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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. |
An excess of liquid mercury is added to an acidicfied solution of `1.0xx10^(-3) M Fe^(3+)` . It is found that `5%` of `Fe^(3+)` remains at equilibrium at `25^(@)C`. Calculate `E^(c-)._((Hg_(2)^(2+)|Hg))` assuming that the only reaction that occurs is `2Hg+2Fe^(3+) rarr Hg_(2)^(2+)+2Fe^(2+)` Given `: E^(c-)._((Fe^(3+)|Fe^(2+)))=0.77V` |
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Answer» For `2Hg+2Fe^(3+)hArrHg_(2)^(2+)+2Fe^(2+)` `{:("before reaction Excess 10"^(-3)),("after reaction Excess 10"^(-3)xx(5)/(100)):}` `{:(" 0 0"),((95)/(2xx100)xx10^(-3)" "(95)/(100)xx10^(-3)):}` For cell at equilibrium `E_(cell)=0=E_(OP_(Hg//Hg_(2)^(2+)))+E_(RP_(Fe^(3+)//Fe^(2+)))` `0=E_(OP_(Hg//Hg_(2)^(2+)))^(0)-(0.059)/(2)log_(10)[Hg_(2)^(2+)]+E_(RP_(Fe^(3+)//Fe^(2+)))^(0)+(0.059)/(2)log_(10).([Fe^(3+)]^(2))/([Fe^(2+)]^(2))` `0=E_(OP_(Hg//Hg_(2)^(2+)))^(0)+0.77+(0.059)/(2)log_(10).([Fe^(3+)])/([Fe^(2+)]^(2)[Hg_(2)^(2+)])` `(becauseE_(OP_(Fe^(2+)//Fe^(3+)))^(0)=-0.77VthereforeE_(RP_(Fe^(3+)//Fe^(2+)))^(0)=+0.77V)` or `E_(OP_(Hg//Hg_(2)^(2+)))^(0)=-0.771-(0.059)/(2)` `([(5)/(100)xx10^(-3)])/([(95xx10^(-3))/(100)]^(2)[(95xx10^(-3))/(2xx100)])=-0.793V` |
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| 152. |
The electrochemical equivalent of an element is `0.001118`gm/coulomb. Its equivalent weight isA. `10.7`B. `53.5`C. 1007D. 107 |
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Answer» Correct Answer - D `e=(E)/(F)rArrE=eF` |
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| 153. |
In the electrochemical cell `H_(2)(g),"1 atm"|H^(+)(1M)"||"Cu^(2+)(1M)|Cu(s)`, which one of the following statements is true?A. `H_(2)` is cathode, Cu is anodeB. Oxidation occurs at Cu electrodeC. Reduction occurs at `H_(2)` electrodeD. `H_(2)` is anode, Cu is Cathode |
| Answer» Correct Answer - D | |
| 154. |
Which graph correctly correlates `E_(cell)` as a function of concentration for the cell `Zn(s)+2Ag^+(aq)toZn^(2+)(aq)+2Ag(s), E_(cell)^(@)=1.56V` y-axis: `E_(cell)`, X-axis: `log_(10)"([Zn^(2+)])/([Ag^+]^2)`A. B. C. D. |
| Answer» Correct Answer - B | |
| 155. |
Given the cell: `Cd(s)|Cd(OH)_2(s)|NaOH(aq,0.01M)|H_2(g,1"bar")|Pt(s)` with `E_(cell)=0.0V."if"E_(Cd^(2+)|Cd)^(@)=-0.39V,"then"K_(sp)"of"Cd(OH_2)`is:A. `0.1`B. `10^(-13)`C. `10^(-15)`D. `10^(-18)` |
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Answer» Correct Answer - C `E_(cell)=E_(H^(+)//H_(2))-E_(OH^(-)|Cd(OH)_(2)|Cd)` `=E_(H^(+)|H_(2))-E_(OH^(-)|Cd(OH)_(2)|Cd)=E_(H^(+)|H_(2))-E_(OH_(Cd^(2+)|Cd))` `E_(cell)=E^(@)-(0.06)/(2)log.([cd^(2+)])/([H^(+)]^(2))` `E_(cell)="0 , E"_(cell)^(0)=0.31log.([cd^(2+)][OH^(-)]^(2))/(K_(w)^(2))` `log.(K_(sp))/(K_(w)^(2))=(0.09)/(0.03)=13,K_(sp)=10^(13)xx(10^(-14))^(2)=10^(-15)` |
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| 156. |
You are given the followin cell at `298 K, Zn|(Zn^(+ +) ._((aq.))),(0.01M)||(HCl_((aq.))),(1.0lit)||(H_(2)(g)),(1.0atm)|Pt` with `E_(cell)=0.701` and `E_(Zn^(2+)//Zn)^(0)=-0.76V`. Which of the following amounts of `NaOH (` equivalent weight `=40 )` will just make the `pH` of cathodic compartment to be equal to `7.0 :`A. `0.4 g`B. `4g`C. `10g`D. `2g` |
