3195 + Interview Questions in ELECTROCHEMISTRY IN GENERAL AWARENESS Page 58 InterviewSolution

2851.	Calculate molar conductance for `NH_(4)OH`, given that molar conductance for `Ba(OH)_(2), BaCI_(2)` and `NH_(4)CI` are `523.28, 280.0` and `129.8 ohm^(-1) cm^(2) mol^(-1)` respectively.
Answer» `mu_(Ba(OH)_(2))^(prop) = lambda_(Ba^(2+))^(prop) +2lambda_(OH^(-))^(prop) = 523.28` ...(i) `mu_(BaCI_(2))^(prop) = lambda_(Ba^(2+))^(prop) +2lambda_(CI^(-))^(prop) = 280.00` ...(ii) `mu_(NH_(4)CI)^(prop) = lambda_(NH_(4)^(+))^(prop) +lambda_(CI^(-))^(prop) = 129.80` ...(iii) `mu_(NH_(4)OH)^(prop) = lambda_(NH_(4)^(+))^(prop) +lambda_(OH^(-))^(prop) = ?` Eq.(iii) `+(Eq(i))/(2) -(Eq.(ii))/(2)` will gives `lambda_(NH_(4)^(+))^(prop) +lambda_(OH^(-))^(prop) = lambda_(NH_(4)OH)^(prop) = (502.88)/(2) = 251.44 ohm^(-1) cm^(2) mol^(-1)`

Discussion

2852.	Consider an electrochemical cell: `A(s)\|A^(n+)(aq,2M)\|\|B^(2n+)(aq,1M)\|B(s)`. The value of `DeltaH^(theta)` for the cell reaction is twice that of `DeltaG^(theta)` at 300K. If the emf of the cell is zero, the `DeltaS^(theta)` (in `KJ^(-1)mol^(-1)`) of the cell reaction per mole of B formed at 300 K is ________ (Given: ln (2)=0.7, R (universal gas constant)=8.3 `JK^(-1)mol^(-1)`. H,S and G are enthalpy, entropy an Gibs energy, respectively).
Answer» Correct Answer - A::B `(-11.62)` `underset(DeltaH^(o)=2DeltaG_(O)^(o))(A(s)\|A^(+n)(aq,2M))\|\|B^(+2n)underset(E_(cell)=0)((aq,1M)\|B(s))` Cell Rx `[AtoA^(+n)+ n e^(-)]xx2` `B^(+2n)+2n e^(-)toB(s)` `2A(s)+underset(1M)(B^(+2n))(aq)tounderset(2M)(2A^(+n))(aq)+B(s)` `DeltaG=DeltaG^(o)+RT" ln "([A^(+n)]^(2))/([B^(+2n)])` `DeltaG^(o)=-RT" ln "([A^(+n)]^(2))/([B^(+2n)])=-RT.ln(2^(2))/(1)=-RT.ln4` `DeltaG^(o)=DeltaH^(o)-TDeltaS^(o)` `DeltaG^(o)=2DeltaG^(o)-TDeltaS^(o)` `DeltaS^(o)=(DeltaG^(o))/(T)=-(RTln4)/(T)=-8.3xx2xx0.7=-11.62J//K.mol`

Discussion

2853.	How many minutes will it take to plate out 5.2g of cr from a `Cr_2(SO_4)_3` solution using a current of 9.65 A ? (Atomic mass:Cr=52.0)A. 200B. 50C. 100D. 103
Answer» Correct Answer - B

Discussion

2854.	A current is passed through two cells connected in series. The first cell contians `X(NO_(2))_(2)`(aq). The relative atomic masses of X and Y are in the ratio 1:2. What is the ratio of the liberated mass of X to that of Y ?A. `3:2`B. `1:2`C. `1:2`D. `3:3`
Answer» Correct Answer - C (c ) The oxidation states of X and Y in their respective nitrates are 3 and 2 respectively. The equivalent masses of the metals can be calculated by dividing the respective atomic masses with oxidation states or valences. These are in the ratio `1//3 : 2//2` or `1//3 : 1` or `1:3`

Discussion

2855.	How many coulombs of electricity are c onsumed when a 100mA current is passed through a solution of `AgNO_(3)` for half an hour during an electrolysis experiment?A. 108B. 180C. 1800D. 18000
Answer» Correct Answer - B `Q = ixxt` `i= 100 mA = (100)/(1000) A` `t = 30 ` min = `30 xx 60` seconds `therefore Q = (100)/(1000) xx 30 xx 60 = 180` coulombs

Discussion

2856.	A current of 2.0A passed for 5 hours through a molten metal salt deposits 22.2 g of metal (At. Wt. =177). The oxidation state of the metal in the metal salt isA. `+1`B. `+2`C. `+3`D. `+4`
Answer» Correct Answer - C `E = (W)/(Q) xx 96500 = (22.2)/(2 xx 5 xx 60 xx 60) xx 96500 = 59.5` Ox.state = `("At.mass")/("Eq. mass") = (177)/(59.5) = 3`.

