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.
| 1351. |
How long will it take for a uniform current of 6.0 ampere to deposite 78.0 g gold from asolution of `AuCl_(4)^(-)` ? What mass of chlorine gas will be formed simultaneously at the anode in the electrolytic cell ?A. t = 3010 sec, w = 35.50 gB. t = 20306 sec, w = 45.54 gC. t = 19500 sec, w = 54.5 gD. t = 19139 sec, w = 42.24 g |
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Answer» Correct Answer - D Reactions take place at electrodes are : At cathode : `AuCl_4^(-) + 3e^(-) to Au + 4Cl^(-)` At anode : `Cl^(-) to 1/2Cl_2 + e^(-)` For the deposition of 197 g ( 1 mole ) of Au = 3F of charge is required thus, for the deposition of 78 g of Au , Charge required =`3/197xx78`=1.19 F = 1.19 x 96500 coulombs From Q=I x t `t=Q/t=(1.19xx96500)/6`=19136.16 sec By 1 F charge , 35.5 g of `Cl_2` gas is formed , thus from 1.19 F, mass of chlorine gas formed = 35.5 x 1.19 = 42.24 g |
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| 1352. |
What a lead stroange is discharged:-A. `PbSO_(4)` is formedB. `Pb` is fomedC. `SO_(2)` is consumedD. `H_(2)SO_(4)` is consumed |
| Answer» Correct Answer - A::D | |
| 1353. |
At `25^@C`, Nernst equation isA. `E_"cell"=E_"cell"^@-"0.0591"/n "log" "[ion]"_"RHS"/"[ion]"_"LHS"`B. `E_"cell"=E_"cell"^@-"0.0591"/n "log" "[M]"_"RHS"/"[M]"_"LHS"`C. `E_"cell"=E_"cell"^@+"0.0591"/n "log" "[ion]"_"RHS"/"[ion]"_"LHS"`D. `E_"cell"=E_"cell"^@-"0.0591"/n "log" "[ion]"_"LHS"/"[ion]"_"RHS"` |
| Answer» Correct Answer - A | |
| 1354. |
For `Zn^(2+) //Zn, E^(@) =- 0.76`, for `Ag^(+)//Ag, E^(@) = -0.799V`. The correct statement isA. the reaction `Zn` getting reduced `Ag` getting oxidized is spontaneousB. `Zn` undergoes reduction and `Ag` is oxidizedC. `Zn` undergoes oxidation and `Ag^(+)` gets reducedD. No suitable answer |
| Answer» Correct Answer - C | |
| 1355. |
Given below are few reactions with some expressions. Mark the expressions which is not correctly matched.A. For concentration cell, `Ag|Ag^(+)(C_1)||Ag^(+)(C_2)|Ag, E_"cell"=-0.0591/1"log"C_1/C_2`B. For the cell , `2Ag^(+) +H_2 "(1 atm)" to 2Ag + 2H^(+) (1M) , E_"cell"=E_"cell"^@-0.0591/2"log"([Ag^+]^2)/[H^+]^2`C. For an electrochemical reaction, at equilibrium `aA + bB overset("ne"^-) hArr cC + dD, E_"cell"^@=0.0591/n "log"([C]^c [D]^d)/([A]^a [B]^b)`D. For the cell , `M_((aq))^(n+) + "ne"^(-) to M_((s)) , E=E^@-0.0591/n "log"1/([M^(n+)])` |
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Answer» Correct Answer - B For `2Ag^(+) + H_2` (1 atm) `to 2Ag + 2H^(+)` (1 M ) `E_"cell"=E_"cell"^@-0.0591/2"log"([H^+]^2)/([Ag^+]^2)` |
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| 1356. |
`E^@` values of three metals are listed below. `{:(Zn_((aq))^(2+) + 2e^(-) to Zn_((s)),, E^@=-0.76V),(Fe_((aq))^(2+) + 2e^(-) to 2Fe(s) , , E^@=-0.44 V),(Sn_((aq))^(2+)+2e^(-) to Sn_((s)) ,, E^@=-0.14 V) :}` Which of the following statements are correct on the basis of the above information ? (i)Zinc will be corroded in preference to iron if zinc coating is broken on the surface. (ii)If iron is coated with tin and the coating is broken on the surface then iron will be corroded . (iii)Zinc is more reactive than iron but tin is less reactive then iron.A. (i) and (ii) onlyB. (ii)and (iii) onlyC. (i),(ii) and (iii)D. (i) and (iii) only |
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Answer» Correct Answer - C Iron coated with zinc does not get rusted even if cracks appear on the surface because Zn will take part in redox reaction not Fe as Zn is more reactive than Fe. If iron is coated with tin and cracks appear on the surface, Fe will take part in redox reaction because Sn is less reactive than Fe . |
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| 1357. |
