InterviewSolution
Saved Bookmarks
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.
| 2201. |
Assertion : Electrolysis of NaCl solution gives chlorine at anode instead of `O_(2)`. Reason : Formation of oxygen at anode requires overvoltage.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. |
|
Answer» Correct Answer - A Reason is the correct explanation for assertion. |
|
| 2202. |
At 298 K the resistance of a 0.1M KCl solution is found to be 39.0ohm. If the conductivity (k) of this solution is 1.29`xx10^(-2)ohm^(-1)cm^(-1)` at 298K, what is cell contantA. `5.03xx10^(-1)cm^(-1)`B. `10.06xx10^(-1)cm^(-1)`C. `15.09xx10^(-1)cm^(-1)`D. `2.51xx10^(-1)cm^(-1)` |
|
Answer» Correct Answer - A The cell cosntnat =l/a=R.k =39 ohm 1.29`xx10^(-2) ohm^(-1)cm^(-1)` `=5.03xx10^(-1)cm^(-1)` |
|
| 2203. |
At 298K the resistance of a 0.5N NaOH solution is 35.0 ohm. The cell constant is 0.503 `cm^(-1)` the electrical conductivity of the solution isA. `1.437xx10^(-2)ohm ^(-1)cm^(-1)`B. `1.473ohm^(-1)cm^(-1)`C. `1.06ohm^(-1)cm^(-1)`D. `3.5ohm^(-1)cm^(-1)`. |
|
Answer» Correct Answer - A The electrical conductivity, `k=(1)/(R)xx(l)/(a)` `=(1)/(35)xx0.503=1.437xx10^(-2)ohm^(-1)cm^(-1)` |
|
| 2204. |
Which of the following is concentration cellA. `Cu_((s))//Cu_((aq.1M))^(2+)||||Cu_((aq.1M))^(2+)//Cu_((s))`B. `Cu_((s))//Cu_((aq.0.5M))^(2+)////Cu_((aq.0.5M))^(2+)//Cu_((s))`C. `Zn_((s))//Zn_((aq.0.5M))^(2+)//Cu_((aq.1M))^(2+)//Cu_((s))`D. `.^(Theta)Pt//H_(2_("g 1 bar"))//HCl_(aq" "0.002M)////HCl_((aq.0.005M))//H_(2(g.1"bar"))//Pt^(o+)` |
|
Answer» Correct Answer - D `.^(Theta)Pt//H_(2_(g" 1 bar"))//HCl_((aq" 0.002M")////HCl_((aq." 0.005M"))//H_(2(g.1"bar"))//Pt^(o+)` The above cell represents similar electrode concentration but the concentration similar electrolytes (HCl) are different. This is electrolyte concentration cell. |
|
| 2205. |
At 298K the electrolytic conductivity of a 0.2 M KCl solution is `2.5xx10^(-2) ohm^(-1)cm^(-1)` compute its molar c onductivity.A. 62.5`ohm^(-1) cm^(2)"mol"^(-1)`B. `125ohm^(-1)cm^(2)"mol"^(-1)`C. `250ohm^(-1)cm^(2)"mol"^(-1)`D. `175ohm^(-1)cm^(2)"mol"^(-1)` |
|
Answer» Correct Answer - B Molar conducitivity `=(1000K)/(C) ohm^(-1)cm^(2)"mol"^(-1)` `=(1)/(100)xx0.037=3.7xx10^(-4)ohm^(-1)cm^(-1)` `therefore `Molar conductivity `=(1000xx3.7xx10^(-4))/(0.05)` `=7.4ohm^(-1)cm^(2) "mol"^(-1)` |
|
| 2206. |
What are the factors that affect the conductivity of an ionic (electrolytic) solution? |
|
Answer» The conductivity of anionic (electrolytic) solution depends upon the following factors.