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Answer» Correct Answer - A `{:("Anode",,ZnrarrZn^(2+)+2e^(-)),("Cathode",,2H^(+)+2e^(-)rarrH_(2)),("Cell:",,Zn+2H^(+)hArrH_(2)+Zn^(2+)" "E_(cell)^(@)=0-(-0.76)=0.76V):}` `:." "0.701=0.76-(0.059)/(2)log.(0.01xx1)/([H^(+)]^(2))` `:." "[H^(+)]=10^(-2)M` `:. " "NaOH` required is `0.01` mole `=0.4 ` grams. |
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| 157. |
Determine the potential of the following cell: `Pt|H_2(g,0.1 "bar")|H^+(aq,10^(-3)M"||"MnO_4^-)(aq),0.1M)` `Mn^(2+)(aq,0.01M),H^+(aq,0.01M)|Pt` Given : `E_(MnO_4^(-)|Mn^(2+))^(@)=1.51V`A. 1.54 VB. 1.48 VC. 1.84 VD. 1.68 V |
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Answer» Correct Answer - B Anode : `H_(2)(g)rarr2H^(+)(aq)+2e^(-),E^(@)=0.0V` Cathode : `MnO_(4)^(-)(aq)+8H^(+)(aq)+5e^(-)rarrMn^(2+)(aq)+4H_(2)O(l),E^(@)+1.51V` `E_(cell)=E_("RP(RHS))-E_(RP(LHS))` `rArr[1.51-(0.06)/(5)log.([Mn^(2+)])/([MnO_(4)^(-)][H^(+)]^(8))]` `-[0-(0.06)/(2)log.(P_(H_(2)))/([H^(+)]^(2))]` `=1.51-[(0.06)/(5) log. (0.001)/(0.1xx(10^(-2))^(8))]+(0.06)/(2)log.[(0.1)/((10^(-3))^(2))]` `=1.48V` |
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| 158. |
Calculate the e.m.f. of the cell `Pt|H_(2)(1.0atm)|CH_(3)COOH (0.1M)||NH_(3)(aq,0.01M)|H_(2)(1.0atm)|Pt` `K_(a) (CH_(3)COOH) =1.8 xx 10^(-5), K_(b),(NH_(3)) = 1.8 xx 10^(-5)`A. `-0.92 V`B. `-0.46 V`C. `-0.35 V`D. `-0.20 V` |
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Answer» Correct Answer - B Pt, `H_(2)(1atm)//CH_(3)COOH(0.1M)"//" NH_(3)(aq,0.01M)//H_(2)(1atm)Pt` `K_(a)(CH_(3)COOH)=1.8xx10^(-5)` `K_(b)(CH_(3)=1.8xx10^(-5)` `{:(A:(1)/(2)H_(2)-e-rarrH^(+)" "E^(@)=0),(C:H^(+)+e-rarr(1)/(2)H_(2)(g)" "E_("cell")^(@)=0),(-------------),(E_("cell")=E_("cell")^(@)-(0.06)/(1)"log"([H^(+)]_(A))/([H^(+)]_(C ))):}` `[OH^(-)]^(2)=0.01xx1.8xx10^(-5)` `[OH^(-)]=4.2xx10^(-4),[H^(+)]_(C )=(10^(-14))/(4.2xx10^(-4))` Similarly `[H^(+)]^(2)=1.8xx10^(-5)xx0.1` `[H^(+)]_(A)=sqrt(1.8xx10^(-6))` `E_("cell")=-0.0591xx7.78=-0.46 v` |
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| 159. |
The specific conductance of saturated solution os silver chloride is `k(ohm^(-1)cm^(-1))`. The limiting ionic conductance of `Ag^(+)` and `Cl^(-)` ions are x and y respectively. The solubility of AgCl in `"gram liter"^(-1)` is : (Molar mass of `AgCl=143.5"g mol"^(-1)`)A. `kxx(1000)/(x-y)`B. `(k)/(x+y)xx143.5`C. `(kxx1000xx143.5)/(x+y)`D. `(x+y)/(k)xx(1000)/(143.5)` |
| Answer» Correct Answer - C | |
| 160. |
`E^(c-)` of `Mg^(2+)|Mg,Zn^(2+)|Zn`, and `Fe^(2+)|Fe` are `-2.37V,-0.76V`, and `-0.44V`, respectively. Which of the following is correct ?A. Mg oxidizes FeB. Zn oxidizes FeC. Zn reduces `Mg^(2+)`D. Zn reduces `Fe^(2+)` |
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Answer» Correct Answer - D Low SRP metal reduces metal with high SRP. |
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| 161. |
Calculate the cell potential of half cell having at the reaction : `M_(2)S+2e^(-)rarr2M+S^(2-)` at `27^(@)C` in a solution of pH = 3 and saturated with `0.1 MH_(2)S` for `H_(2)SK_(1)=10^(-8)` and `K_(2)=10^(-13)Ksp (M_(2)S)=1.0 ,10^(-50)E_(M^(+)//M)^(@)=1V` assume R = 10 J/K/mol and `(2.303RT)/(nF)=(0.07)/(n)` Express the magnitued of your answer after multiplication with 25 and as nearest highest whole number. |