Discussion

2857.	A current is passed through 2 voltmeters connected in series. The first voltmeter contians `XSO_4)(aq)` and second has `Y_(2)SO_(4)`(aq). The relative atomic masses of X and Y are in the ratio `2:1`. The ratio of the mass of X liberated to the mass of Y liberated is:A. `1 : 1`B. `1 : 2`C. ` 2 : 1`D. None of these
Answer» Correct Answer - A A/c. to `II^(nd)` law Eq. of X = Eq. of Y `(W_(1))/(2 M//2) = (W_(2))/(M//1) " " therefore W_(1) = W_(2)`

Discussion

2858.	When an electric current is passed through acidified water, `112ml` of `H_(2)` gas at `NTP` is collected at the cathode is `965` seconds. The current passed in amperes isA. `1.0`B. `0.5`C. `0.1`D. `2.0`
Answer» Correct Answer - A

Discussion

2859.	When an electric current is passed through acidified water, `112ml` of `H_(2)` gas at `NTP` is collected at the cathode is `965` seconds. The current passed in amperes isA. 2.0 amperesB. 1.5ampereC. 1 ampereD. 0.11ampere.,
Answer» Correct Answer - C 22,400 mL of hydrogen at STP=2g 1 mL of hydrogen at STP=`(2)/(22,400)` 112 mL of hydrogen at STP=`(2)/(22.400)` 112 mL of hydrogen at STP`=(2)/(22,400)xx112` From first law of Faraday, W=Zit `0.1=0.00001xx1xx965` `therefore 1=(0.01)/(0.00001xx965)` `[because "Z of hydrogen"=0.00001]` =1 ampere

Discussion

2860.	When an electric current is passed through acidified water, `112ml` of `H_(2)` gas at `NTP` is collected at the cathode is `965` seconds. The current passed in amperes isA. 1B. `0.5`C. `0.1`D. 2
Answer» Correct Answer - A 1 g eq. of `H_(2)` , i.e., 1 g or 11200 cc is obtained from 96500 C `therefore` 112 mL will be obtained from 965 C . `I(A) = (Q)/(t) = (965)/(965) = 1A`.

Discussion

2861.	Calculate the concentration of silver ions in the cell constructed by using `0.1` M concentration of `Cu^(2+) and Ag^(+)` ions. Cu and Ag metals are used as electrodes. The cell potential is `0.422` V. `[E _(Ag^(2+)//Ag)=0.80V,E_(Cu^(2+)//Cu)=+0.34V]`
Answer» Given `E^(0)of Ag^(+)//Ag =0.80V` `E^(0) of Cu^(+2)//Cu=0.34V` `E^(0)` of cell `E_(RHS) -E_(LHS)=0.80-0.34=0.46V` Cell is `Cu\|Cu_(""(0.1M))^(+2)\|\|Ag^(+)\|Ag` ? `E_(cell ) =E_(cell)^(0) -(0.059)/(2)log ""(0.1)/([Ag^(+)]^(2))=-0.038=-0.295log ""(0.1)/([Ag ^(+)]^(2))` `log ""(0.1)/([Ag ^(+)]^(2))=(0.038)/(0.0295)=1.288` `[Ag+]^(2)=(0.1)/(19.32)=5. 176 xx10^(-3)` `[Ag^(+)]=sqrt(51. 76 xx10^(-4))=7.16xx10^(-2)M`

Discussion

2862.	Calculate the emf of the cell with the cell reaction `Ni _((s))+2Ag ^(+)(0.002M)to Ni^(2+)(0.160M) +2 Ag_((s))` `E_(cell)^(0)=1.05V.`
Answer» From the given cell reaction and Nernst equation `E_(cell) =E_(cell)^(0) -(0.0591)/(n )log ""([Ni^(2+)])/([Ag^(+)]^(2))` `=1.05 V- (0.0591)/(2)log ""([0.160])/([0.002]^(2))` `=1.05 V-(0.0591)/(2)log (4xx 10^(4))` `=1.05 -(0.0591)/(2)(4.6021)` `=1.05 -0.14=0.91V=0.91V`

Discussion

2863.	Assertion Molar conductance of an electrolyte increases with dilution Reason Ions move fast in dilute solutions.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 - C

Discussion

2864.	Assertion : Molar conductance of an electrolyte increases upon dilution. Reason : Ions move faster in dilute solution.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.
Answer» Correct Answer - B (b) Correct explanation. Molar conductance increases because with dilution in a weak electrolyte, dissociation increases while in a strong electrolyte, inter-ionic attractions decrease.