|
|
| 2207. |
The amount of substance liberated when `1 ` ampere of current is passed for 1 second through an electrolytic solution is called`…………………` . |
|
Answer» Correct Answer - Electrochemical equivalent Electrochemical equivalent |
|
| 2208. |
Define molar conductivity of a solution. |
|
Answer» The molar conductivity of a solution at a given concentration is the conductance of volume ‘V’ of a solution containing 1 mole of the electrolytic kept between two electrodes with cross-sectional area ‘A’ and distance of unit length. Or, Δm = \(\frac{\triangle}{1}\) K Now, 1 = 1 and Δ = V (volume containing 1 mole of the electrolyte) ∴ Δm = KV |
|
| 2209. |
Which of the following reaction occurs at the cathode of a common dry cellA. `MntoMn^(2+)+2e^(-)`B. `2MnO_(2)+Zn^(2+)+2e^(-)toZnMn_(2)O_(4)`C. `2ZnO_(2)+Mn^(2+)+2e^(-)toMnZn_(2)O_(4)`D. `ZntoZn^(2+)+2e^(-)` |
|
Answer» Correct Answer - B In common dry cell. Anode: `ZntoZn^(++)+2e^(-)` Cathode: `2MnO_(2)+Zn^(++)+2e^(-)toZnMn_(2)O_(4)`. |
|
| 2210. |
The amount of charge carried by `N^(3-)` ion is `…………………………` |
|
Answer» Correct Answer - `(4.8xx10^(-19)C)` `(4.8 xx 10^(-19)C)` Charge of `e^(-) =1.60xx10^(-19)C` Charge of `N^(3-)=3xx1.6xx10^(-19)C=4.8xx10^(-19)C` |
|
| 2211. |
Which of the following reaction occurs at the cathode during the working of alkaline dry cell ?A. Zn is converted into ZnOB. `MnO_(2)` is converted into `Mn_(2)O_(3)`C. `Zn` is converted into `Zn^(2+)`D. `Mn_(2)O_(3)` is converted into `MnO_(2)` |
|
Answer» Correct Answer - B During alkaline dry cell `MnO_(2)` is converted into `Mn_(2)O_(3)` at cathode . |
|
| 2212. |
The current is carried through metallic conductor by `………………………..` and in electrolytic substance by `…………………………` . |
|
Answer» Correct Answer - Electrons, ions Electrons, ions |
|
| 2213. |
Fill in the blanks with appropriate words. The electrolytic solution is always neutral because the total charge on __(i)___ is equal to __(ii)__ on ___(iii)___ .Unlike the metallic conductor, the electrolyte conducts the electric current by virtue of movement of its ___(iv)___ . The property due to which a metal tends to go into solution in term of positive ions is known as ___(v)__. (i),(ii),(iii),(iv) and (v ) respectively areA. cations, partial charge , anions, electrons , reductionB. cations , total charge , anions , ions , oxidationC. cations , ionic charge , anions , atoms , dissolutionD. cations , partial charge , anions , molecules, electrolysis . |
| Answer» Correct Answer - B | |
| 2214. |
Standard electrode potenital data are useful for understanding the suitablity of an oxidant in a redox titration. Some half cell reactions and their standard potential are given below: `MnO_(4(aq))^(-) + 8H_((aq.))^(+) + 5e rarr Mn_((aq.))^(2+) + 4H_(2)O_(l),E^(@) = 1.51 V` `Cr_(2)O_(7(aq.))^(2-) + 14H_((aq.))^(+) + 6e rarr 2Cr_((aq.))^(3+) + 7H_(2)O_(l), E^(@) = 1.38 V` `Fe_(aq.)^(3+) 2e^(-) rarr Fe_((aq.))^(2), E^(@) = 0.77 V` `Cl_(2(g)) + 2e^(-) rarr 2Cl_((aq.))^(-), E^(@) = 1.40 V` Identify the only incorrect statement regarding the quantitative estimation of aqueous `Fe(NO_(3))_(2)`:A. `MnO_(4)^(-)` can be used in aqueous `HCl`B. `Cr_(2)O_(7)^(2-)` can be used in aqueous `HCl`C. `MnO_(4)^(-)` can be used in aqueous `H_(2)SO_(4)`D. `Cr_(2)O_(7)^(2-)` can be used in aqueous `H_(2)SO_(4)` |
|
Answer» Correct Answer - A `MnO_(4)^(-)` will oxidise `Cl^(-)` ion according to equation. `Mn^(7+) + 5e rarr Mn^(2+)` `2Cl^(-) rarr Cl_(2) + 2e` Thus `E_(cell)^(@) = E_(OP_(Cl^(-)//Cl_(2)))^(@) + E_(RP_Mn^(7+)//Mn^(2))^(@)` `= -1.40 + 1.51 = 0.11 V` or reaction is feasible. `MnO_(4)^(-)` will oxidise `Fe^(2+)` to `Fe^(3+)` `Mn^(7+) + 5e rarr Mn^(2+)` `Fe^(2+) rarr Fe^(3+) + e` `E_(cell)^(@) = E_(OPFe^(2+)//Fe^(3+))^(@) + E_(RPMn^(7+)//Mn^(2+))^(@)` `= -0.77 + 1.51 = 0.74 V` or reaction is feasible. Thus `MnO_(4)^(-)` will not oxidise only `Fe^(2+)` to `Fe^(3+)` in aqueous `HCl` but it will also oxidise `Cl^(-)` to `Cl_(2)`. Suitable oxidant shoould not oxidise `Cl^(-)` to `Cl_(2)` and should oxidise only `Fe^(2+)` to `Fe^(3+)` in redox titration. |
