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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.
| 151. |
For the reaction, `C+O_(2)rarrCO_(2)`, `DeltaH=-393 J` `2Zn+O_(2) rarr 2ZnO`, `DeltaH=-412J`A. carbon can oxidise zincB. oxidation of carbon is not possibleC. oxidation of zinc is not fessibleD. zinc can oxidise carbon. |
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Answer» Correct Answer - A Zinc can lose electrons (O.S. of Zn chagnes from 0 to+2) and carbon can gain electrons (O.S. of C change from 0 to-4). Hence zinc is oxidised by carbon . |
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| 152. |
Standard free energies of formation (in kJ/mol) at 298 K are 237.2, -394.4 and -8.2 for `H_(2)O(l),CO_(2)(g)` and pentane (g), respectively. The value of `E_(cell)^(o)` for the pentane-oxygen fuel cell isA. 1.0968VB. 0.0968C. 1.968VD. 2.0968V |
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Answer» Correct Answer - C `C_(5)H_(12)+8O_(2)to5CO_(2)+6H_(2)O` `DeltaG^(@)=-5xx394.4-6xx237.2+8.2` `DeltaG^(@)=-nFE_(Cell)^(@)` `-3387kJ//`mole |
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| 153. |
Standard free energies of formation in kJ/mol) at 298 K are -237.2, -394.4 and -8.2 for `H_(2)O(l),CO_(2)(g)` and pentane (g) respectively. The value of `E_(cell)^(@)` for the pentane-oxygen fuel cell isA. 1.968 VB. 2.0968VC. 1.0968VD. 0.0968V |
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Answer» Correct Answer - C The balanced equation for pentane-oxygen cell reaction will be `C_(5)H_(12)+8O_(2)to5CO_(2)+6H_(2)O,n=32` `Delta_(r)G^(@)=[5xxDelta_(f)G^(@)(CO_(2))+6Delta_(f)G^(@)(H_(2)O)]-[Delta_(f)G^(@)(C_(5)H_(12))+8Delta_(f)G^(@)(O_(2))]` `=[5(-394.5)+6(-237.2)]-[(-8.2)+0]` `=-1972-1423.2+8.2=-3387" kJ "mol^(-1)` `Delta_(f)G^(@)=-nFE_(ceLL)^(@)` `therefore-3387000=-32xx96500xxE_(cell)^(@)` or `E_(cell)^(@)=1.0968V` |
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| 154. |
The potential of the Daniell cell, `Zn|(ZnSO_4),((1M))||(CuSO_4),((1M))|Cu` was reported by Buckbee, Surdzial, and Metz at `E^@ = 1 . 1028 - 0. 41 xx 10^(-3) T + 0. 72 xx 10^(-5) T^2` where T is the Celsius temperature. Calculate `Delta S^@` for the cell reaction at `235^@ C`,.A. ` - 45 . 32`B. ` - 34. 52`C. ` - 25 . 43`D. ` - 55. 39 ` |
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Answer» Correct Answer - D `(dE^@)/(dT)=- 0.647 xx 10^(-3) + 2 xx 0. 72 xx 10^(-5) T` `Delta S = 2 xx 96500 xx [-0.647 xx 10^(-3) +2 xx 0. 72 10^(-5) xx 250]` ` =- 55 . 39`. |
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| 155. |
Standard free energies of formation (I `kJ //`mol ) at `298 K` are ` -237 .2 , - 394 .4` and `- 8.2 ` for `H_2 O(1), CO_2 (g)` and pentange (g) , respectively . The value of `E_(cell)^@` for the pentane-oxygen fuel cell is .A. `1.968V`B. `2.0968V`C. 1.0968VD. 0.0968V |
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Answer» Correct Answer - C The balanced equation for the pentane oxygen cell reaction will be `C_(5)H_(12)+8O_(2)rarr5CO_(2)+6H_(2)O,n=32` `(because 16O+32e^(-)rarr16O^(2-))` `DeltaG=5xxDelta_(f)G^(@)(CO_(2))+6 xxDelta _(f)G^(@)(H_(2)O)` `-Delta_(f)G^(@)(C_(5)H_(12))+8xxDelta_(f)G^(@)(O_(2))` `= 5xx(-394.4)+6(-237.2)-(-8.2)-0` `=-1972-1423.2+8.2` `=-3387 kJ "mol"^(-1)` `DeltaG_(f)=-nFE_("cell")^(@)` `therefore -3387000=-32xx96500xxE_("cell")^(@)` or `E_("cell")^(@)=1.0968V` |
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| 156. |
A zinc rod is dipped in 0.1 M solution of `ZnSO_(4)`. The salt is 95% dissociated at this dilution at 298K. Calculation the electrode potential `(E_(Zn^(2+)//Zn)^(@)=-0.76V)` |