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Answer» Correct Answer - 6 `M_(2)S hArr 2M^(+)+S^(2)Delta G_(1)^(@)` `{:(" "2M^(+)+2e^(-)rarr2M Delta G_(2)^(@)),(bar(M_(2)S+2e^(-)rarr2M+S^(2-)Delta G_(3)^(@))):}` `Delta G_(3)^(@)=Delta G_(1)^(@)+Delta G_(3)^(@)` `Delta G_(1)^(@)=-2.303xx10xx300 log 10^(-50)` `=2.303xx3000xx50` `=34.545xx10^(4)J` `Delta G_(2)^(@)=-2xx96.500xx1=-193000` `Delta G_(3)^(@)=-19300+345450` `Delta G_(3)^(@)=152450-nFE_(3)^(@)=152450` `E_(3)^(@)=(-152450)/(2xx96,500)=-(15245)/(19300)` `=-0.79` `~~-0.8` `E_("cell")=E_("cell")^(@)-(0.07)/(n)log [S^(2-)]` `=-0.8-(0.07)/(2)log 0.2` `K_(1)K_(2)=([H^(+)]^(2)[S^(2-)])/([H_(2)S])` `=(10^(-6)[S^(2-)])/(0.1)=[S^(2-)]=10^(-16)` `E_("cell")=-0.8-(0.07)/(2)log 10^(-16)` `=-0.8+(0.07)/(2)xx-16` `=-0.8+0.07xx8 = -0.24V` After multiplication with 25, the magnitude of above answer becomes 6 |
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| 162. |
Calculate the quantity of electricity that would be required to reduce 12.3g of nitrobenzene to aniline, if current efficiency is 50% . If the potential drops across the cell is 3.0V, how much energy will be consumed ? |
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Answer» `C_(6)H_(5)NO_(2)+6H^(+)+6e rarrC_(6)H_(5)NH_(2)+2H_(2)O` `N^(3+)+6e rarr N^(3-)` `therefore` Eq. wt. of nitrobenzene `=(M)/(6)=(123)/(6)` Since current efficiency is 50% `therefore i=(50i_(0))/(100)` Now `w=("E.i.t")/(96500), 12.3=(123xxi_(0)xxtxx50)/(6xx100xx96500)` `therefore i_(0)xx t=115800` coulomb Now energy used `QxxV=115800xx3=347.4kJ` |
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| 163. |
The Edison storage cell is represented as, `Fe_((s))|FeO_(S)||KOH_((aq.))||Ni_(2)O_(3(S))|No_((S))` the half-cell reactions are : `Ni_(2)O_(3(S))+H_(2)O_((l))+2e^(-)rarr 2NiO_((S))+2OH^(-) , E^(@)=+0.40V` `FeO_((S))+H_(2)O_((I))+2e^(-)rarr Fe_((S))+2OH^(-) , E^(@)=-0.87V` (i) What is the cell reaction ? (ii) What is the cell e.m.f. ? How does ir depend on the concentration of KOH ? (iii) What is the maximum amount of electrical energy that can be obtained from one mole of `Ni_(2)O_(3)` ? |
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Answer» Given, `E_(FeO //Fe)^(@)=-0.87V,E_((Ni_(2)O_(3)//NiO))=+0.40V` `thereforeE_(FeO //Fe)^(@)=+0.887V,E_(NiO //Ni_(2)O_(3))^(@)=-0.40V` Since, `E_(OP)^(@)" for Fe"//FeOgtE_(Op)^(@)" for NiO/Ni"_(2)O_(3)`, and thus, redox change are, At anode `Fe_((S))+2OH^(-)rarrFeO_((S))+underset(("oxidation"))(H_(2)O_((l)))+2e` At cathode `Ni_(2)O_(3(S))+H_(2)O_((l))+2erarrunderset(("reduction"))(2NiO_((S))+2OH^(-))` Redox reaction: `Fe_((S))+Ni_(2)O_(3(S))rarrFeO_((S))+2NiO_((S)` i) `E_(cell)==E_(OP_(Fe//FeO))^(@)-(0.059)/(2)log_(10).([H_(2)O])/([OH^(-)]^(2))+E_(RP_(Ni_(2)O_(3)//NiO))^(@)+(0.059)/(2)log_(10).([H_(2)O])/([OH^(-)]^(2))` `E_(RP_(Ni_(2)O_(3)//NiO))^(@)=0.87+0.40=1.27V` ii) The `E_(cell)` is independent of `OH^(-)` ion concentration. iii) `{:(-DeltaG^(0)=nE^(@)F,=2xx1.27xx96500),(,=245.11J),(,=245.11kJ):}` |
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| 164. |