|
| 2215. |
The efficiency of a hypothetical cell is about `84%` which involves the following reactions: `A(s) + B^(2+)(aq) rightarrow A^(2+)(aq) _ B(s)` `DeltaH = -285kJ` Then, the standard electrode potential of the cell will be: (Asume `DeltaS = 0)A. 1.20 VB. 2.40 VC. 1.10 VD. 1.24 V |
|
Answer» Correct Answer - D Efficiency `= (Delta G^(@))/(Delta H^(@)) = (-nFE^(@))/(Delta H^(@))` `0.84 = - (2 xx E^(@) xx 96500)/(-285 xx 1000)` `E^(@) = + 1.24 V` |
|
| 2216. |
Potential for some half cell reactions are given below. On the basis of these mark the correct answer. (i) `H^(+)(aq) + e^(-) rightarrow (1/2)H^(2)(g)` `E_(cell)^(@) = 0.00V` (ii) `2H_(2)O(l) rightarrow O_(2)(g) + 4H^(+)(aq) + 4e^(-)`, `E_(cell) = 1.23V` (iii) `2SO_(4)^(2-)(aq) rightarrow S_(2)O_(8)^(2-)(aq) + 2e^(-), E_(cell)^(@) = 1.96V`A. In dilute sulphuric acid solution, hydrogen will be reduced at cathodeB. In concentrated sulphuric acid solution, water will be oxidised at anodeC. In dilute sulphuric acid solution water will be oxidised at anodeD. In dilute sulphuric acid solution `SO_(4)^(2-)` ion will be oxidised to tetrahionate ion at anode. |
|
Answer» During the electrolysis of dilute sulphuric acid above given three reaction occurs each of which represents particular reaction either oxidation half cell reaction or reduction half cell reaction. Oxidation half cell reactions occur at anode are as follows `2SO_(4)^(2-)(aq)rarrS_(2)O_(8)^(2-)+2e^(-) E_("cell")^(@)=1.96V` `2H_(2)O^(+)(l)rarrO_(2)(g)+4H^(+)(aq)+4e^(-) E_("cell")^(@)=1.23V` Reaction having lower value of `E_("cell")^(@)` will undergo faster oxidation. Hence, oxidation of water occur preferentially reduction half cell reaction occurs at cathode Hence, oxidation of water occur preterentially reduction half cell reaction occurs at cathode `H^(+)(aq)+e^(-)rarr(1)/(2) H_(2 )(g)` at cathode Hence option (a) aned (b) are correct. |
|
| 2217. |
Conductors allow the passage of electric current through them. Metallic and electrolytic are the two type of conductors. Current carriers in metallic and electrolytic conductors are free electrons and free ions respectively. Specific conductance or consuctivity of the electrolyte solution is given by the following relation : `kappa=c xx l/A` where, `c=1//R` is the conductance and `l//A` is the cell constant. Molar conductance `(Lambda_(m))` and equivalence conductance `(Lambda_(e))` of an electrolyte solution are calculated using the following similar relations : `Lambda_(m)= kappa xx 1000/M` `Lambda_(e)= kappa xx 1000/N` Where, M and N are the molarity and normality of the solution respectively. Molar conductance of strong electrolyte depends on concentration : `Lambda_(m)=Lambda_(m)^(@)-b sqrt(c)` where, `Lambda_(m)^(@)=` molar conductance at infinite dilution c= concentration of the solution b= constant The degrees of dissociation of weak electrolytes are calculated as : `alpha=Lambda_(m)/Lambda_(m)^(@)=Lambda_(e)/Lambda_(e)^(@)` Which of the following decreases on dilution of electrolyte solution ?A. Equivalent conductanceB. Molar conductanceC. Specific conductanceD. Conductance |
| Answer» Correct Answer - C | |
| 2218. |