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Answer» The electrode reaction written as reduction reaction is: `Zn^(2+)+2e^(-)Zn(n=2)` Applying nernst equation, we get `E_(Zn^(2+)//Zn)=E_(Zn^(2+)//Zn)^(@)-(0.0591)/(2)"log"(1)/([Zn^(2+)])` As 0.1 M `ZnSO_(4)` solution is 95% dissociated, this means that in the solution, `[Zn^(2+)]=(95)/(100)xx0.01M=0.095M` `thereforeE_(Zn^(2+)//Zn)=-0.76-(0.0591)/(2)"log"(1)/(0.095)=0.76-0.02955(log1000-log95)` `=0.76-0.02955(3-1.9777)=-0.76-0.03021=-0.79021` volt |
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| 157. |
`C_(4)H_(10)+(13)/(2)O_(2)(g)rarr4CO_(2)(g)+5H_(2)O(l), DeltaH =- 2878 kJ` `DeltaH` is the heat of……….of butane gas.A. `+4.74 V`B. `+0.547 V`C. `+1.09V`D. `+4.37 V` |
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Answer» Correct Answer - C In the reaction `overset(-10 + 10)(C_(4) H_(10)) (g) + (13)/(2) O_(2) (g) rarr overset(4 CO_(2))(g) + 5H_(2)O (l)` Change in oxidation number of carbon `= + 16 - (-10) = + 26` `:.` Number of electrons involved in cell process will be 26 `E^(@) = (-DeltaG^(@))/(nF) = - ((-2746) xx 1000)/(26 xx 96500) = + 1.09 V` |
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| 158. |
Find out the molar conductivity of an aqueous solution of `BaCl_(2)` at infinite dilution when ionic conductances of `Ba^(2+)` and `Cl^(-)` ion are 127.30 S `cm^(2)mol^(-1)` and 76.34 S `cm^(2)mol^(-1)` respectively. |
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Answer» Correct Answer - `279.99" S "cm^(2)mol^(-1)` `wedge_(m)^(@)(BaCl_(2))=lamda_(Ba^(2+))^(@)+2lamda_(Cl^(-))^(@)=127.30+2(76.34)" S "cm^(2)mol^(-1)=279.99" S "cm^(2)mol^(-1)` |
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| 159. |
At `25^(@)C`, the molar conductances at infinite dilution for the strong electrolytes `NaOH, NaCl " and " BaCl_(2) " are " 248 xx 10^(-4), 126 xx 10^(-4)" and " 280 xx 10^(-4) S m^(2) mol^(-1)` respectively. `Lamda_(m)^(@) Ba (OH)_(2) " in " S m^(2) mol^(-1)`A. `52.4 xx 10^(-4)`B. `524 xx 10^(-4)`C. `402 xx 10^(-4)`D. `262 xx 10^(-4)` |
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Answer» Correct Answer - B `Lamda_(m)^(@) Ba(OH)_(2) = Lamda_(m)^(@) BaCl_(2) + 2 Lamda_(m)^(@) NaOH - 2 Lamda_(m)^(@) NaCl` `= 280 xx 10^(-4) + 2 xx 249 xx 10^(-4) - 2 xx 126 xx 10^(-4)` `= 524 xx 10^(-4) S m^(2) mol^(-1)` |
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| 160. |
`AgNO_(3)(aq.)` was added to an aqeous `KCl` solution gradually and the conductivity of the solution was measured. The plot of conductance `(Lambda)` versus the volume of `AgNO_(3)` is : A. PB. QC. RD. S |
| Answer» Correct Answer - d | |
| 161. |
The solubility product `(K_(sp) , mol^(3) dm^(-9)` ) of `MX_(2)` at 298 K based on the information available for the given concentration cell is (take `2.303 xx R xx 298 // F = 0.059 `V)A. `1 xx 10^(-15)`B. `4 xx 10^(-15)`C. `1 xx 10^(-12)`D. `4 xx 10^(-12)` |
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Answer» Correct Answer - B The given concentration cell is `M | M^(2+) ` (saturated = `C_(1) ) || M^(2+) (0.001 M = C_(2))| M` EMF of concentration cell is : `E_(cell) = (2.303 RT)/(nF) "log" (C_(2))/(C_(1)) = (0.059)/(n) "log" (C_(2))/(C_(1))` `therefore 0.059 = (0.059)/(2) "log" (0.001)/(C_(1))` `or 1 = (1)/(2) "log" (10^(-3))/(C_(1)) or log 10^(-3) - log C_(1) = 2` `or -3-log C_(1) = 2 or log C_(1) = -5` or `C_(1) = 10^(-5)` M For the salt `MX_(2) , underset(S)(MX^(2)) to underset(S)(M^(2+)) + 2 underset(2 S)(X^(-))` `K_(sp) = (S) (2S)^(2) = 4 S^(3) = 4 xx (10^(-5))^(3) = 4 xx 10^(-15)` |