Copper sulphate solution `(250 ML)` was electrolyzed using a platinum anode and a copper cathode. A constant current of `2mA` was passed for `16 mi n`. It was found that after electrolysis the absorbance of the solution was reducted to `50%` of its original value . Calculate the concentration of copper sulphate in the solution to begin with. |
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Answer» We known, Equivalent of `Cu^(2+)` lost during electrolysis `=(ixxt)/(96500)=(2xx10^(-3)xx16xx60)/(96500)=1.989xx10^(-5)` or mole of `Cu^(2+)` lost during electrolysis `= (1.989xx10^(-5))/(2)` This value is 50% of the intial concentration of solution. Thus, initial mole of `CuSO_(4)=(2xx1.989xx10^(-5))/(2)` `=1.989xx10^(-5)` Thus, initial concentration of `CuSO_(4)=(1.989xx10^(-5)xx1000)/(250) [CuSO_(4)]=7.95xx10^(-5)M` |
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| 165. |
The standard reduction potential of the `Ag^(o+)|Ag` electrode at `298K` is `0.799V`. Given that for `AgI,K_(sp)=8.7xx10^(-17)`, evaluate the potential of the `Ag^(o+)|Ag` electrode in a saturated solution of `AgI`. Also calculate the standard reduction potential of the `I^(c-)`|Agl|Ag` electrode. |
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Answer» We know, `becauseE_(Ag^(+)//Ag)=E_(Ag^(+)//Ag)+(0.059)/(1)log_(10)[Ag^(+)]`...(1) Also `K_(SP_(Agl))=[Ag^(+)][I^(-)]` `because[Ag^(+)]=[I^(-)]` (for a saturated solution) `therefore[Ag^(+)]=sqrt(K_(SP_(Agl)))sqrt(8.7xx10^(-17))=9.32xx10^(-9)`...(2) `therefore` By Eq. (1) `E_(Ag^(+)//Ag)=0.799+(0.059)/(1)log_(10)(9.32xx10^(-9))` `=0.799-0.474=0.32V` Also `AgrarrAg^(+)+e,E_(OP)^(0)=-0.799V` `(AgI_((S))+erarrAg+I^(-))/(AgIrarrAg^(+)+I^(-))` `thereforeE_(cell)=E_(OP_(Ag//Ag^(+)))^(0)-(0.059)/(1)log[Ag^(+)]+E_(RP_(I^(-)//AgI//Ag))^(0)+(0.059)/(1)log.(1)/([I^(-)])`...(3) `becauseE_(cell)=0` at equilibrium, thus, from Eq. (3) `E_(O_(Ag//Ag^(+)))^(0)+E_(RP_(I^(-)//AgI//Ag))^(0)=(0.059)/(1)log[Ag^(+)][I^(-)]` `=(0.059)/(1)logK_(SP_(AgI))` `-0.799+E_(RP_(I^(-)//AgI//Ag))^(0)=(0.059)/(1)log8.7xx10^(-17)` `orE_(RP_(I^(-)//AgI//Ag))^(0)=-0.948+0.799=-0.149V` |
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| 166. |
A quantity of electrcity required to reduce `12.3` g of nitrobenzene to aniline arising `50%` current efficiency isA. 115800CB. 579000CC. 231600CD. 289500C |
| Answer» Correct Answer - A | |
| 167. |
The cathode reaction in electrolysis of dilute sulphuric acid with Platinum electrode isA. OxidationB. ReductionC. Both oxidation and reductionD. Neutralization |
| Answer» Correct Answer - B | |
| 168. |
The potential across the metal and the aqueous solution of its ions of unit activity at 298K is known asA. Electrode potentialB. Standard electrode potentialC. Formal electrode potentialD. Oxidation potential |
| Answer» Correct Answer - B | |
| 169. |
Resistance of ` 0.2 M` solution of an electrolue is ` 50 Omega`. The specific conductance of the solution is ` 1.4 S m^^(-1)`. The resistance of `0.5` M solution of the same electrolyte is `280. Omega`. The molar conducitivity of ` 0.5 M` solution of the electrolyte is ` S m^2 "mol"^(-1)` is.A. `5xx10^(3)`B. `5xx10^(2)`C. `5xx10^(-4)`D. `5xx10^(-3)` |
| Answer» Correct Answer - C | |
| 170. |