Conductors allow the passage of electric current through them. Metallic and electrolytic are the two type of conductors. Current carriers in metallic and electrolytic conductors are free electrons and free ions respectively. Specific conductance or consuctivity of the electrolyte solution is given by the following relation : `kappa=c xx l/A` where, `c=1//R` is the conductance and `l//A` is the cell constant. Molar conductance `(Lambda_(m))` and equivalence conductance `(Lambda_(e))` of an electrolyte solution are calculated using the following similar relations : `Lambda_(m)= kappa xx 1000/M` `Lambda_(e)= kappa xx 1000/N` Where, M and N are the molarity and normality of the solution respectively. Molar conductance of strong electrolyte depends on concentration : `Lambda_(m)=Lambda_(m)^(@)-b sqrt(c)` where, `Lambda_(m)^(@)=` molar conductance at infinite dilution c= concentration of the solution b= constant The degrees of dissociation of weak electrolytes are calculated as : `alpha=Lambda_(m)/Lambda_(m)^(@)=Lambda_(e)/Lambda_(e)^(@)` The correct order of equivalent conductances at infinite dilution of LiCl, NaCl and KCl is :A. `LiCl gt NaCl gt KCl`B. `KCl gt NaCl gt LiCl`C. `NaCl gt KCl gt LiCl`D. `LiCl gt KCl gt NaCl` |
| Answer» Correct Answer - B | |
| 2219. |
Distinguish between electronic and electrolytic conductors. |
|
Answer» Electronic conductors : 1. The flow of electricity takes place by direct flow of electrons through the conductor. 2. The conduction does not involve the transfer of a matter. 3. No chemical change is involved during conduction. 4. The resistance of the conductor increases and conductivity decreases with the increase in temperature. 5. The conductance of metallic conductors is very high. 6. Examples are solid or molten metals, such as Al, Cu, etc. Electrolytic conductors : 1. The electron transfer takes place by the migration of ions (cations and anions) of the electrolyte. 2. The conduction involves the transfer of a matter. 3. Chemical changes are always involved during the passage of an electric current. 4. The resistance decreases and the conductivity increases with the increase in temperature. 5. The conductance of the electrolytes is comparatively low. 6. Examples are aqueous solutions of acids, bases or salts. |
|
| 2220. |
What are the types of electric conductors? On what basis are they classified? |
|
Answer» The electric conductors are classified according to the mechanism of the transfer of electrons or charge. There are two types of conductors as follows : (i) Electrons (or metallic) conductors : The electric conductors through which the conduction of electricity takes place by a direct flow of electrons under the influence of applied potential are called electronic conductors. In this case, There is no transfer of matter like atoms or ions. For example, Solid and molten metals such as Al, Cu, etc. (ii) Electrolytic conductors : The conductors in which the conduction of electricity takes place by the migration of positive ions (cations) and negative ions (anions) of the electrolyte are called electrolytic conductors. In this, The conduction involves the transfer of matter and it is accompanied with chemical changes. For example, Solutions of electrolytes (strong and weak), molten salts. |
|
| 2221. |
What is a flow of electricity or a transfer of electric charge? |
|
Answer» The flow of electricity or a transfer of electric charge through a conductor involves the transfer of electrons from one point to the other point. This takes place under the influence of applied electric potential. |
|
| 2222. |
What are the electric conductors? |
|
Answer» The substances that allow the flow of electricity or electric charge transfer through them are called the electric conductors. |
|
| 2223. |
What is electric conduction? |
|
Answer» The transfer of electric charge or electrons from one point to another is called electric conduction which results in an electric current. |
|
| 2224. |
What is electrochemistry? |
|
Answer» Electrochemistry : It is the branch of physical chemistry which involves the study of the inter-relation between chemical changes and electrical energy and also concerned with the electrical properties of electrolytic solutions such as resistance and conductance. |