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| 162. |
A cell contains two hydrogen electrodes. The negative electrode is in contact with a solution of `10^(-6)M` hydrogen ions. The `EMF` of the cell is `0.118V` at `25^(@)C`. Calculate the concentration of hydrogen ions at the positive electrode. |
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Answer» Anode `H_(2) rarr 2H^(+) + 2e` (negative polarity) `[H^(+)] = 10^(-6)M` Cathode: `2H^(+) + 2e rarr H_(2)` (positive polarity) `[H^(+)] rarr aM` `:. E_(cell) = E_(OP_(H^(+)//H)) + E_(RP_(H^(+)//H))` `E_(OP_(H^(+)//H))^(@) - (0.059)/(2)log_(10)[H^(+)]_("Anode")^(2)` `+ E_(RP_(H^(+)//H))^(@) + (0.059)/(2)log_(10)[H^(+)]_("Cathode")^(2)` `= (0.059)/(2)log_(10).([H^(+)]_("Cathode")^(2))/([H^(+)]_("Anode")^(2))` `0.118= (0.059)/(2)log_(10).([H^(+)]_("Cathode")^(2))/(10^(-6))^(2)` `= (0.059)/(1)log_(10).([H^(+)]_("Cathode")^(2))/(10^(-6))` `:. [H^(+)]_("Cathode") = 10^(-4) M` |
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| 163. |
Consider the following four electrodes: `P=Cu^(2+)(0.0001M)//Cu(s)" "Q=Cu^(2+)(0.1M)//Cu(s)` `R=Cu^(2+)(0.01M)//Cu(s)" "S=Cu^(2+)(0.001M)//Cu(s)` If the standard electrode potential of `Cu^(2+)//Cu` is +0.34V, the reduction potentials in volts of the above electrodes follow the order:A. PgtSgtRgtQB. SgtRgtQgtPC. RgtSgtQgtPD. QgtRgtSgtP |
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Answer» Correct Answer - D `Cu^(2+)+2e^(-)toCu,E_(red)=E_(red)^(@)+(0.0591)/(2)log[Cu^(2+)]` Thus, greater is `Cu^(2+)` ion concentration, greater will be the reduction potential. |
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| 164. |
Consider the following four electrodes: `A=Cu^(2+)(0.0001M)//Cu_((s))` `B=Cu^(2+)(0.1M)//Cu_((s))` `C= Cu^(2+)(0.01M)//Cu_((s))` `D=Cu^(2+)(0.001M)//Cu_((s))` If the standard reduction potential of `Cu^(+2)//Cu` is `+0.34V`, the reduction potentials (in volts) of the above electrodes follow the orderA. `PgtSgtRgtQ`B. `SgtRgtQgtP`C. `RgtgtRgtP`D. `QgtRgtSgtP` |
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Answer» Correct Answer - D `Cu^(2+)+2e ^(-)rarrCu` `E_("red")=E_("red")^(@)+(0.0591)/(2)"log"[Cu^(2+)]` `therefore` greater is `Cu^(2+)` ion concentration, greater will be the reduction potential. |
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| 165. |
A cell contains two hydrogen electrodes. The negative electrode is in contact with a solution of `10^(-6)` M hydrogen ions. The emf of the cell is 0.118 V at `25^(@)C` calculate the concentration of hydrogen ions at the positive electrode. |
| Answer» Correct Answer - `10^(-4)M` | |
| 166. |
What is the potential of the cell containing two hydrogen electrodes as represented below ? `Pt | (1)/(2) H_(2) (g) | H^(+) (10^(-8)M) ||H^(+) (0.001M) | (1)/(2) H_(2) (g) | Pt`A. `-0.295` VB. `-0.0591` VC. 0.295D. 0.0591 V |
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Answer» Correct Answer - C `E_(cell) = (0.0592)/(1) log ([H^(+)]_(RHS))/([H^(+)]_(LHS))` ` = 0.0592 log_(10) (10^(-3))/(10^(-8)) = 0.0592 xx 5 = 0.295` V |
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| 167. |
In two vessels each containing 500 ml water, 0.5m mol of aniline (`K_(b)=10^(-9))` and 25 m mol of HCl are added separately. Two hydrogen electrodes are constructed using these solutions. Calculate the emf of cell made by connecting them appropriately. |