During the electrolysis of aqueous solution of sodium chloride, pH of the electrolyteA. Remains constantB. Gradually incresesC. Gradually decreasesD. Decreases first and then increases. |
| Answer» Correct Answer - B | |
| 171. |
In the electrolysis of an aqueous potassium sulphate solution, the `Ph` of the solution in the space near an electrode increased. Which pole of the current source is the electrode connected to ?A. The positive poleB. Could be either poldeC. The negative poleD. Cannot be determined |
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Answer» Correct Answer - C `K_(2)SO_(4)hArr2K^(+)+SO_(4)^(3+)` `H_(2)OhArr H^(+)+OH^(-)` `(pHdarr)`Anode`2OH^(-)rarrH_(2)O+(1)/(2)O_(2)+2e^(-)` `(pHuarr)`Cathode`2H^(+)+2e^(-)rarrH_(2)O` |
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| 172. |
The electrode potential measures theA. tendency of the electrode to gain or lose elctronsB. electron affinity of elementsC. difference in the ionization potential of electrode and metal ionD. heat of combustion |
| Answer» Correct Answer - A | |
| 173. |
The metal that cannot obtained by electrolysis of an aqueous solution of its salts is `:`A. CuB. CrC. AgD. Ca |
| Answer» Correct Answer - D | |
| 174. |
Assertion (A): The extent of dissociation is different for different electrolytes Reason (R): The extent of dissociation is dependent on nature of electrolyte and independent on nature of electrolyte.A. A and R are correct R is the correct explanation of AB. A and R are correct R is not the correct explanation of AC. A is correct, but R is worngD. A is wrong, but R is correct |
| Answer» Correct Answer - C | |
| 175. |
The metal which cannot liberate `H_(2)` gas from hydrochloric acidA. ZnB. CuC. MgD. Al |
| Answer» Correct Answer - B | |
| 176. |
Assertion A: Copper does not liberate hydrogen from the solution of dilute hydrochloric acid. Reason (R): Hydrogen is below copper in the electrochemical series.A. A and R are correct R is the correct explanation of AB. A and R are correct R is not the correct explanation of AC. A is correct, but R is worngD. A is wrong, but R is correct |
| Answer» Correct Answer - C | |
| 177. |
Assertion (A): Equivalent weights of NaCl, NaOH, KCl, KBr etc. are equal to their molecular weights Reason (R): Only one electron take part in electrode reaction.A. A and R are correct R is the correct explanation of AB. A and R are correct R is not the correct explanation of AC. A is correct, but R is worngD. A is wrong, but R is correct |
| Answer» Correct Answer - A | |
| 178. |
Assertion(A): The absolute value of the electrode potential cannot be determined experimentally Reason (R): The electrode potentials are generally determined with respect to standard hydrogen electrodes.A. A and R are correct R is the correct explanation of AB. A and R are correct R is not the correct explanation of AC. A is correct, but R is worngD. A is wrong, but R is correct |
| Answer» Correct Answer - A | |
| 179. |
Assertion (A): `E^(@)` value of single electrode is determined experimentally by combing the electrode with SHE Reason (R): SHE is taken as a reference electrodeA. A and R are correct R is the correct explanation of AB. A and R are correct R is not the correct explanation of AC. A is correct, but R is worngD. A is wrong, but R is correct |