|
| 2225. |
A lead storage battery has been used for one month (30 days) at the rate of one hour per day by drawing a constant current of 2 amperes. `H_(2)SO_(4)` consumed by the battery is:-A. 1.12 moleB. 2.24 moleC. 3.36 moleD. 4.48 mole |
|
Answer» Correct Answer - B The reactions occurring during discharge are: Anode: `Pb(s)+SO_(4)^(2-)(aq)toPbSO_(4)(s)+2e^(-)` `underset("At cathode:"PbO_(2)(s)+4H^(+)(aq)+SO_(4)^(2-)(aq)+2e^(-)toPbSO_(4)(s)+2H_(2)O(l))` Overall reaction reaction: `Pb(s)+PbO_(2)(s)+2H_(2)SO_(4)to2PbSO_(4)(s)+2H_(2)O(l)` Quantity of electricity consumed `=(2A)(30xx3600s)=216000C` `2xx96500` coulombs consume `H_(2)SO_(4)=2` moles `therefore216000` coulombs will consume `H_(2)SO_(4)` `=(1)/(96500)xx216000=2.24` moles. |
|
| 2226. |
A solution of a salt of a metal was electrolysed for 150 minutes by passing 0.15 A current. The weight of the metal deposited was 0.783 g. The specific heat of the metal is `0.057" cal"//gK`. The atomic mass X of the metal is :A. `111.80" g"//mol`B. `52.2" g"//mol`C. `200" g"//mol`D. `250" g"//mol` |
|
Answer» Correct Answer - A (a) According to Dulong & Petits Law, Approximate Atomic mass`=(6.4" Cal"//K)/(0.057"Cal"//gK)=112.28" g"` Quantity of charge passed `=(0.5" amp")xx(150xx60 s)` `=1350" amp"-s=1350" C"` 1350 C charge deposit metal =0.738 g 96500 C charge deposit metal `=((0.738g))/((1350c))xx(96500c)=55.9 g` `:.` Equivalent mass of metal=55.97 g Approximate valency`=("App.Atomic mass")/("Equivalent mass")` `=((112.28g))/((55.97 g))=2.006` Since valency is a whole number, Exact valency =2 Exact atomic mass`="Eq.mass"xx"valency"` `55.97xx2=111.94 g` |
|
| 2227. |
In which of the following pair, first specie is a better oxidising agent than second species under standard conditions? (1) Br2 & Au+3 (2) H2 & Ag+ (3) Cd+2 & Cr+3 (4) O2 in acidic medium & O2 in basic medium |
|
Answer» Correct option (1) Br2 & Au+3 Explanation: Halogens act as an oxidising agent and in Au3+ the oxidation state of Au is maximum i.e., +3. So, Au3+ also acts as oxidizing agent. |
|
| 2228. |
The specific conductances of four electrolytes in `ohm^(-1)cm^(-1)` are given below. Which one offers higher resistance to passage of electric current?A. `7.0 xx 10^(-5)`B. `9.2 xx 10^(-9)`C. `6.0 xx 10^(-7)`D. `4.0 xx 10^(-8)` |
|
Answer» Correct Answer - B `k = (1)/(R) xx (l)/(a)` , i.e., `R prop (1)/(k)`. |
|
| 2229. |
Starting with `250 ml` of `Au^(3+)` solution `250 ml` of `Fe^(2+)` solution, the following equilibrium is established `Au^(3+) (aq) +3Fe^(2+)(aq) hArr 3Fe^(3+) (aq) +Au(s)` At equilibrium the equivalents of `Au^(3+), Fe^(2+), Fe^(3+)` and `Au` are x,y,z and w respectively. Then:A. `K_(c) = (Z^(3))/(6xy^(3))`B. `K_(c) = (3Z^(3))/(2xy^(3))`C. `K_(c) = (4Z^(3))/(9xy^(3))`D. `K_(c) = (8Z^(3))/(9 xy^(3))` |
|
Answer» Correct Answer - B n factor of `Au^(+3), Fe^(+3), Fe^(+3) & Au` are 3,1,1,& 3 respectively. Total volume `500 ml = 0.5 lit` `K_(c) =((Z)^(3)(0.5))/(((x)/(3))((y)/(1))^(3))` `K_(c) = (3Z^(3))/(2xy^(3))` |
|
| 2230. |
Accidentally chewing on a stray fragmet of aluminium foil can causes a sharp tooth path if the aluminium comes in contact with an amalgam filing. The filing, an alloy of silver, tin and mercury, acts as the cathode of a tiny galvanic cell, the aluminium behaves as the anode, and salivas serves as the electrolyte. when the aluminium and the filling come in contact, and electric current passage from the aluminium to the filling which is sensed by a nerve in the tooth. Aluminium is oxidized at the anode, and `O_(2)` gas is reduced to water at the cathode. `E_(AI^(3+)//AI)^(@) =- 1.66 E_(O_(2)H^(+)//H_(2)O)6^(@) = 1.23 V` Standard `E.M.F` experienced by the person with dental filing is:A. `+2.89V`B. `-2.89V`C. `-0.93V`D. `+0.43V` |
|
Answer» Correct Answer - C `{:(AIrarrAI^(+3)+3e^(-),E^(@)=1.66V),(O_(2)+4H^(+)+4e^(-)rarr2H_(2)O,E^(@)=1.23V),( :.E_(cell)^(@)=2.89V,):}` |