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Answer» Correct Answer - `E = 0.395V` For aniline `K_(b) = calpha^(2)` `10^(-9) = 10^(-3) alpha^(2)` `alpha^(2) = 10^(-6)` `alpha = 10^(-3)` `[OH]^(-) = 10^(-3) xx 10^(-3) = 10^(-6)` `[H^(+)] = 10^(-8)` `[H^(+)] = (25)/(500)` `[H^(+)] rArr 5 xx 10^(-2)` |
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| 168. |
What is the potential of a cell containing two hydrogen electrodes the negative one in contact with `10^(-8)M" "H^(+)` and positive one in contact with 0.025 M `H^(+)`A. 0.18 VB. 0.28 VC. 0.38VD. 0.48V |
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Answer» Correct Answer - C `(1)/(2)H_(2)toH^(+)(10^(-8)M)+e^(-)` (oxidation) `H^(+)(0.025M)+e^(-)to(1)/(2)H_(2)` (reduction) Cell reaction is: `H^(+)(0.025M)toH^(+)(10^(-8)M),E_(cell)=0.38V` |
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| 169. |
An example of a simple furl cell is :A. lead storage batteryB. ` H_2 -O_2` cellC. Deaniel cellD. Ldeclanche cell |
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Answer» Correct Answer - B ` H_2 -O_2` cell. |
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| 170. |
During discharge of a lead storage cell the density of sulphuric acid in the cell:A. InceasingB. decreasingC. remains unchangedD. initially increases but cecrease subsequentily |
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Answer» Correct Answer - B Discharging reaction `Pb(s) + PhO_2 (s) + 2H_2SO_4 rarr 2 PbSO_4 2H_2O`. |
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| 171. |
During discharge of a lead storage cell the density of sulphuric acid in the cell:A. InceasingB. decreasingC. remains unchanged D. initially increases but cecrease subsequentily |
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Answer» Correct Answer - B Discharging reaction `Pb(s) + PhO_2 (s) + 2H_2SO_4 rarr 2 PbSO_4 2H_2O`. |
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| 172. |
The density of `Cu` is `8.94 g cm^(-3)`. The quantity of electricity neede to plate an area `10 cm xx 10 cm` to a thickenss of `10^(-2) cm` using `CuSO_4` solution qould be .A. ` 13586 C`B. ` 27172 C`C. ` 40758C`D. ` 20348 C` |
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Answer» Correct Answer - B `(axx b xx c) xx rho = (Eit)/(96500)` `(10 xx 10xx 10^(-2)) xx (8.94) = (63.5)/2 xx (It)/(96500)` (It) `= 27172 C`. |
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| 173. |
The reduction potential diagram for `Cu` in acid solution is : Calculate `X`. Does `Cu^(+)` disproportionate in solution ? |
| Answer» Correct Answer - `X = +0.325` volt, Yes | |
| 174. |
In a hydrogen oxygen uel cell, combustion of hydrogen occurs toA. Produce high purity waterB. remove adsorbed oxygen from electrode surfaceC. generate heatD. create potential between two electrodes |
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Answer» Correct Answer - D Fuel cell works when potential difference is developed. |
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| 175. |
Predict whether the following reaction (s) is (are) feasible or not (i) `Fe+Zn^(2+)toFe^(2+)+Zn,E_(Zn)^(@)=-0.76,E_(Fe)^(@)=-0.44V` (ii) `Zn+2Ag^(+)toZn^(2+)+2Ag,E_(Zn)^(@)=-0.76V,E_(Ag)^(@)=-0.80V` |
| Answer» (i) `E_(cell)^(@)=E_(Fe//Fe^(2+))^(@)+E_(Zn^(2+)//Zn)^(@)=+0.44+(-0.76)=-0.32V` ltbRgt (ii) `E_(cell)^(@)=E_(Zn//Zn^(2+))^(@)+E_(Ag^(+)//Ag)^(@)=+0.76+0.80=1.56V`. | |
| 176. |
Electrode potential of `Zn^(2+)//Zn` is -0.76V and that of `Cu^(2+)//Cu` is +0.34V. The EMF of the cell constructued between these two electrodes isA. 1.10VB. 0.42VC. `-1.1V`D. `-0.42V` |