Discussion

2865.	How does the molar conductance of an electrolyte vary with dilution ?
Answer» Molar conductance of an electrolyte increases upon dilution because its degree of ionsation(a) increases.

Discussion

2866.	Why has the molar conductance of an electrolyte the maximum value at infinite dilution?
Answer» As the dilution of an electrolytic solution increases or concentration decreases, the dissociation of an electrolyte increases. At infinite dilution, the dissociation of an electrolyte is complete (100% dissociation). Hence all the ions from one mole of an electrolyte are available to carry electricity. Therefore the molar conductance at infinite dilute (∧₀) for a given electrolyte has the highest or limiting value. It is always constant for the given electrolyte at constant temperature.

2866.

Why has the molar conductance of an electrolyte the maximum value at infinite dilution?

Answer»

As the dilution of an electrolytic solution increases or concentration decreases, the dissociation of an electrolyte increases.
At infinite dilution, the dissociation of an electrolyte is complete (100% dissociation). Hence all the ions from one mole of an electrolyte are available to carry electricity.

Therefore the molar conductance at infinite dilute (∧₀) for a given electrolyte has the highest or limiting value. It is always constant for the given electrolyte at constant temperature.

Discussion

2867.	Breathalyzer is used to detect the alcohol content in the suspected drunk drivers. The ethanol in the exhaled breath is oxidized to ethanoic acid with an acidic solution of `K_(2)Cr_(2)O_(7)` as follows `:` `underset(Ehanol)(3CH_(3)CH_(2)OH(aq))+2Cr_(2)O_(7)^(2-)(aq) +16H^(o+)(aq) rarr underset(Ethanoic ac i d)(3CH_(3)COH(aq))+4Cr^(3+)(aq)+11H_(2)O(l)` The breathalyzer measures the colour change and produces a metre reading calibrated in the terms of blood alcohol content. The `EMF` of the reaction when the concentration of all the species are `1.0M` and `pH` is `4.0` isA. `1.64`B. `0.31`C. `-1.01V`D. `0.95V`
Answer» Correct Answer - d `pH=4.0implies[H^(o+)]=10^(-4)M`. `28H^(o+)+2Cr_(2)O_(7)^(2-)+12e^(-) rarr underset((Reduction at cathode))( 4Cr^(3+)+14H_(2)O)` `:. n_(cell)=12` `[Take2.303(RT)/(F)=0.06]` `E=E^(c-)._(cell)-(0.06)/(12)log .([CH_(3)COOH]^(3)[Cr^(3+)]^(4)[H_(2)O]^(11))/([CH_(3)CH_(2)OH]^(3)[Cr_(2)O_(7)^(2-)]^(2)[H^(o+)]^(16))` `=1.27V-(1)/(200)log.((1)^(3)(1)^(4)(1)^(11))/((1)^(3)(1)^(2)(10^(-4))^(16))` `=1.27V-(1)/(200)log[10^(64)]` `=1.27-(1)/(200)xx64=1.27-0.32=0.95V`

Discussion

2868.	How long (approximate) should water be electrolysed by passing through 100 amperes current so that the oxygen released can completely burn 27.66 g of diborane (Given atomic mass of B=10.8u)A. 0.8 hourB. 3.2 hoursC. 1.6 hoursD. 6.4 hours
Answer» Correct Answer - B (b) `B_(2)H_(6)+3O_(2) to B_(2)O_(3)+3H_(2)O` `2xx10.8+6xx1` `=27.6 g(1mol)` 3 moles of oxygen are produced by the electrolysis of water. `6H_(2)Ooverset("Electrolysis") rarr 6H_(2)+O_(2)` or `6H_(2)O to 12H^(+)+3O_(2)+underset((12F))(12e^(-))` Now `Q=lxxt=(12xx96500" C")=(100" amp")xxt` `t=((12xx96500" C"))/((100" amp"))=11580" s"` `=(11580)/(3600)=3.21" h"`.

Discussion

2869.	How long (approximate) should water be electrolysed by passing through 100 amperes current so that the oxygen released can completely burn 27.66g of diboraneA. 0.8 hoursB. 3.2 hoursC. 1.6 hoursD. 6.4 hous
Answer» Correct Answer - B `B_(2)H_(6)+3O_(2)toB_(2)O_(3)+3H_(2)O` 1 mol 3 mol 3 mol `O_(2)` is required for burning 1 mol `B_(2)H_(6)` `H_(2)O overset("Electrolysis")toH_(2)+(1)/(2)O_(2)" (V.F. of "O_(2)=4)` `("Equivalent of "O_(2))/("V.F. of "O_(2))=" mol of "O_(2)=3` `[((100A)xxtsec)/(96500)]xx(1)/(4)=3` `thereforet=(3xx96500)/(100xx3600)hr.=3.22hrs`.