|
| 2231. |
The amount of an ion liberated on an electrode during electrolysis does not depend upon:A. conductance of the solutionB. current strengthC. timeD. electrochemical equivalent of the element |
|
Answer» Correct Answer - A |
|
| 2232. |
A silver wire dipped in `0.1M HCI` solution saturated with `AgCI` develops a potential of `-0.25V`. If `E_(Ag//Ag^(+))^(@) =- 0.799V`, the `K_(sp)` of `AgCI` in pure water will beA. `2.95 xx 10^(-11)`B. `5.1 xx 10^(-11)`C. `3.95 xx 10^(-11)`D. `1.95 xx 10^(-11)` |
|
Answer» Correct Answer - B `Ag +CI^(-) rarr AgCI +e^(-)` `E = E^(@) + 0.0591 log [CI^(-)]` `-0.25 = E^(@) +0.0591 log 0.1 E^(@) =- 0.1909` Now for reaction `AgCI +CI^(-)rarrAgCI +e^(-)` `{:(Ag^(+)+e^(-)rarrAg),(bar(Ag^(+)+I^(-)rarrAgCI)):}` `E = E_(Ag//AgCI//CI^(-))^(@) +E_(Ag^(+)//Ag)^(@) +0.0591 log K_(sp)` `O =- 0.1909 +0.799 +0.0591 log K_(sp)` `K_(sp) = 5.13 xx 10^(-11)` |
|
| 2233. |
Consider the reaction fo extraction of gold from its ore `Au +2CN^(-) (aq)+(1)/(4)O_(2) (g)+(1)/(2)H_(2)O rarr Au(CN)_(2)^(-) +OH^(-)` Use the following data to calculate `DeltaG^(@)` for the reaction `K_(f) [Au(CN)_(2)^(-)] = X` `{:(O_(2)+2H_(2)O +4e^(-)rarr4OH^(-),:,E^(@) =+0.41 "volt"),(Au^(3+)+3e^(-)rarr Au,:,E^(@) = +1.5 "volt"),(Au^(3+)+2e^(-)rarr Au^(+),:,E^(@) =+1.4 "volt"):}`A. `-RT In X +1.29 F`B. `-RT In X - 2.11 F`C. `-RT In(1)/(X) +2.11F`D. `-RT In X -1.29 F` |
|
Answer» Correct Answer - A Given `Au^(+3) +3e^(-) rarr E_(1)^(@) = 1.5` `Au^(+3) +2e^(-) rarr Au^(+) rarr E_(2)^(@) = 1.4` Ao for reaction `Au rarr Au^(+) +e^(-) E^(@) = 2E_(2)^(@) - 3E_(1)^(@) rArr E^(@) - 1.7` `Au +2CN^(-) rarr Au(CN)_(2)^(-) +e^(-) E^(@) = x` `Au^(+) +e^(-) rarr E^(@) = 1.7` `O = x +1.7 -(RT)/(F) In X - 1.7` `x = (RT)/(F) In - 1.7` so for reaction `Au +2CN^(-) +(1)/(4)O_(2) +(1)/(2)H_(2)O rarr Au(CN)_(2)^(-) +OH^(-)` `E^(@) = 0.41 +(RT)/(F) In x - 1.7 =- 1.29 +(RT)/(F) In X` `DeltaG^(@) =- nFE^(@) (n =1)` `DeltaG^(@) = 1.29 F - RT In x` |
|
| 2234. |
How many coulombs are required for the reduction.12.55 g of nitrobenzene to aniline in acidic medium. |
|
Answer» Correct Answer - 57900 C The reaction reaction in the acidic medium is : `C_(6)H_(5)NO_(2)+6H^(+)+6e^(-) toC_(6)H_(5)NH_(2)+2H_(2)O` Molar mass of nitrobenzene `(C_(6)H_(5)NO_(2))=6xx12+5xx1+14+2xx16` `=72+5+14+32=123 g mol^(-1)` 123 g of nitrobenzene for reduction require charge `=6xx96500" C"` 12.3 g of nitrobenzene for reduction for reduction require charge `=(6xx(96500" C")xx(12.3 g))/((123 g))=57900" C"` |
|
| 2235. |
Assertion Sodium chloride undergoes hydrolysis in its solution in water. Reason When the ionic product of a salt in a solution exceeds its solubility product at a giv en temperature, the salt precipitates out.A. Both A and R are true and R is the correct explanation of AB. Both A and R are true but R is not a correct explanation of AC. A is true but R is falseD. Both A and R are false. |
| Answer» Correct Answer - D | |
| 2236. |
Assertion On passing HCl gas through a saturated solution of common salt, NaCl precipitates out. Reason The stronger an acid , the weaker must be its base and vice-versa.A. Both A and R are true and R is the correct explanation of AB. Both A and R are true but R is not a correct explanation of AC. A is true but R is falseD. Both A and R are false. |
| Answer» Correct Answer - A | |
| 2237. |
In a hydrogen oxyge fuel cell, electricity is produced. In this process `H_2`(g) is oxided at anode and `O_2`(g) reduced at cathode Given: Cathode `O_2(g)+2H_2O(l)+4e^(-)to4OH^(-)(aq)` Anode `H_2(g)+2OH^(-)(aq)to2H_2O(l)+2e^(-)` 4.48 litre `H_2` at 1atm and 273 k oxidised in 9650 sec. The mass of water produced is :A. 7.2gB. 3.6gC. 1.8gD. 0.9g |
|
Answer» Correct Answer - B |
|
| 2238. |