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Answer» Correct Answer - A `E_(cell)^(@)=E_("cathode")^(@)-E_("anode")^(@)=0.34-(0.76)=1.10V` |
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| 177. |
Write the equation which ralates th rate constants `k_(1)and k_(2)` at temperatures `T_(1) and T_(2)` of a reaction. |
| Answer» `log ""((k_(2))/(k_(1)))=(E_(a))/(2.303R)[(1)/(T_(1))-(1)/(T_(2))],E_(a)=` activation energy , R = Universal gas constant. | |
| 178. |
Chemical reaction involve interaction of atoms and molecules. A large number of atoms/molecules (approximately `6.022xx10^(23)`)are present in a few grams of any chemical compound varying with their atomic/molrcular mass. To handle such a large numbers conveniently, the mole concept was introduced. This concept has implications in diverse areas such as analytical in diverse areas such as analytical chemistry, biochemistry, electrochemistry and radiochemistry. The following example illustrates a typical case, involving chemical/ electrochemical reaction, which requires a clear understanding of the mole concept. A 4.0 molar aqueous solution of NaCl is prepared and 500 mL of this solution is electrolysed. This leads to the evolution of chlorine gas at one of teh electrodes (atomic mass: Na=23, Hg=200, 1F=96500 coulombs) The total charge in couloms required to complete the electrolysisA. 24125B. 48250C. 96500D. 193000 |
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Answer» Correct Answer - D 2 mol electrons will be required, therefore, required charge will be 2 faraday or 193000 coulomb. |
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| 179. |
Derive the intergrate rate equation for a zero order rection. |
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Answer» Zerto order reaction is the reaction in which rate of reaction does not depends on the concentratin of reactants. `R to P` Rate `=(-d[R])/(dt)=k [R]^(@)` Rate `=(-d[R])/(dt),d[R]=-k. dt` Intergration on both sides `[R]=-kt +I- (1)` I = Intergration constant At `=0 to R= [R]_(0)` initial concentration `I = [R]_(0)` Substituting `I=[R]_(0)` in the above equation (1) `[R] =-kt +[R]_(0)` `k=([R_(0)]-[R])/(t)` This is the intergrated rate equation for a zero order reaction. |
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| 180. |
The equivalent conductances of two strong electrolytes at infinite dilution in `H_(2)O` (where ions move freely through a solution) at `25^(@)C` are given below : `Lambda_(CH_(3)COONa)^(@) = 91.0 S cm^(2)//"equi v"`. `Lambda_(HCl)^(@) = 426.2 S cm^(2)//"equiv"`. What additional information//quantity one need to calculate `Lambda^(@)` of an aqueous solution of acetic acid ?A. `Lambda^(@)` of chloroacetic (`ClCH_(2)COOH`)B. `Lambda^(@)` of NaClC. `Lambda^(@)" of " CH_(3)COOK`D. the limiting equivalent conductance of `H^(+)` ions. |
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Answer» Correct Answer - B (b) `Lambda^(@)` of NaCl is needed. `Lambda_(CH_(3)COOH)^(@)=Lambda_(CH_(3)COOH)^(@)+Lambda_(HCl)^(@)-Lambda_(NaCl)^(@)` |
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| 181. |
Chromium plating is applied by electrolysis to objects suspended in a dichlromate solution , according to following `(` unbalanced `)` hald reaction `:` `Cr_(2)O_(7)^(2-)(aq) +e^(-) +H^(o+)(aq) rarr Cr(s)+H_(2)O(l)` How many hours would it take to apply a chromium plating of thickness `2.0xx10^(-2)mm` to a car bumper of suface area `0.25m^(2)` in an electrolysis cell carrying a current of `75.0A?` `[` Density of chromium is `7.19g cm^(-3)]`A. `2.2h`B. `1.5h`C. `3.0h`D. `0.25h` |