Discussion

2870.	If `3F` of electrictiy is passed through the solutions of `AgNO_3, CuSO_4` and `Auc=CL_3`, the molar ration of the cations deposited at the cathode is .A. `1:1:1`B. `1:2:3`C. `3:2:1`D. `6:3:2`
Answer» Correct Answer - D Since `Ag^+ + e^_ rarr Ag, Cu^(2+) + 2 e^(-) rarr Cu, Au^(3+) +3e^(-) rarr Au`, ` 3F` of electricity will depostit `3` moles of Ag, `1.5 ` moles of coppre, and `1` mole of gole. Therefpre, the molar ration is ` 3:1.5:1` or `6: 3:2`.

Discussion

2871.	The metal that cannot obtained by electrolysis of an aqueous solution of its salts is `:`A. ` Ag`B. ` Cr`C. ` Cu`D. `Mg`
Answer» Correct Answer - D The reduction potential of Mg is less than that of water `(E^@=- 0. 83 V)`. Hence their ions in the aqueous solution cannot be reduced instead qater will be reduced `2 H_2 O + 2e^(-_ rarr H_2+ 2 oH^(-)`.

Discussion

2872.	Which of the follwing will not conduct electricity in aqueous solution ?A. Copper sulphateB. Common saltC. SugarD. None of these
Answer» Correct Answer - C Sugar solutin does not form ion , hencr does not conduct electricity in solution .

Discussion

2873.	Which of the following is //are function (s) of salt bredge ?A. It completes the electrical circuit with electrons flowing from one electrode to the other through extermal wires and a flow of ions between the two compartments through salt bridgeB. it minimised the liquid-liquid junction potentialC. both are correctD. none of there
Answer» Correct Answer - C Salt bridge completes the electrical circuit and minimises the liquid-liquid junction potential.

Discussion

2874.	In which of the follwing cells net cell reaction is :A. ` Ag\|AfCl(s)\|KCl(aq)\|\|NH_4NO_3AgNO_3\|Ag`B. `Ag\|AfCl(s) \|Cl(a)\|Ag`C. `Ag\|AfNO_3\|\|NH_4NO_3\|\|KCl(aq)\|AgCl\|Ag`D. All of the above
Answer» Correct Answer - A ` Ag\|AfCl(s) \|LCl(aq)\|\|NH_4NO_3\|\|AgNO_3\|Ag` Salt bridge LHE reaction : `Ag+Cl^- rarr AgCl(s) +e` RHE reaction : `Ag^+ +e rarr Ag` Net reaction :`Ag^+ + Cl^- rarr AgCl(s)`.

Discussion

2875.	The `E_(M^(3+)//M^(2+)^@` values for ` Cr`, Mn, Fe and ` Co` are ` 0.41, + 1.57 , + 0.77` and `+1, 97 V` respectively. For which one of these metals the change ub oxidation state from ` =2 ` to ` 3` is easiest :A. `Co`B. `Mn`C. `Fe` and `Au`D. `Cr`
Answer» Correct Answer - D More is `E_(RP)^(@)`, more is the tendency to get reduced or lesser is tendency to get oxidised. `E_(RP_(Cr^(3+)//Cr^(2+)))^(@)` is lowest or `E_(OP_(Cr^(2+)//Cr^(3+)))^(@)` is maximum among all.

Discussion

2876.	Enf of a cell in terms of reduction potential of its left and right electrodes is :A. `E = E_("left") - E_(right)`B. `E= E_("left") + E_(right)`C. `E = E_(right) - E_("left")`D. `E = -[E_(right) + E_("left)"]`
Answer» Correct Answer - C `E_(cell) = E_(OP_(L)) + E_(RP_(R))` `E_(cell) = -E_(RP_(L)) + E_(RP_(R))`

Discussion

2877.	Enf of a cell in terms of reduction potential of its left and right electrodes is :A. `E=E_("left") - E_("right")B. `E=E_("left") + E_("right")C. `E=E_("left") - E_("right")D. `E=E_("left") + E_("right")`
Answer» Correct Answer - C ` E_("cell") = E_(OP_L) +E_(RP_R)` E_("cell") = E_(OP_L) +E_(RP_R)`

Discussion

2878.	Chemical reactions involve interaction of atoms and molecules. A large number of atoms/molecules (approximately 6.023 × 1023) are present in a few grams of any chemical compound varying with their atomic/ molecular masses. To handle such large numbers conveniently, the mole concept was introduced. This concept has implications 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 the electrodes (atomic mass: Na = 23, Hg = 200, 1 faraday = 96500 coulomb).The total charge (coulomb) required for complete electrolysis is (1) 24125 (2) 48250 (3) 96500 (4) 193000
Answer» Correct option (4) 193000 Explanation: Total charge required = 2F = 2 x 96500 = 193000 C.

2878.