In a hydrogen oxyge fuel cell, electricity is produced. In this process `H_2`(g) is oxided at anode and `O_2`(g) reduced at cathode Given: Cathode `O_2(g)+2H_2O(l)+4e^(-)to4OH^(-)(aq)` Anode `H_2(g)+2OH^(-)(aq)to2H_2O(l)+2e^(-)` 4.48 litre `H_2` at 1atm and 273 k oxidised in 9650 sec. The current produced is (in amp):A. 1AB. 2AC. 4AD. 8A |
|
Answer» Correct Answer - C |
|
| 2239. |
The amount of ion discharged during electrolysis is not directly proprtional toA. CurrentB. TimeC. ResistanceD. Chemical equivalent of the ion |
|
Answer» Correct Answer - C ` w=zit, Q=it` |
|
| 2240. |
When the sample of copper with zinc impurityn is to be purified by electrolysis, the appropriate electrode are .A. Cathode =Pure zine , Anode=Pure copperB. Cathode=mpure sample, Anode= Pure copperC. Cathode = Pure copper,Anode, Impure sampleD. Cathode =Impure zinc, Anode= Impure ,sample |
|
Answer» Correct Answer - C Impure metal mad anode while pure metal made cathode. |
|
| 2241. |
The standard electrode potential for the following reaction is `-0.57` V. What is the potential at pH=12.0 ? `TeO_(3)^(2-)(aq,1M)+3H_(2)O(l)+4e^(-)toTe(s)+6OH^(-)(aq)`A. `-017 V`B. `+0.21 V`C. `-0.39 V`D. ``+1.95 V` |
|
Answer» Correct Answer - C |
|
| 2242. |
What is the souce of electrical energy in a galvanic cell ? |
| Answer» In a galvanic cell, the redox reaction is of spontaneous nature. The energy as free energy (`DeltaG`) is released in the reaction. This gets converted into the elctro into the electrical energy. | |
| 2243. |
In a zinc manganese dioxide dry cell, the anode is made up of zinc and cathode of a carbon rod surrounded by a mixture of `MnO_(2)`, carbon `NH_(4)Cl and ZnCl_(2)` in aqueous base. The cathodic reaction may be represented as: `2MnO_(2) (s) + Zn^(2+) + 2e^(-) rarr ZnMn_(2) O_(4) (s)` Let there be 8g `MnO_(2)` in the cathodic compartment. How many days will the dry cell continue to give a current of `4 xx 10^(-3)` ampere ? |
|
Answer» When `MnO_(2)` will be used up in cathodic process, the dry cell will stop to produce current. Cathodic process: `overset(+4)(2MnO_(2))(s) + Zn^(2+) + 2e^(-) rarr overset(+3)(ZnMn_(2)O_(4))` Equivlanet mass of `MnO_(2) = ("Molecular mass")/("Change in oxidation state")` `= (87)/(1) = 87` From first law of electrolysis, `W = (I tE)/(96500)` `8 = ( 4 xx 10^(-3) xx t xx 87)/(96500)` `t = 2218390.8` second `= (2218390.8)/(3600 xx 24) = 25.675` days |
|
| 2244. |
An aqueous solution of `NaCl` on electrolysis gives `H_(2)(g), Cl_(2)(g),` and `NaOH` accroding to the reaction `:` `2Cl^(c-)(aq)+2H_(2)Orarr2overset(c-)(O)H(aq)+H_(2)(g)+Cl_(2)(g)` A direct current of `25A` with a current efficiency of `62%` is passed through `20L` of `NaCl` solution `(20%` by weight`)`. Write down the reactions taking place at the anode and cathode. How long will it take to produce `1 kg ` of `Cl_(2)`? What will be the molarity of the solution with respect to hydroxide ion ? `(` Assume no loss due to evaporation . `)` |
|
Answer» `a.` At cathode`: 2Cl^(c-) rarr Cl_(2)+2e^(-)` At anode `: 2H_(2)O+2e^(-) rarr 2 overset(c-)(O)H +H_(2)` `b. [2e^(-)+2F=2xx96500C-=1 mol `of `Cl_(2)=71g]` or `1F=96500C=1Eq` or `Cl_(2)=35.5g` Since current efficiency is `50%` `:.` Current `=96.5xx(50)/(100)=(96.5)/(2)` Use direct relation, `W_(Cl_(2))=(Ew_((Cl_(2)))xxIxxt)/(96500C)` `35.5g=(35.5gxx96.5//2xxt)/(96500C)` `:.t=(96500xx35.5xx2)/(35.5xx96.5)=2000s` `2000s=(2000)/(3600)=0.55h` `c.` `Eq` of `overset(c-)(O)H` formed`=Eq` of `Cl_(2)` `=(35.5g)/(35.5g)=1Eq` `[overset(c-)(O)H]=(Equival ent)/(V_(1))=(1)/(100L)=10^(-2)N` `d.` `pOH=-log [10^(-2)]=2impliespH=14-2=12` |
|
| 2245. |
In a zinc manganese dioxide dry cell, the anode is madeup of `Zn1` and cathode of carbon rod surrounded by a mixture of `MnO_(2)`, carbon, `NH_(4)Cl`, and `ZnCl_(2)` in aqueous base. The cathodic reaction is `: ` `ZnMnO_(2)(s)+Zn^(2+)+2e^(-) rarr ZnMn_(2)O_(4)(s)` `8.7g` of `MnO_(2)` is taken in the cathodic compartment. How many days will the dry cell continue to give a current of `9.65xx10^(-3)A` ? `(` Atomic weight of `Mn=55)(Mw` of `MnO_(2)=87g mol^(-1))` |