Electrical conductance through metals is called metallic or electronic conuctance and is due to the movement of electrons. The electronic conductance depends onA. the nature and structure of the metalB. the number of valence electrons per atomC. change in tempertureD. all of these. |
| Answer» Correct Answer - D | |
| 1358. |
`2HgtoHg_(2)^(++),E^(@)=+0.855V` `HgtoHg^(2+),E^(@)=0.799V` Equilibrium constant for the reaction `Hg+Hg^(2+)toHg_(2)^(++)` at `27^(@)C` isA. 89B. 82.3C. 79D. none of these |
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Answer» Correct Answer - C `E^(@)` for the given reaction `=0.855-0.799=0.056V` `E_(cell)^(@)=(0.0591)/(2)"log"K_(c)` or `logK_(c)=(0.056xx2)/(0.059)=1.8983` `K_(c)`=Antilog `1.8983=79:12=79` |
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| 1359. |
The most convenient method to protect the bottom of the ship made of iron isA. coating it with red lead oxideB. white tin platingC. connecting it with `Mg` blockD. connecting it with `Pb` block |
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Answer» Correct Answer - C For the sacrificial protection of pipelines, piers and ships against corrosion, ingots of `Mg` are connected to the steel structures. Being more electropositive they corrode preferentially. Their periodical replacement is cheaper and more convenient than painting the whole structure at frequent intervals. |
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| 1360. |
Assertion: `H_(2)-O_92)` fuel cell gives a constant ions, yet the overall concentration of `OH^(-)` ions does not chagne. Reason: In this cell, `H_(2)` reacts with `OH^(-)` ions, yet the overall concentration of `OH^(-)` ion does not change.A. If both assertion and reason are true, and reason is the true explanation of the assertionB. if both assertion and reason are true, but reason is not the true explanation of the assertion.C. if assertion is true, but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - A Both A and R are correct and R is the correct explanation of A. |
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| 1361. |
In a galvanic cell, the salt bridge(1) Does not participate chemically in the cell reaction (2) Stops the diffusion of ions from one electrode to another (3) Is necessary for the occurrence of the cell reaction (4) Ensures mixing of the two electrolytic solutions |
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Answer» Correct option (1, 2) Explanation: In a galvanic cell, the salt bridge does not participate in the cell reaction, stops diffusion of ions from one electrode to another and is not necessary for the occurrence of the cell reaction. |
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| 1362. |
The overall reactions of a hydrogen -oxygen fuel cell isA. `2H_(2(g)) + O_(2(g)) to 2H_2O_((l))`B. `2H_(2(g)) + 4OH_((aq)) to 4H_2O_((l)) + 4e^(-)`C. `O_(2(g)) + 2H_2O_((l)) + 4e^(-) to 4OH_((aq))^-`D. `4OH_((aq))^(-) + 4e^(-) to 2H_2O_((l))` |
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Answer» Correct Answer - A `{:("At anode:",2H_(2(g))+4OH_((aq))^(-) to 4H_2O_((l))+4e^(-)),("At cathode:",O_(2(g))+2H_2O_((l))+4e^(-) to 4OH_((aq))),("Overall reaction:",2H_(2(g)) +O_(2(g)) to 2H_2O_((l))):}` |
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| 1363. |
Which of the following reactions does not take place during rusting?A. `H_2CO_3 hArr 2H^(+) + CO_3^(2-)`B. `4Fe^(2+) + O_(2(dry)) to Fe_2O_3`C. `4Fe^(2+) + O_2+ 4H_2O to 2Fe_2O_3 + 8H^(+)`D. `Fe_2O_3 + xH_2O to Fe_2O_3. xH_2O` |
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Answer» Correct Answer - B `Fe^(2+)` ion does not react with dry oxygen |
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| 1364. |
Zinc displaces copper from the solution of its salt becauseA. Atomic number of zinc is more than that of copperB. Zinc salt is more soluble in water tha the copper saltC. Gibbs free energy of zinc is less than that of copperD. Zinc is placed higher than copper in electro-chemical series |