Chemical reactions involve interaction of atoms and molecules. A large number of atoms/molecules (approximately 6.023 × 1023) are present in a few grams of any chemical compound varying with their atomic/ molecular masses. To handle such large numbers conveniently, the mole concept was introduced. This concept has implications 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 the electrodes (atomic mass: Na = 23, Hg = 200, 1 faraday = 96500 coulomb).The total charge (coulomb) required for complete electrolysis is (1) 24125 (2) 48250 (3) 96500 (4) 193000

Answer»

Correct option (4) 193000

Explanation:

Total charge required = 2F = 2 x 96500 = 193000 C.

Discussion

2879.	Conductance (Siemens, S) is directly proportional to the area of the vessel and the concentration of solution in it and is inversely proprtional to the length of the vessel, then the unit of constant of proportionlity is :A. `S m mol^(-1)`B. `S m^(2) mol^(-1)`C. `S^(-2) m^(1) mol`D. `S^(2) m^(2) mol^(-2)`
Answer» Correct Answer - B `c prop a` and `c prop` molarity `prop (1)/(l)` `:.` Unit of `k = S(m)/(m^(2)) xx (m^(3))/(mol.) = S m^(2) mol^(-1)`

Discussion

2880.	Conductivity (unit siemens) is directly propotional to area of the vessel and the concentration of the solution it and is inversely proportional to the length of the vessel then the unit of constant of proportionality isA. `"Sm mol"^(-1)`B. `Sm^(2) mol^(-1)`C. `S^(-2) m^(2) mol`D. `S^(2)m^(2) mol^(-2)`
Answer» Correct Answer - B `Sm^(2) mol^(-1)`

Discussion

2881.	Assertion Sodium ions are discharged at the Hg electrode during electrolysis is preference to `H^(+)` ions. Reason The nature of electrode also affect the order or discharge of cations.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

Discussion

2882.	We can construct innumerable number of galvanic cells on the pattern of Daniell cell by taking combi nation of different half cells.(a) What is a galvanic cell?(b) Name the cathode and anode used in the Daniell cell.(c) Name the cell represented by Pt(S), H2(g)/H+(aq).(d) According to convention what is the potential of the above cell at all temperatures?(e) Write the use of the above cell.
Answer» (a) It is a device for converting chemical energy into electrical energy. The decrease in free energy in a spontaneous chemical process appears as elec trical energy. e.g., Daniell cell, (b) A zinc rod dipped in 1 M solution of ZnSO₄ acts as the anode. Here oxidation takes place. A copper rod dipped in 1 M solution of CuSO₄ acts as the cathode. Here reduction takes place.(c) This represents the andard Hydrogen Electrode (S.H.E), when it acts as the anode. (d) According to convention, S.H.E is assigned a zero potential at all temperatures (e) It is used as a primary reference electrode for determining the standard electrode potential of an unknown electrode. The electrode whose standard potential is to be determined is coupled with a reference electrode of known potential i.e., S.H.E to get a galvanic cell. The potential of the resulting galvanic cell is determined experimentally. E = E – E Knowing the potential of one electrode that of the other can be calculated.

2882.

We can construct innumerable number of galvanic cells on the pattern of Daniell cell by taking combi nation of different half cells.(a) What is a galvanic cell?(b) Name the cathode and anode used in the Daniell cell.(c) Name the cell represented by Pt(S), H2(g)/H+(aq).(d) According to convention what is the potential of the above cell at all temperatures?(e) Write the use of the above cell.

Answer»

(a) It is a device for converting chemical energy into electrical energy. The decrease in free energy in a spontaneous chemical process appears as elec trical energy. e.g., Daniell cell,

(b) A zinc rod dipped in 1 M solution of ZnSO₄ acts as the anode. Here oxidation takes place. A copper rod dipped in 1 M solution of CuSO₄ acts as the cathode. Here reduction takes place.(c) This represents the andard Hydrogen Electrode (S.H.E), when it acts as the anode.

(d) According to convention, S.H.E is assigned a zero potential at all temperatures

(e) It is used as a primary reference electrode for determining the standard electrode potential of an unknown electrode. The electrode whose standard potential is to be determined is coupled with a reference electrode of known potential i.e., S.H.E to get a galvanic cell. The potential of the resulting galvanic cell is determined experimentally. E = E – E Knowing the potential of one electrode that of the other can be calculated.

Discussion

2883.	Identification of anode and cathode in an Galvanic cell is made by the use ofA. galvanometerB. salt bridgeC. voltmeterD. none
Answer» Correct Answer - A The direction of deflection decides the nature of electrode .