| Answer» Correct Answer - A | |
| 180. |
During the working of a galvanic cell and with the passage of time : a) spontaneity of the cell reaction decreases, `E_(cell)` decreases b) reaction quotient Q decreases, `E_(cell)` decreases c) Reaction quotient Q increases, `E_(cell)` decreases d) At equilibrium, `Q =K_(e),E_(cell)=0` Corrent statements areA. b,c,dB. a,dC. a,c,dD. All |
| Answer» Correct Answer - C | |
| 181. |
To the Daniel cell `ZnSO_(4)` is added to the left hand side electrode. Then cell emfA. IncreasesB. DecreasesC. Doest not changeD. First increases & then decreases |
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Answer» Correct Answer - B In Daniel cell `ZnSO_(4)` is added to LHE, EMF, decreases. |
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| 182. |
The life span of a Daniel cell may increased byA. Large Cu electrodeB. Lowering of `CuSO_(4)` concentrationC. Lowering of `ZnSO_(4)` concentrationD. Large zinc electrode |
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Answer» Correct Answer - D Life span can be increased by increasing the size of anode electrode |
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| 183. |
In fuel cell oxidants used areA. `O_(2)`B. `H_(2)O_(2)`C. `HNO_(3)`D. All |
| Answer» Correct Answer - D | |
| 184. |
Following are some of the facts about a dry cell I : It is alos called Leclanche cell II : It is also called Daniel cell III : Electrolyte is a moist paste of `NH_(4)Cl` and `ZnCl_(2)` in starch IV : Cathodic reaction is `2MnO_(2)(s)+2NH_(4)^(+)(aq)+2e^(-)` `rarrMn_(2)O_(3)(s)+2NH_(3)(g)+H_(2)O(l)` Select correct facts :A. I, II,IIIB. I, IVC. I, III, IVD. II, III, IV |
| Answer» Correct Answer - C | |
| 185. |
In Leclanche cell, Zinc rod is placed inA. `10%NH_(4)Cl`B. `20%NH_(4)Cl`C. `30% NH_(4)Cl`D. `40% NH_(4)Cl` |
| Answer» Correct Answer - B | |
| 186. |
Cathode is made of …………in mercury batteryA. ZnB. ZnOC. Carbon in contact with HgOD. Zn in contact with HgO |
| Answer» Correct Answer - C | |
| 187. |
W.r.t. Ni - Cd storage cell, the incorrect statement isA. anode is cadmium metalB. it is a primary is `1.4`VC. cell potential is `1.4V`D. electrolyte used is KOH |
| Answer» Correct Answer - B | |
| 188. |
Alkali storage cell is commonly calledA. lead accumulatorB. Edison batteryC. fuel cellD. Leclanche shell |
| Answer» Correct Answer - B | |
| 189. |
Which of the following metals acts as a sacrificial anode for iron articles?A. CuB. ZnC. AgD. Sn |
| Answer» Correct Answer - B | |
| 190. |
The standard emf for the cell cell reaction ` Zn + Cu^(2+) rarr Zn^(2+) + Cu ` is 1.10 volt at ` 25^@ C`. The emf for the cell reaction when ` 0.1 M Cu^(2+)` and ` 0.1 M ZN^(2+)` solutions are used at `25^@ =C` is .A. 1.10VB. 0.110 VC. `-1.10V`D. `-0.110V` |
| Answer» Correct Answer - A | |
| 191. |
The equivalent conductivity of a solution containing 2.45 g of `CuSO_(4)` per litre, is `91.0 Omega^(-1) cm^(2) eq^(-1)`. Its conductivity would beA. `1.45xx10^(-3)Omega^(-3)cm^(-1)`B. `2.17xx10^(-3)Omega^(-3)cm^(-1)`C. `2.90xx10^(-3)Omega^(-1)cm^(2)`D. `2.9xx10^(-3)Omega^(-1)cm^(-1)` |