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Answer» Correct Answer - b Volume to be covered `=(0.25xx10^(4)cm^(2))xx(2xx10^(-3)cm)=5cm^(3)` `implies` Mass of `Cr` to be deposited `=5xx7.19g` `=35.95g` `-=(35.95)/(52)mol` `=0.69mol` `Cr_(2)O_(7)^(2-)+14H^(o+) +12e^(-) rarr 2Cr+7H_(2)O` Now, `2 mol Cr` deposited `-=12 F` electricity `implies 0.59 mol Cr` deposited `-=(12)/(2)xx0.069F` `=(It)/(96500)=(75xxt)/(96500)t=1.48h~~1.5h` |
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| 182. |
When an acid cell is charged, then:A. voltage of cell increasesB. resistance of cell increasesC. eletrolyte of cell dilutesD. None of these |
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Answer» Correct Answer - A |
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| 183. |
Conductivity of an electrolytic solution depends on ____________.(i) nature of electrolyte.(ii) concentration of electrolyte.(iii) power of AC source.(iv) distance between the electrodes. |
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Answer» (i) nature of electrolyte. (ii) concentration of electrolyte. |
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| 184. |
Conductivity of an electrolytic solution depends on:A. nature of electrolyteB. concentration of electrolyteC. power of AC sourceD. distance between the electrodes. |
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Answer» Correct Answer - A::B (a, b) are both correct options. Conductivity of the electrolytic solution does not depend upon power of AC sourece and distance between the electrodes. |
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| 185. |
Why is it not possible to determine `Lambda_(m)^(oo)` for weak electrolytes graphically ? Explain. |
| Answer» Because the dissociation of a weak electrolyte is never complete. | |
| 186. |
What is the effect of decreasing concentration on the molar conductivity of weak electrolyte ? |
| Answer» Correct Answer - It increases. | |
| 187. |
For the fuel cell reaction `2H_(2)(g) +O_(2)(g) rarr 2H_(2)O(l), Deta_(f)H_(298)^(@) (H_(2)O,l) =- 285.5 kJ//mol` What is `Delta_(298)^(@)` for the given fuel cell reaction? Given: `O_(2)(g) +4H^(+)(aQ) +4e^(-) rarr 2H_(2)O(l) E^(@) = 1.23 V`A. `-0.322 J//K`B. `-0.635 kJ//K`C. `3.51 kJ//K`D. `-0.322 kJ//K` |
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Answer» Correct Answer - D `{:(2H_(2) +O_(2) rarr H_(2)O ,DeltaH = 2X - 285.5 kJ),(O_(2)+4H^(+) +4e^(-) rarr 2H_(2)O, E^(@) = 1.23),(2H_(2) rarr 4H^(+) +4e^(-),E^(@) = 0),("for reaction":,),(O_(2) +2H_(2) rarr 2H_(2)O, E^(@) = 1.23):}` `DeltaG^(@) = -nFE^(@) =- 4 xx 96500 xx 1.23 =- 474780` `DeltaG^(@) = DeltaH^(@) - T DeltaS^(@) - 474780 =- 571000 - 298 DeltaS^(@)` `DeltaS^(@) =- 322.8 J//K` |
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| 188. |
For the equilibrium, `2H_(2)(g)+O_(2)(g)hArr2H_(2)O(l)` at `25^(@)C,DeltaG^(@)` is -474.78kJ `mol^(-1)`. Calculate log K for it (R=8.314 `JK^(-1)mol^(-1)`) |
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Answer» Correct Answer - 83.2 `-DeltaG^(@)=2.303RT" log K"," "474780=2.303xx8.314xx298" log K or log K"=83.2` |
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| 189. |
Given: (i) `Cu^(2+)+2e^(-)toCu," "E^(o)=0.337V` (ii) `Cu^(2+)+e^(-)toCu^(+)," "E^(o)=0.153V` Electrode potential, `E^(o)` for the reaction, `Cu^(+)+e^(-)toCu`, will beA. 0.52 VB. 0.90 VC. 0.30 VD. 0.38 V |