Discussion

2884.	(i) For a weak electrolyte, molar conductance in dilute solution increases sharply as its concentration in solution is decreased. Give reason.
Answer» Upon dilution, the degree of dissociation (a) of a weak electrolyte increases. As a result, the number of ions in solution increases. Molar conductivity of the solution sharply increases. For more details, consult section 19

Discussion

2885.	Calculate the charge in coloumbs required for oxidation of 2 moles of water to oxygen ? (Given 1 F=96,500 C `mol^(-1)`)
Answer» The oxidation reaction is : `underset(2"mol")(2H_(2)O)to4H^(+)+O_(2)+underset(4F)4e^(-)` Change in coloumbs (Q)`=4F=4xx(96500" C")=386000" C"`

Discussion

2886.	Assertion (A) `E_("cell")` should have a positive value for the cell to function, Reason(R) `E_("cathode")ltE_("anode")`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 - C Reason is false. `E_("Cathode")gtE_("anode")`

Discussion

2887.	Assertion(A) Cu is less reactive than hydrogen. Reason(R) `E_(Cu^(2+)//Cu)^(o+)` is negative.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 - C ( C) Reason is false. `E_(CU^(2+)//CU)^(Ï´)` is positive.

Discussion

2888.	In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (i) Both assertion and reason are true and the reason is the correct explanation of assertion. (ii) Both assertion and reason are true and the reason is not the correct explanation of assertion. (iii) Assertion is true but the reason is false. (iv) Both assertion and reason are false. (v) Assertion is false but reason is true.Assertion : Cu is less reactive than hydrogen.Reason : EΘCu2+ /Cu is negative.
Answer» (iii) Assertion is true but the reason is false.

Discussion

2889.	Assertion(A) Cu is less reactive than hydrogen. Reason(R) `E_(Cu^(2+)//Cu)^(o+)` is negative.A. Both assertion and reason are true and the reason is the correct explanation of assertion.B. Both assertion and reason are true and the reason is not the correct explanation of assertion.C. Assertion is true but the reason is false.D. Both assertion and reason are false.
Answer» Correct Answer - C Correct Reason: `E_(Cu^(2+)//Cu)^(@)` is positive `(+0.34V)`

Discussion

2890.	Assertion(A) Cu is less reactive than hydrogen. Reason(R) `E_(Cu^(2+)//Cu)^(o+)` is negative.A. Both assertion and reason are true and the reason is the correct explanation of assertion.B. Both assertion and reason are true and reason is not the correct explanation of assertion.C. Assertion is true but the reason is false.D. Both assertion and reason are false.
Answer» Assertion is true but the reason is false Electrode potneial of `Cu^(2+)//CU` is +0.34V and Electrode potential of `2H^(+)//H_(2)` is 0.00V. Hence, correct reason is due to positive value of `Cu^(2+) //Cu` it looses electron to `H^(+)` and get reduces, while `H_(2)` gas evolves out.

Discussion

2891.	Match the followings . `{:(,"List - I" ,, ,"List- II") , (,(A) "Potential of" ,, (I), 0.76 V), (, "Hydrogen electrode at pH = 10 " ,, (II) , 0.0592 V), (, (B) Cu^(2+)\|Cu ,, (III) , -0.592 V), (, (C) Zn\|Zn^(2+) ,, (IV) , 0.337 V), (, (D) (2.303RT)/(F) ,, (V) , -.76V):}` The correct is -A. `{:( A, B , C , D) , (V , I , IV , II):}`B. `{:( A, B , C , D) , (III , I , II , V):}`C. `{:( A, B , C , D) , (II , V , I , IV):}`D. `{:( A, B , C , D) , (III , IV , I , II):}`
Answer» Correct Answer - D (A) pH = 10 `therefore [H^(+)] = 0 - (0.0592)/(1) "log" (1)/(10^(-10))` = `-0.0592 xx log 10^(10) = -0.0592 xx 10 = -0.592 V ` (B) `Cu^(2+) \| Cu , E_("Red")^(@) = +0.337 V` (C) `Zn^(2+) \| Zn = E_(Red)^(@) = -0.76 V` (D) `(2.303 RT)/(F) = 0.0592 V `