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Answer» Correct Answer - D The reduction potential of Zn is very higher than Cu. |
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| 1365. |
The molar conductivities of KCl, NaCl and `KNO_(3)` are 152,128 and `111" S cm"^(2) mol^(-1)` respectively. What is the molar conductivity of `NaNO_(3)` ?A. 101 S `cm^(2)mol^(-1)`B. 87 S `cm^(2)mol^(-1)`C. `-101`S `cm^(2)mol^(-1)`D. `-391`S `cm^(2)mol^(-1)` |
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Answer» Correct Answer - B `wedge_(K^(+))^(@)+wedge_(Cl^(-))^(@)=152S" "cm^(2)mol^(-1)` `wedge_(Na^(+))^(@)+wedge_(Cl^(-))=128S" "cm^(2)mol^(-1)` `wedge_(K^(+))^(@)+wedge_(NO_(3)^(-))^(@)=111S" "cm^(2)mol^(-1)` Now, `wedge_(NaNO_(3))=wedge_(NaCl)+wedge_(KNO_(3))-wedge_(KCl)` `wedge_(NaNO_(3))=87S" "cm^(2)mol^(-1)`. |
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| 1366. |
Resistance of a conductvity cell filled with a solution of an electrolyte of concentration 0.1 M is 100 `Omega`. The conductivity of this solution is 1.29 `Sm^(-1)`. Resistance of the same cell when filled with 0.02M of the same solution is `520 Omega`. the molar conductivity of 0.02M solution of the electrolyte will be:A. `124xx10^(-4)Sm^(2)mol^(-1)`B. `1240xx10^(-4)Sm^(2)mol^(-1)`C. `1.24xx10^(-4)Sm^(2)mol^(-1)`D. `1.24xx10^(-4)Sm^(2)mol^(-1)` |
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Answer» Correct Answer - A Cell constant`=("Conductivity (I)")/("Conductance (I)")=("Conductivity (II)")/("Conductance (II)")` Conductivity (II)=`(129)/(520)Sm^(-1)` 0.2M=200 mol `m^(-3)`. Molar conductivity of 0.2 M solution`=(129//520)/(200)` `=12.4xx10^(-4)S" "m^(2)mol^(-1)`. Note: From given it is not possible to calculate molar conductivity of 0.02M solution exactly. however, it is sure that its molar conductivity will be slightly greater than `12.4xx10^(-4)Sm^(2)mol^(-1)`. Assuming linear inverse variation of molar conductivity with concentration we can say that `therefore`Molar conductivyt of 0.02M solution`=1.24xx10^(-3)xx10=124xx10^(-4)S" "m^(2)mol^(-1)`. |
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| 1367. |
Assertion: Copper liberates hydrogen from a solution of dilute hydrochloric acid. Reason: Hydrogen is above copper in the electro- chemical series.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 - B | |
| 1368. |
Assertion: Copper liberates hydrogen from a solution of dilute hydrochloric acid. Reason: Hydrogen is above copper in the electro- chemical series.A. If both assertion and reason are correct and reason is correct explanation for assertion.B. If both assertion and reason are correct but reason is not correct explanation for assertion.C. If assertion is correct but reason is incorrect.D. If assertion and reason both are incorrect. |
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Answer» Correct Answer - D (d) Correct assertion : Copper does not liberate hydrogen from dilute HCl. Correct reason : Hydrogen is placed below copper in the activity series. Only the metals placed below hydrogen can liberate hydrogen from dilute HCl |
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| 1369. |
Assertion: Copper sulphate solution can be kept in a zinc vessel. Reason: The position of copper is higher than zinc is the electrochemical series.A. If both assertion and reason are true, and reason is the true explanation of the assertionB. if both assertion and reason are true, but reason is not the true explanation of the assertion.C. if assertion is true, but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - D Correct A. copper sulphate solution cannot be kept inzinc vessel. ltBrgt Correct R. The position of zinc is higher than copper in the electrochemical series. |
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| 1370. |