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Answer» Correct Answer - A `DeltaG^(@)=nFE^(@)` For, `Cu^(2+)+2e^(-)toCu,DeltaG_(1)^(@)=-2F(0.337)` . . . . (i) `Cu^(2+)+e^(-)toCu^(+),DeltaG_(2)^(@)=-F(0.153)` . . . (ii) Subtracting (ii) from (i), `Cu^(+)+e^(-)toCu` `DeltaG^(@)=-0.674F+0.153F` . . . (iii) `nFE^(@)=-0.521F` `because`For (iii), n=1, `thereforeE^(@)=0.521V`. |
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| 190. |
For the reduction of silver ions with copper metal, the standard cell potential was found to be `+0.46`V at `25^(@)C`. The value of standard Gibbs energy,` DeltaG^(@)` will be `(F=96500Cmol^(-1))`A. `-98.0kJ`B. `-89.0kJ`C. `-89.0J`D. `-44.5kJ` |
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Answer» Correct Answer - B `2Ag^(+)+Cuto2Ag+Cu^(2+)` `n=2` `DeltaG=-nFE_(cell)` `DeltaG=-2xx96500xx0.46` joul `DeltaG=-88.78kJ=-89kJ`. |
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| 191. |
`E_(cell)=0.78` volt for the following cell.`underset(Fe_((s))|Fe_((aq))^(2+))||underset((0.01M)(Cu_((aq))^(2+)|Cu_((s)))` `E_(Fe//Fe^(2+)(aq))^(@)=0.44V,E_(Cu//Cu^(2+)(aq))^(@)=-0.34V`A. x cannot be predictedB. x=0.01MC. `xgt0.01M`D. `xlt0.01M` |
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Answer» Correct Answer - B If a solution contains two or more catios, the cation to be deposited on the cathode is the one which has higher reduction potential. Here the reaction will be `Fe+Cu^(2+)toFe^(2+)+Cu` `E_(cell)=E_(cell)^(o)-(0.0591)/(n)"log"([Fe^(2+)])/([Cu^(2+)])` `0.78=0.78-(0.591)/(n)"log"[(Fe^(2+))/(Cu^(2+))]` or `0="log"(x)/(0.01)implies(x)/(0.01)=1impliesx=0.01`. |
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| 192. |
The extent to which an electrolyte can dissociate inot ions is called _____ |
| Answer» Correct Answer - degree of dissociation | |
| 193. |
Electrolytes when dissolved in water dissociate into ions because:A. they are unstableB. tge water dussolves it.C. the force of repulsion increasesD. the force of electrostatic attraction is broken down by water. |
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Answer» Correct Answer - d |
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| 194. |
What is the volume of `O_(20` liberated at anode at `STP` in the electrolysis of `CdSO_(4)` solution when a current of `2 A` is passed for `8 m i n ?` |
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Answer» Aqueous `CdSO_(4)overset(el ectrolysis)rarr Cd^(2+)+SO_(4)^(2-)` Number of Faradays `=(Ixxts)/(96500C)=(2Axx8xx60s)/(96500C)` `=9.0099~~0.001 F` In aqueous solution, oxidation of `H_(2)O` takes place than that of `SO_(4)^(2-)` ions, since oxidation potential of `H_(2)Ogt` oxidation potential of `SO_(4)^(2-)`. So oxidation of `H_(2)O` at anode occurs. `2H_(2)O rarr O_(2)+4H^(o+)+4e^(-)` Frist method `4e^(-)=4F-= 1 mol O_(2)=22.4 L `at `STP` `:. 0.001F=(22.4Lxx0.001F)/(4F)` `=0.056L of O_(2)` Second method `1F=1` equivalent of `O_(2)=(22.4L)/(4)O_(2)` `[{:(Equivalent of O_(2)),(=(Volume of 1 mol of a gas)/(n fact o r)),(=(22.4)/(4)=5.6LO_(2)),(n fact o r f o r O_(2)=4) :}]` `2O^(2-) rarr O_(2)+4e^(-)` `:. 1 F =5.6 L `or `O_(2)`at `STP` `0.001 F=5.6 xx 0.001 =0.056 L O_(2)` |
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| 195. |
What is the amount of `Al` deposited on the electrolysis of molten `Al_(2)O_(3)` when a current of `9.65 A` is passed for `10.0 s` . |