Discussion

2892.	Calculate the standard cell potentials of galvanic cells in which the following reactions take place: (i) `2Cr(s)+3Cd^(2+)(aq)to2Cr^(3+)(aq)+3Cd(s)` (ii) `Fe^(2+)(aq)+Ag^(2+)(aq)toFe^(3+)(aq)+Ag(s)` Given `E_(Cr^(3+),Cr)^(@)=-0.74V,E_(Cd^(2+),Cd)^(@)=-0.40V,E_(Ag^(+),Ag)^(@)=0.80V,E_(Fe^(3+),Fe^(2+))^(@)=0.77V` Also calculate `Delta_(r)G^(@)` and equilibrium constant of the reaction.
Answer» (i) `E_(cell)^(@)=E_(cathode)^(@)-E_(anode)^(@)=-0.40V-(-0.74V)=+0.34V` `Delta_(r)G^(@)=-nFE_(cell)^(@)=-6molxx96500" C "mol^(-1)xx0.34V` `=-196860" CV "mol^(-1)=-196860J" "mol^(-1)=-196.86" kJ "mol^(-1)` `-Delta_(r)G^(@)=2.303" RT "logK` `196860=2.303xx8.314xx298logK` or `logK=34.5014` K=Antilog 34.5014=3.192`xx10^(34)` (ii) `E_(cell)^(@)=+0.80V-0.77V=+0.03V` `Delta_(r)G^(@)=-nFE_(cell)^(@)=-(1mol)xx(96500" C "mol^(-1))xx(0.03V)` `=-2895" CV "mol^(-1)=-2895" J "mol^(-1)` ltBrgt `=-2.895" kJ "mol^(-1)` `Delta_(r)G^(@)=-2.303" RT "logK` `-2895=-2.303xx8.314xx298xxlogK` or log K`=0.5974` or K=Antilog (0.5974)=3.22.

Discussion

2893.	Using the data given below: `E_(Cr_(2)O_(7)^(2-)\|Cr^(3+))^(@)=1.33V E_(Cl_(2)\|Cl^(-))^(@)=1.36V` `E_(MnO_(4)^(-)\|Mn^(2+))^(@)=1.51V E_(Cr^(3+)\|Cr)=-0.74V` Mark the strongest reducing agent.A. `Cl^-`B. `Cr`C. `Cr^(3+)`D. `Mn^(2+)`
Answer» Correct Answer - B Lower the value of reduction potential , stronger is the reducing agent i.e., undergoes oxidation most easily. `Cr to Cr^(3+) + 3e^(-)` (oxidation) Hence, Cr is ther strongest reducing agent .

Discussion

2894.	Calculate `E_(cell)^(@)` for the following reaction at `25^(@)C`. `A+B^(2+)(0.001 M) to A^(2+)(0.0001 M)+B` (Given. `E_(cell)=2.6805 V, 1 F=96500 C mol^(-1)`
Answer» According to Nernst equation `E_(cell)=E_(cell)^(@)-(0.0591)/(n)"log"([Anode])/([Cathode])` `E_(cell)^(@)=E_(cell)+(0.0591)/(2)"log"([0.0001])/([0.001])=2.6805+0.02955" log"10^(-1)"` =2.6805-0.02955=2.6510 V

Discussion

2895.	The following chemical reaction occurring in an electrochemical cell : `Mg(s)+2Ag^(+)(0.0001 M) to Mg^(2+)(0.10 M)+2Ag(s)` Given `E_(Mg^(2+)//Mg)^(@)=-2.36" V", E_(Ag^(+)//Ag)^(@)=+0.81" V"` For the cell, calculate/write : (i) `E^(@)` value for the electrode `2Ag^(+)//2Ag` (ii) Standard cell potential `(E^(@))` (iii) Cell potential (E) (iv) Give the symbolic representation of the above cell (v) Will the above cell reaction be spontaneous ?
Answer» (i) `E^(@)` value of the electrode , `2Ag^(+)//2Ag` will remain the same i.e., 0.81 V. It does not change. (ii) `" " E_(cell)^(@)=E_(cathode)^(@)-E_(anode)^(@)=0.81-(-2.36)=3.17" V"` (iii) `" " E_(cell)=E_(cell)^(@)=((0.0591" V"))/(2)"log"([Mg^(2+)(aq)])/([Ag^(+)(aq)]^(2))` `=(3.17" V")-((0.0591" V"))/(2)"log"((0.1))/((0.0001)^(2)` `=(.17" V")-(0.2068" V")=2.9632" V" ` (iv) Cell notation :` Mg(s)//Mg^(2+)(0.10" M") \|\| 2Ag^(+)(0.0001" M")//2Ag(s)` (v) Since Mg is placed below Ag in the activity series, it is a stronger reducing agent. Therefore, the cell reaction is spontaneous.

Discussion

2896.	Using the data given below: `E_(Cr_(2)O_(7)^(2-)\|Cr^(3+))^(@)=1.33V E_(Cl_(2)\|Cl^(-))^(@)=1.36V` `E_(MnO_(4)^(-)\|Mn^(2+))^(@)=1.51V E_(Cr^(3+)\|Cr)=-0.74V` In which option the order of reducing power is correct?A. `Cr^(3+) lt Cl^(-) ltMn^(2+) ltCr`B. `Mn^(2+) lt Cl^(-) lt Cr^(3+) lt Cr`C. `Cr^(3+) lt Cl^(-) lt Cr_(2)O_(7)^(2-) lt MnO_(4)^(-)`D. `Mn^(2+) lt Cr^(3+) lt Cl^(-) lt Cr`
Answer» Correct Answer - B (b) It is the correct order of increasing reducing strength.