Assertion: I Daniell cell, if concentrations of `Cu^(2+)` ions and `Zn^(2+)` ions are doubled, emf of the cell will be doubled. Reason: If concentration of ions in contact with the metal is doubled, the electrode potential is doubled.A. If both assertion and reason are true, and reason is the true explanation of the assertionB. if both assertion and reason are true, but reason is not the true explanation of the assertion.C. if assertion is true, but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - D Correct A. `Zn+Cu^(2+)toZn^(2+)+Cu,E_(cell)=E_(cell)^(@)-(RT)/(nF)"ln"([Zn^(2+)])/([Cu^(2+)])` Thus, if `[Zn^(2+)]` and `[Cu^(2+)]` are doubled, there is no effect of emf. Correct R. `Cu^(2+)+2e^(-)toCu,E=E^(@)-(RT)/(nF)"ln"(1)/([Cu^(2+)])` Thus, if `[C^(2+)]` is doubled, E is not doubled. |
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| 1371. |
Assertion `(A): Cu` liberates `H_(2)(g)` from a dilute solution of `HCl`. Reason `(R):` Hydrogen is below `Cu` in the electrochemical series.A. If both `(A)` and `(R)` are correct, and `(R)` is the correct explanation of `(A)`.B. If both `(A)` and `(R)` are correct, but `(R)` is not the correct explanation of `(A)`.C. If `(A)` is correct, but `(R)` is incorrect.D. If both `(A)` and `(R)` are incorrect. |
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Answer» Correct Answer - e Both `(A)` and `(R)` are incorrect. Correct `(A): Cu` does not liberate `H_(2)(g)` from dilute `HCl` solution. Correct `(R):Cu` is below hydrogen in the electrochemical series. |
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| 1372. |
What is collision frequency (Z) of a reaction ? How is rate related to it for the reaction ` A+B to ` Products. |
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Answer» The number of collision per second per unit volume of the reaction mixture is called collision frequency (Z). For a bimolecular elementary reactions. `A + B to` products, Rate `=A_(AB) e - E_(a)//RT.` |
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| 1373. |
A piece of Cu is added to an aqeuous solution of `FeCl_(3)`A. No iron will be precipitated from the solution.B. Copper will not dissolve in the solution.C. Copper will not dissolve in the solution.D. Iron will be precipitated from the solution. |
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Answer» Correct Answer - A,C `Cu+FeCl_(3)rarrFeCl_(2)+CuCl_(2)` |
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| 1374. |
Which of the following displaces `Br_(2)` brom an aqeuous solution containing bromide ionsA. `Cl_(2)`B. `Cl^(-)`C. `I_(2)`D. `I_(3)^(-)` |
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Answer» Correct Answer - A According to electrochemical series. |
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| 1375. |
Conductivity of 0.12 M `CuSO_(4)` solution at 298 K is `1.8 xx 10^(-2) S cm^(-1)` Calculate its equivalent conductivty |
| Answer» 75 S `cm^(2)eq^(-1)` | |
| 1376. |
Each of the two platinum electrodes having area 64 `mm^(2)` of a conducitvity cel are separted by 8mm The resistance of the cell cont aining `7.5xx10^(-3)` M KCI solution at 298 K is 1250 ohm calculate (i) Cell constant (ii) Conductivity and (iii) Molar conductivity |
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Answer» (i)`1.25 cm^(-1)` (ii) `1xx120^(-3) S cm^(-1)` (iii) 133.33 S `cm^(2) mol^(-1)` |
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| 1377. |
The ionization constant of a weak acid is `1.6xx10^(-5)` and the molar conducitvity at infinite dilutionis `380xx10^(-4)" S " m^(2) mol^(-1)`. If the cell constant is `0.01 m^(-1)`, then the conductance of 0.01 M solution is :A. `1.52xx10^(-5)" S "`B. `1.52" S "`C. `1.52xx10^(-3)" S "`D. `1.52xx10^(-4)" S "` |
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Answer» Correct Answer - D (d) For a weak acid `K_(a)=C alpha^(2)` `K_(a)=C((lambda_(m)^(c ))/(lambda_(m)^(alpha)))^(2)` `1.6xx10^(-5)=0.01((lambda_(m)^(c ))/(380xx10^(-4)))^(2)` `16xx10^(-6)=10^(-2)((lambda_(m)^(c ))/(380xx10^(-4)))^(2)` `(lambda_(m)^(c ))/(380xx10^(-4))=(16xx10^(-4))^(1//2)=4xx10^(-2)` `lambda_(m)^(c )=(380xx10^(-4))xx(4xx10^(-2))` `=152xx10^(-5)=1.52xx10^(-3)` |
