, higher |
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| 92. |
Dry cell is a `…………………` cell and lead storage cell is a `…………………….` cell. |
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Answer» Correct Answer - Primary, secondary Primary, secondary |
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| 93. |
In the electrolysis of aqueous NaCl solution, `Cl_(2)` is produced at the anode and not `O_(2)`. This is due to____shown by water for oxidation to `O_(2)`. |
| Answer» Correct Answer - Overvoltage | |
| 94. |
What is the role of `ZnCl_(2)` in dry cell? |
| Answer» `ZnCl_(2)` combines with the `NH_(3)` produced to form the complex salt `[Zn(NH_(3))_(2)Cl_(2)]` as otherwise the pressure developed due to `NH_(3)` would crack the seal of the cell. | |
| 95. |
Nernst equation halps us to understand the effect of `"______"` on the electrode of the half-cell and `emf` of the voltiv cellA. temperatureB. concentrations of solultesC. parical pressures of gasesD. All of these |
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Answer» Correct Answer - D Electrode potentials and cell potentials and cell potentials, like Gibbs-energy chages, depends on temperture and on the compo-sition of the reacion mixture, i.e., on the concentrations of solutes and the partial pressures of gases. The Nernst equation is used to calculate electrode potentials and cell potentials for temperature, concentrations and parical pressures other than standard state values. |
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| 96. |
The Nernst equation gives`……………. EMF` of the cell and in a Deniell cell current flows from `…………………` to `……………….` |
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Answer» Correct Answer - Concentration dependence, cathode, anode Concentration dependence, cathode, anode |
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| 97. |
Why fluorine cannot be obtained by electrolysis of aqueous HF solution, though it is a good conductor of electricity? |
| Answer» `H_(2)O` has higher oxidation potential than `F^(-)` ions. Hence, in aqueous solution, as compared to `F^(-)` ions, `H_(2)O` is more easily oxidized to give `O_(2)` gas. Therefore, `F_(2) ` cannot be obtained by electrolysis of aqueous HF solution. | |
| 98. |
Following reactions occur at cathode during electrolysis of aqueous silver chloride solution : `Ag^(+)(aq)+e^(-) to Ag(s),E^(@)=+0.80" V"` `H^(+)(aq)+e^(-) to 1//2 H_(2)(g), E^(@)=0.00" V"` On the basis of standard reduction potential (`E^(@)` value), which reaction is feasible at cathode and why ? |
| Answer» Reaction with higher `E^(@)` value will be feasible since more then `E^(@)` value more will be the magnitudeof `-DeltaG^(@)(DeltaG^(@)=-nFE_(cell)^(@)`). | |
| 99. |
Following reactions occur at cathode during the electrolysis of aqueous silver chloride solution: `Ag^(+)(aq)+e^(-)toAg(s),E^(@)=+0.80V,H^(+)(aq)+e^(-)to(1)/(2)H_(2)(g),E^(@)=0.00V` On the basis of their standard reduction electrode potential `(E^(@))` values, which reaction is feasible at the cathode and why? |
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Answer» Higher the standard reduction potential of a species, more easily it is reduced at the cathode. As `Ag^(+)(aq)` has greater standard reduction potential, therefore, the reaction that will occur at the cathode is `Ag^(+)(aq)+e^(-)toAg(s)`. |
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| 100. |
If `E_(Red)^(@)` for `Ag^(+)+e^(-)toAg` is 0.80 V, then for the reaction 2 `Ag^(+)+2e^(-)to2Ag,E_(Red)^(@)` will be_____. |
| Answer» Correct Answer - 0.80 V | |