Discussion

2897.	Calculate equilibrium constant for the reaction : `Mg(s) \|Mg^(2+)(0.001 M) \|\| Cu^(2+)(0.0001 M)\|Cu(s)` Given `E_((Mg^(2+)//Mg))^(@)=-2.37" V " , E_((Cu^(2+)//Cu))^(@)=0.34" V "`
Answer» Correct Answer - `4.9xx10^(90)` Step I. Calculation of emf of cell Cell reaction : `Mg(s)+Cu^(2+)(0.0001" M")toMg^(2+)(0.01" M")+Cu(s)` `E_(cell)=E_(cell)^(@)-(0.0591)/(2)"log"([Mg^(2+)])/([Cu^(2+)])=[0.34-(-2.37)]-(0.0591)/(2)"log"((0.001))/((0.0001))` `=2.71-0.02955" log "(10)=2.71-0.02955=2.68" V "` Step II. Calculation of equilibrium constant `logK_(C)=(nE_(cell)^(@))/(0.0591)=(2xx2.68)/(0.0591)=90.69,K_(C)="Antilog "90.69=4.9xx10^(90)`.

Discussion

2898.	Calculate the e.m.f. of the cell, `Mg//Mg^(2+)(0.1 M)\|\|Ag^(+)(1.0xx10^(-3)M)//Ag` The values of `E_(Mg^(2+)//Mg)^(@)` and `E_(Ag^(+)//Ag)^(@)` are -2.37`" V"` and +0.80`" V "` respectively.
Answer» Correct Answer - 3.02`" V "` Cell reaction : `Mg(s)+2Ag^(+)(1.0xx10^(-3)M) to Mg^(2+)(0.1 M)+2Ag(s)` `E_(cell)=E_(cell)^(@)-(0.0591)/(2)"log"([Mg^(2+)])/([Ag^(+)]^(2))=[+0.8-(-2.37)]-(0.0591)/(2)"log"(0.1)/(1xx10^(-3))^(2)` `=3.17-0.02955" log "10^(5)=3.17-0.02955xx5` `=3.17-0.14775=3.02225" V "=3.02 "V "`.

Discussion

2899.	Calculate emf and `DeltaG` for the following reaction at 298 K `Mg(s)\|Mg^(2+)(0.01 M)\|\|Ag^(+)(0.0001 M)\|Ag(s)` Given `E_((Mg^(2+)//Mg))^(@)=-2.37" V ", E_((Ag^(+)//Ag))^(@)=+0.80" V "`
Answer» Correct Answer - `"emf"=2.9930" V ";DeltaG=-577.649" kJ mol^(-1)` Step I. Calculation of `E_(cell)` `E_(cell)^(@)=E_(cathode)^(@)-E_(anode)^(@)=(+0.80" V")-(-2.37" V")=3.17" V"` According to Nernst equation `E_(cell)=E_(cell)^(@)-(0.0591)/(n)"log"(["Anode"])/(["Cathode"])` `=3.17V-(0.0591(V))/(2)"log"((0.01))/((0.0001)^(2))` `=3.17" V"-(0.02955" V") log10^(6)=3.17" V"-(0.02955" V")+6` `=3.17" V"-0.1770" V"=2.9930" V"` Step II. Calculation of `DeltaG` `DeltaG=-nFE_(cell)=(-2)xx(96500" C mol"^(-1))xx(2.9930" V")` `=-(577649"V")mol^(-1)=-577649" J mol"^(-1)=-577.649" kJ mol^(-1)`

Discussion

2900.	Calculate the emf of the cell. `Mg(s)\|Mg^(2+) (0.2 M)\|\|Ag^(+) (1xx10^(-3))\|Ag` `E_(Ag^(+)//Ag)^(@)=+0.8` volt, `E_(Mg^(2+)//Mg)^(@)=-2.37` volt What will be the effect on emf If concentration of `Mg^(2+)` ion is decreased to `0.1 M` ?
Answer» Correct Answer - 3.0134`" V "`; increases Step I. Calculation of EMF of the cell Cell reaction : `Mg(s)+2Ag^(+)(1xx10^(-3)" M")to Mg^(2+)(0.2" M") +2Ag(s)` `E_(cell)=E_(cell)^(@)-(0.0591)/(n)"log"([Mg^(2+)])/([Ag^(+)]^(2))=[0.8-(-2.37)]-(0.0591)/(2)"log"(0.1)/((1xx10^(-4))^(2)` `=3.17-0.02955" log "(2xx10^(5))=3.17-0.02955xx5.3010` `=3.17-0.15664=3.0134" V "` Step II. Effect of decreasing concentration of `Mg^(2+)` ions on emf of cell The EMF of cell wil increase on decreasing the concentration of `Mg^(2+)` ions in solution.

Discussion

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