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| 1378. |
which of the following statement is correct?A. cathode is -ve terminal in both ,galvanic and electrolytic cellsB. Anode is +ve terminal in both,galvanic and electrolytic cellsC. cathode and node are -ve terminal in electrolytic and galvanic cell.D. Cathode and node are +ve terminal in electroytic and galvanic cell. |
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Answer» Correct Answer - c |
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| 1379. |
The element indium is to be obtained by electrolysis of a molten halide of the element. Passage of a current of 3.20 A for a period of 40.0 min results in formation of 3.05 g of In. what is the oxidation state of indium in the halide melt? (Atomic mass of In=114.8)A. 3B. 2C. 5D. 1 |
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Answer» Correct Answer - A |
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| 1380. |
The gas`X` at `1 atm` is bubbled through a solution containing a mixture of `1MY^(c-)` and `1MZ^(c-)` at `25^(@)C` . If the order of reduction potential is `ZgtYgtX`, then |
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Answer» Correct Answer - `(X` and ``Z)` `(X` and `Z)` Since `E^(c-)._(red)` fo `YgtX`, therefore, `Y` will oxidize `X.` Similarly , `E^(c-)._(red)` of `YltZ`, thereofre `Y` will not oxidize `Z`. |
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| 1381. |
Redox reaction play a pivotal role in chemistry and biology. The values of standard redox potential `(E^(@))` of two half-cell reactions decide which way the reaction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper gets deposited. give below are a set of half-cell reaction (acidic medium) along with their `E^(@)` (V with respect to normal hydrogen electrode) values. using this data obtain the correct explanation to folloing Questions. `I_(2)+2e^(-)to2I^(-)" "E^(0)=0.54` `Cl_(2)+2e^(-)to2Cl^(-)" "E^(0)=1.36` `Mn^(3+)+e^(-)toMn^(2+)" "E^(0)=1.50` `Fe^(3+)+e^(-)toFe^(2+)" "E^(0)=0.77` `O_(2)+4H^(+)+4e^(-)to2H_(2)O" "E^(0)=1.23` Q. While `Fe^(3+)` is stable, `Mn^(3+)` is not stable in acid solution becauseA. `O_(2)` oxidises `Mn^(2+)` to `Mn^(3+)`B. `O_(2)` oxidises both `Mn^(2+)` to `Mn^(3+) and Fe^(2+)` to `Fe^(3+)`C. `Fe^(3+)` oxidises `H_(2)O` to `O_(2)`D. `Mn^(3+)` oxidises `H_(2)O` to `O_(2)` |
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Answer» Correct Answer - D `4Mn^(3+)+2H_(2)Oto4Mn^(2+)+O_(2)+4H^(+)` `E^(o)=E_(Mn^(3+)//Mn^(2+))^(o)+E_(H_(2)O//O_(2))^(o)=1.50+(-1.23)=0.27V` |
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| 1382. |
Redox reaction play a pivotal role in chemistry and biology. The values of standard redox potential `(E^(@))` of two half-cell reactions decide which way the reaction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper gets deposited. give below are a set of half-cell reaction (acidic medium) along with their `E^(@)` (V with respect to normal hydrogen electrode) values. using this data obtain the correct explanation to folloing Questions. `I_(2)+2e^(-)to2I^(-)" "E^(0)=0.54` `Cl_(2)+2e^(-)to2Cl^(-)" "E^(0)=1.36` `Mn^(3+)+e^(-)toMn^(2+)" "E^(0)=1.50` `Fe^(3+)+e^(-)toFe^(2+)" "E^(0)=0.77` `O_(2)+4H^(+)+4e^(-)to2H_(2)O" "E^(0)=1.23` Q. Among the following, identify the correct statementA. Chloride ion oxidised `O_(2)`B. `Fe^(2+)` is oxidised by iodineC. Iodide ion is oxidised by chlorineD. `Mn^(2+)` is oxidised by chlorine |
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Answer» Correct Answer - C `2I^(-)+Cl_(2)toI_(2)+2Cl^(-)` `E^(o)=E_(I^(-)//I_(2))^(o)+E_(Cl_(2)//Cl^(-))^(o)` `=-0.54+1.36` `E^(o)=0.82V` `E^o` is positive hence iodide ion is oxidized by chlorine. |
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| 1383. |