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Answer» Correct Answer - D `N=(2.54)/(254)" "lambda=(Kxx1000)/(N)` |
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| 192. |
What happens at infinite dilution in a given solution?A. The degree of dissociation is unity for wead electrolytesB. The electrolyte is `100%` ionisedC. All inter ionic attractions disappearD. All the three |
| Answer» Correct Answer - D | |
| 193. |
If `X` is the specific resistance of the solution and `N` is the normality of the solution, the equivalent conductivity of the solution is given byA. `(1000x)/(N)`B. `(1000)/(Nx)`C. `(1000N)/(x)`D. `(Nx)/(1000)` |
| Answer» Correct Answer - B | |
| 194. |
Which of the following solid is an electronic conductorA. NaClB. DiamondC. CuSD. KCl |
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Answer» Correct Answer - C Salts like CuS and CdS act as electronic conductros due to crystal defects. |
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| 195. |
The units of conductivity of the solution areA. `ohm^(-1)`B. ohmsC. `ohm^(-1)cm^(-1)`D. `ohm^(-1)eq^(-1)` |
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Answer» Correct Answer - C units of conductivity of solution is `rarrohm^(-1)cm^(-1)` `k=cxx(l)/(a)=ohm^(-1)xx(cm)/(cm^(2))` |
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| 196. |
Cell constant has unit `m^(-1)`.A. `ohm^(-1)`B. `ohm-cm`C. `cm^(-1)`D. `ohm^(-1) cm^(2) eq^(-1)` |
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Answer» Correct Answer - C Cell constant `=(l)/(a)=(cm)/(cm^(2))=cm^(-1)` |
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| 197. |
The cell constant is the product of resistance andA. conductanceB. molar conductanceC. specific conductanceD. specific resistance |
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Answer» Correct Answer - C Cell constant = resistance X specific conductance. |
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| 198. |
The equation representing Kohlrausch law from the following is `(V^(+)="No. of cations,"V^(-)="No. of anions")`A. `lambda_(m)=(100K)/(C_(m))(V^(+)+V^(-))`B. `lambda_(m)^(0)=v^(+)lambda_(+)^(0)+v^(-).lambda_(-)^(0)`C. `lambda_(eq)=(1000K)/(C_(eq))(V^(+)-V^(-))`D. `lambda_(m)^(0)=lambda_(c)+lambda_(a)` |
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Answer» Correct Answer - B `Lambda_(c)=Lambda_(0)-bsqrtc` |
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| 199. |
If the specific conductance and conductance of a solution are same, then its cell constant is equal to:A. 1B. 0C. 10D. 100 |
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Answer» Correct Answer - A K= conductance X cell constant Here K = C, then cell constant = 1 |
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| 200. |
The expression showing the relationship between equivalent conductance and molar conductance is (z = Total positive (or) negative charge per formula unit of electrolyte)A. `lambda_(m)=Zxxlambda_(eq)`B. `lambda_(eq)=Zxxlambda_(m)`C. `lambda_(m)=(lambda_(eq))/(Z)`D. `lambda_(m)=lambda_(eq)^(2)` |
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Answer» Correct Answer - A `(lambda_(M))/(lambda_(eq))=("Normality")/("Molarity")` |
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