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Answer» Molten `Al_(2)O_(3)overset(El e trolysis)rarr 2Al^(3)+3O^(2-)` Number of Faradays `=(Ixxts)/(96500C)` `=(9.65Axx10s)/(96500C)=10^(-3)F` First method Reduction of `Al^(3+) ` at cathode `Al^(3+)+3e^(-) rarr Al` `3e^(-)=3F=1 mol Al=27g Al` `:. 10^(-3)F=(27)/(3)xx10^(-3)g` `=0.009g Al ` deposited Second method `1F=1` equivalent of `Al=(1 mol)/(Charg e)=(27g)/(3)=9g` `:. 10^(-3)F=9 xx 10^(-3)g Al` deposited |
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| 196. |
An acidic solution of `Cu^(2+)` salt contaning `0.4` of `Cu^(2+)` is electrolyzed untill all the copper is deposited. The electrolysis is continued for seven more minutes with the volume of solution kept at 100mL and the current at `1.2` amp. Calculate the volume of gases evolved at NTP during the entire electrolysis. |
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Answer» Correct Answer - `V(O_(2)) = 99.68mL V(H_(2)) = 58.46 mL`, Total vol `=158.1 mL` In beginning At anode: `2e^(-) +Cu^(+2) rarr Cu` At anode: `2H_(2)O rarr 4H^(+) +O_(2) +4e^(-)` Eq. of `O_(2) = Eq. of Cu^(+2)` `=(0.4 xx 2)/(63.5) = 0.012598` vol of `O_(2) = (0.012598 xx 22400)/(4) = 70.55mL` Later on: At anode: `2H_(2)O +2e^(-) rarr H_(2) +2OH^(-)` At anode: `2H_(2)O rarr 4H^(+) +O_(2) +4e^(-)` Faraday passed `= (1.2 xx 7 xx 60)/(96500) = 0.00522` moles of `O_(2) = (0.00522)/(4)` moles of `H_(2) = (0.00522)/(2)` Volume of `O_(2) = (0.00522)/(4) xx 22400 = 29.24mL` Volume of `H_(2) = (0.00522)/(2) xx 22400 = 58.464 mL` Total volume of `O_(2)` `= 29.24 + 70.55 = 99.79 mL` Total volume of `H_(2)` `= 58.464 mL` |
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| 197. |
Calculate the number of Faradays required to electrolyze `6.35g` of `Cu^(o+)(aq)` ions from an aqueous solution. |
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Answer» Reduction of `Cu^(o+)` at cathode `Cu^(o+)+1e^(-) rarr Cu` `1e^(-)=1F=1 mol `of `Cu=63.5g `of `Cu` `:. 6.35 g ` of `Cu` will be deposited `(1F xx 6.35g)/(63.5g)=0.1F` `=0.1 F` |
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| 198. |
`100mL` of `0.3 M Fe^(3+)(aq)` ions were electrolyzed by a charge of `0.072F.` In electrolysis, metal was deposited and `O_(2)(g)` was evolved. At the end of electrolysis, it is desired to oxidize the un`-`electrolyzed metal ion. `Fe^(3+)+e^(-) rarrFe^(2+)` `Fe^(2+)+2e^(-)rarrFe` The moles of `Fe^(2+)` ions left un`-`electrolyzed in the solution is `a. 0.009" "b.0.021" "c.0.072" "d.0.042` |
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Answer» `a.` mmoles of `Fe^(3+)=100xx0.3=30` Charge `=0.072F=0.072xx10^(3)=72mF` `i. Fe^(3+)+e^(-) rarr Fe^(2+)` `30` mmolees of `Fe^(3+)` requires `30mF` and `30` mmoles of `Fe^(2+)` formed. `ii. mF` left`=72-30=42mF` `iii. Fe^(2+)+2e^(-) rarr Fe(s)` `42mF` will electrolyze `-21 m mol` of `Fe^(2+)` `[` Since `2e^(-)=2F` or `2mF=1 mmol` of `Fe^(2+)=1mmol `o f `Fe]` `:.` mmoles of `Fe^(2+)` left `=30-21` `9 mmol` `=9xx10^(-3)=0.009 mol` |
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| 199. |
The gas evolved at the anode when `K_(2)SO_(4)(aq)` is electrolyzed between `Pt` electrode isA. `O_(2)`B. `H_(2)`C. `SO_(2)`D. `SO_(3)` |
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Answer» Correct Answer - a Electrolysis of `H_(2)O` takes place `:` `2H_(2)O rarr O_(2)+2H^(o+)+4e^(-)` |
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| 200. |
When dilute `H_(2)SO_(4)` is electrolyzed between `Pt` electrodes, the gas liberated at the anode will be`………………………….`A. `H_(2)`B. `SO_(4)^(2-)`C. `SO_(2)`D. `O_(2)` |
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Answer» Correct Answer - D (d) `O_(2)` is liberated at anode. |
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