Consider the following equations for a cell reaction `A+BiffC+D, E^(@)=x"volt",K_(eq)=K_1``2A+2Biff2C+2D, E^(@)=y"volt",K_(eq)=K_2` then:A. `x=y,k_1+k_2`B. `x=2y,K_1=2K_2`C. `x=y,K_1^2=K_2`D. `X^2=y,K_1^2=K_2` |
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Answer» Correct Answer - C |
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| 1384. |
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. |
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Answer» Correct Answer - B |
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| 1385. |
The cell reaction `2Ag^+(aq)+H_2(g)to2H^+(aq)+2Ag(s)`,best represented by :A. `Ag(s)|Ag^+"(aq)"||"H^+(aq)|H_2(g)|Pt(s)`B. `Pt(s)|H_2(g)|H^+(aq)"||"Ag^+(aq)|Ag(s)`C. `Ag(s)|Ag^+"(aq)"||"H_2(g)|H^+(aq)|Pt(s)`D. `Ag^+"(aq)"|"Ag(s)||H_2(g)|H^+(aq)` |
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Answer» Correct Answer - B |
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| 1386. |
The Nernst equation E=`E^(@)`-RT/nF in Q indicates that the Q will be equal to equilibrium constant `K_c` when:A. E=`E^(@)`B. RT/nF=1C. E=zeroD. `E^(@)=1` |
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Answer» Correct Answer - C |
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| 1387. |
Consider the following cell reaction `Zn +2Ag^(o+)rarr Zn ^(2+)+2Ag.` Given that `E^(c-)._(Zn^(2+)(aq)Zn(s))=-0.76V` `E^(c-)._(Ag^(o+)(aq)Ag(s))=0.80V` `a.` Calculate the standard `EMF` fo the cell. `b.` Which ion is more powerful oxidizing agent ? `c.` Which metal is more powerful reducing agent ?A. `x+2y`B. `2x+y`C. `y-x`D. `y-2x` |
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Answer» Correct Answer - c `E^(c-)._(cell)=(E^(c-)._(red))_(c)-(E^(c-)._(red))_(a)` `=y-x` |
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| 1388. |
What is Lead storage battery? |
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Answer» It is a Rechargeable battery. Used in automobiles and inverters, Rechargeable batteries used in house hold appliances etc. |
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| 1389. |
Predict the products of electrolysis in each of the following:(i) An aqueous solution of AgNO3 with silver electrodes. (ii) An aqueous solution of AgNO3 with platinum electrodes. (iii) A dilute solution of H2SO4 with platinum electrodes. (iv) An aqueous solution of CuCl2 with platinum electrodes. |
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Answer» (i) At cathode: The following reduction reactions compete to take place at the cathode Ag+(aq) + e- → Ag(s); Eθ = 0.80 V H+ (aq) + e- → H2 (g); Eθ = 0.00V The reaction with a higher value of Eθ takes place of the cathode. Therefore, deposition of silver will take place at the cathode. At anode: The Ag anode is attacked by NO3- ions. Therefore, the silver electrode at the anode dissolves in the solution to form Ag+. (ii) At cathode: Same as above At anode: Anode is not attackable and hence OH- ions have lower discharge potential than NO3- ions and OH- ions react to give O2 OH- → OH + e- 4OH → 2H2O + O2 (g) (iii) H2SO4 → 2H+ + SO2-4 HO2 ⇌ H+ + OH- At cathode: 2H+ + 2e- → H2 At anode: 4OH- → 2H2O + O2 + 4e- i. e., H2 will be liberated at cathode and O2 at anode. (iv) CuCl2 → Cu2+ + 2Cl- At Cathode: Cu2+ ions will be reduced in preference to H+ ions Cu2+ + 2e → Cu At anode: Cl- ions will be oxidised in preference to OH- ions. 2Cl- → Cl2 + 2e- i.e., Cu will be deposited on the cathode and Cl2 will be liberated at the anode. |
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| 1390. |
Define Fuel cell. |
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Answer» This cell is based on the combustion of fuel like Hydrogen by oxygen is called fuel cell. |
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| 1391. |
Explain the terms specific conductivity and molar conductivity. |
| Answer» For answer, consult section 17. | |