|
Answer» When `MnO_(2)` will be used up in cathodic porcess, the dry cell will stop to produce current. `overset(+2)(Zn)overset(+2)(Mn)overset(=4)(O_(2))(s)+Zn^(2+)+2e^(-) rarroverset(+2)(Zn)overset(+3xx2)(Mn)overset(-2xx4)(O_(2)(s))` `(n` factor for `MnO_(2)=1 mol `of `MnO_(2)=87g` `(1F=96500C=1 mol ` of `MnO_(2)=87g)` `:. 87g `of `MnO_(2)-=96500C` `8.7g ` of `MnO_(2)=9650C=Ixxt` `=9.5xx10^(-3)Axxt(s)` `:. t=(9650C)/(9.65xx10^(-3)A)=10^(6)s=(10^(6))/(60xx60xx24)` `=11.57` days Alternatively, use the direct formula `E_(MnO_(2))=(EwxxIxxt)/(96500C)` |
|
| 2246. |
In a reaction `2A to ` Products, the concentration of A decreases from `0.5 mol I^(-1)` to `0.4 mol L^(-1)` in 10 minutes. Calculate the rate during this interval. |
|
Answer» `2A to` products `0.5 -0.4` mol lit rate `prop [A]^(2)` rate `=(-1)/(2) (d)/(dt) [A]=1/2 [((0.5-0.4))/(10)]=0.5xx10^(-2)` rate `=5xx10^(-3) mol Lit ^(-1) min ^(-1)` |
|
| 2247. |
A solution of `Ni(NO_(3))_(2)` is electrolyzed between platium electrodes using a current of `5A` for `20 mi n`. What mass of `Ni` is deposited at the cathode ? |
|
Answer» `Ni(NO_(3))_(2)+2H^(+)+underset(2" mol")(2e^(-)) to underset(1" mol")(Ni)+2HNO_(3)` The charge, Q on n moles of elctrons is given by Q=nF Thus, charge required to deposit 1 mole of nickel, `Q=2" mol"xx96500" C mol"^(-1)=193000" C"=193xx10^(5)" C"` Quantity of electricity used `="Current in amperes"xx"Time in seconds"` , `=5Axx(20xx60)s=5xx20xx60" As"=600" C" (`:.`"A s"=C)` Molar mass of nickel =58.7 g `mol^(-1)` `1.93xx10^(5)C`of charge produce nickel `=1" mol"=1 " mol"xx58.7"g mol"^(-1)`=58.7 g `:. 6000" C of charge produce nickel" =((58.7g))/((1.93xx10^(5)C))xx(6000" C")=1.825" g"` |
|
| 2248. |
Given that `E^(@)(Zn^(2+)//Zn)=0.76V, E^(@)(H^(+)//H_(2))=0.00" V "`. What is the value of electrode potential of `Mg^(+)//Mg` electrode when it is dipped in a solution in which concentration of `Mg^(+)` is 0.01 M ? (Given `E_((Mg^(2+)//Mg))^(@)=-2.36" V "` |
|
Answer» Correct Answer - `-2.419" V "` According to Nernst equation `E=E^(º)-(0.0591)/(n)"log"(1)/([M^(n+)(aq)])=(-2.36" V ")-((0.0591" V "))/(2)"log"(1)/((0.01))` `=(-2.36" V ")-((0.0591V))/(2)"log "10^(2)=-2.36" V "-0.0591" V "=-2.419" V "` |
|
| 2249. |
A voltaic cell is set up at `25^(@)C`with the following half cells : `Al^(3+)`(0.001 M) and `Ni^(2+)`(0.50 M) Write the equation for the reaction when the cell generates the electric current. Also determine the cell potential (Given `E_(Ni^(2+)//Ni)^(@)=-0.25 V, E_(Al^(3+)//Al)^(@)=-1.66 V)`. |
|
Answer» Correct Answer - 1.4602V The cell reaction is : `2Al(s)to 2Al^(3+)(aq)+6e^(-)` `(3Ni^(2+)(aq)+6e^(-)to3Ni(s)" ")/(2Al(s)+3Ni^(2+)(aq)to Al^(3+)(aq)+3Ni(s))` `E_(cell)^(@)=E_(Ni^(2)(aq)//Ni(s))^(@)-E_(Al^(3+)(aq)//Al(s))^(@)=(-0.25) -(-1.66)=-0.25+1.66=+1.14"V "` According to Nernst equation : `E_(cell)=E_(cell)^(@)-(0.0591)/(n)"log"([Al^(3+)]^(2))/([Ni^(2+)]^(3))=1.41-(0.0591)/(6)"log"((0.001)^(2))/((0.5)^(3))` `=1.41-(0.0591)/(6)((10^(-6)xx10^(3)))/(125)=1.41-(0.0591)/(6)"log"(log8xx10^(-6))` `=1.41-(0.0591)/(6)[log8-6log10]=1.41-(0.0591)/(6)(0.9031-6)` `=1.41-(0.0591)/(6)(-5.0969)=1.41-(0.3012)/(6)=1.41+0.0502=1.4602" V "` |
|
| 2250. |
Calculate the emf of the following cell at `25^(@)C`. `Zn|Zn^(2+)`(0.001 M) || `H^(+)(0.01 M) | H_(2)`(1 bar) | Pt(s). Given that `E_((Zn^(2+)//Zn))^(@)=-0.76V, E_((H^(+)//H_(2)))^(@)=0.00" V "` |
|
Answer» Correct Answer - 0.7305V According to Nernst equation : `E_(cell)=E_(cell)^(@)-(0.0591V)/(n)"log"([Zn^(2+)])/([H^(+)]^(2))` `E_(cell)=(0.76" V")-((0.0591" V"))/(2)"log"(10^(--3))/((10^(-2))^(2))` `=0.76" V "-(0.02955" V ")" log "10=(0.76-0.2955)" V "=0.7305" V "` |
|