| 1392. |
A solution of sodium chloride is a better conductor of electricity at a temperature of 50°C than at room temperature. Why? |
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Answer» A solution of NaCl shows greater conduction of electricity at a temperature of 50°C than at room temperature because the ionic mobility of a strong electrolyte such as NaCl increases with an increase in temperature. |
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| 1393. |
Give the relationship between molar conductivity and specific conductivity. |
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Answer» Molar conductivity and specific conductivity are related to each other by the given equation. \(\triangle_m = K \times \frac{1000}{C}\) Where, Δm = Molar conductivity K = Specific conductivity C = Molar concentration |
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| 1394. |
Write the relationship between conductivity and molar conductivity and hence unit of molar conductivity. |
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Answer» If k is conductivity and ∧m is molar conductivity then, ∧m = \(\frac{k \times 1000}{C}\) Unit of molar conductivity is, Ω-1 cm2 mol or S cm2 mol-1. |
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| 1395. |
Represent the half cells or electrodes for the following reactions :(i) Zn2+(aq) + 2e- → Zn(s)(ii) \(Ag^+_{(aq)}\)+ e- → Ag(s)(iii) \(4H^+_{(aq)}\)+ 4e- → 2H2(g)(iv) \(Sn^{4+}_{(aq)}\)+ 2e- → \(Sn^{2+}_{(aq)}\) (v) \(\frac{1}{2}\)Cl2(g) + e- → Cl-(aq) |
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| 1396. |
Define electrochemical series or electromotive series. |
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Answer» Electrochemical series (Electromotive series) : It is defined as the arrangement in a series of electrodes of elements (metal or non-metal in contact with their ions) with the electrode half reactions in the decreasing order of their standard reduction potentials. |
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| 1397. |
Explain electrochemical series or electromotive series. |
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Answer» The conventions used in the construction of electrochemical series (or electromotive series) are as follows :
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| 1398. |
Why is cathode in a galvanic cell considered to be positive electrode? |
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Answer» (1) According to IUPAC conventions, the electrode of the galvanic cell where electronation or reduction takes place is called cathode. In this, the electrons from the metal electrode are removed by cations required for their reduction. \(Cu^{2+}_{(aq)}\) + 2e- → Cu(s) (2) Since the electrons are lost, the metal electrode acquires a positive charge. Hence cathode in the galvanic cell is considered to be positive. |
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| 1399. |
Why is anode in a galvanic cell considered to be negative? |
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Answer» 1. According to IUPAC conventions, The electrode of a galvanic cell where de-electronation or oxidation takes place releasing electrons is called anode. Zn(s) → \(Zn^{2+}_{(aq)}\) + 2e- 2. The electrons released due to oxidation reaction are accumulated on the metal electrode surface charging it negatively. Hence anode in the galvanic cell is considered to be negative. |
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| 1400. |
The Zn acts as sacrified of cathodic protection to prevent rusting of iron because:A. `E_(OP)^(@)` of `ZnltE_(OP)^(@)` of FeB. `E_(OP)^(@)` of `ZngtE_(OP)^(@)` of FeC. `E_(OP)^(@)` of `Zn=E_(OP)^(@)` of FeD. Zn is cheaper than iron |
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Answer» Correct Answer - B |
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