3195 + Interview Questions in ELECTROCHEMISTRY IN GENERAL AWARENESS Page 5 InterviewSolution

201.	State the products of electrolysis obtained on the cathode and the anode in the following cases: (i) A dilute solutoin of `H_(2)SO_(4)` with platinum electrodes (ii) An aqueous solution of `AgNO_(3)` with silver electrodes.
Answer» (i) `H_(2)SO_(4)(aq)to2H^(+)(aq)+SO_(4)^(2-)(aq)," "H_(2)OhArrH^(+)+OH^(-)` At cathode: `H^(+)+etoH,H+HtoH_(2)(g)` At anode: `OH^(-)toOH+e^(-),4OHto2H_(2)O(l)+O_(2)(g)` Thus, `H_(2)` is liberated at the cathode and `O_(2)` at anode. (ii) `AgNO_(4)(s)+aqtoAg^(+)(aq)+NO_(3)^(-)(aq)," "H_(2)OhArrH^(+)+OH^(-)`. At cathode. `Ag^(+)` ions ahve lower discharge potential than `H^(+)` ions. hence, `Ag^(+)` ions will be deposited on the cathode `(Ag^(+)+e^(-)toAg)` At anode. Ag anode will be attached by `NO_(3)^(-)` ions. hence, Ag anode will dissolve to form `Ag^(+)` ions in the solution `(AgtoAg^(+)+e^(-))`

Discussion

202.	In electrolysis of dilute `H_(2)SO_(4)` what is liberated at anode?A. `H_(2)`B. `SO_(4)^(2-)`C. `SO_(2)`D. `O_(2)`.
Answer» Correct Answer - D Durrin g electrolysis of dil `H_(2)SO_(4),O_(2)` is liberated at anode (see A-Level information).

Discussion

203.	Degree of dissociation of pure water is `1.9xx10^(-9)`. Molar ionic conductances of `H^(+)` and `OH^(-)` ions at infinite diluton are 200 S `cm^(2)mol^(-1) and 250" S "cm^(2)mol^(-1)` respectively. Molar conductance of water isA. `3.8xx10^(-7)" S "cm^(2)mol^(-1)`B. `5.7xx10^(-7)" S "cm^(2)mol^(-1)`C. `9.5xx10^(-7)" S "cm^(2)mol^(-1)`D. `1.045xx10^(-6)" S "cm^(2)mol^(-1)`
Answer» Correct Answer - D Degree of dissociation `(alpha)=(wedge_(m)^(c))/(wedge_(m)^(0))` `wedge_(m)^(c)=alphaxxwedge_(m)^(0)=(1.9xx10^(-9))xx(200+350)" S "cm^(2)mol^(-1)` `=1.9xx550xx10^(-9)=1.045xx10^(-6)" S "cm^(2)mol^(-1)`

Discussion

204.	When an electric cell is charged thenA. voltage of cell increasesB. electrolyte of cell dilutes.C. resistance of cell increaseD. None of these.
Answer» Correct Answer - A When an electric cell is charged its voltage increases.

Discussion

205.	The half cell reaction for rusting of iron are: `2H^(+)+2e^(-)+(1)/(2)O_(2)rarrH_(2)O(l), E^(@)=+1.23V` `Fe^(2+)+2e^(-)rarrFe(s), E^(@)=-0.44V` `DeltaG^(@)` (in KJ) for the reaction isA. `-76`B. `-322`C. `-161`D. `-152`.
Answer» Correct Answer - B `E^(@)=1.23+0.44=1.67V` `DeltaG^(@)=-nFE^(@)` `=02xx96500xx1.67=-322310J` `=-322.31kJ`

Discussion

206.	The half cell reaction for rusting of iron are: `2H^(+)+2e^(-)+(1)/(2)O_(2)rarrH_(2)O(l), E^(@)=+1.23V` `Fe^(2+)+2e^(-)rarrFe(s), E^(@)=-0.44V` `DeltaG^(@)` (in KJ) for the reaction isA. `-76`B. `-322`C. `-122`D. `-176`
Answer» Correct Answer - B For e.m.f. to be _ve, oxidation should occur at iron electrode, `E_("cell")=1.23+1.67V` `DeltaG=-nFE_("cell")^(@)=2xx 96500xx1.67`. =-322kJ.

Discussion

207.	How long a current of `3A` has to be passed through a solution of silver nitrate to coat a metal surface of `80cm^(2)` with a `0.005-mm-` thick layer ? The density of silver is `10.5g cm^(-3)`.
Answer» Correct Answer - `125.09s` Volume of the surface `(V)=` Area `xx` Thickness Given, Area `=80cm^(2),` thickness `=0.005mm=0.00005cm` Mass of `Ag(W)=Vxx` Density `=0.04xx10.5=0.42g` `Ag^(o+)+e^(-) rarr Ag` `:. W_(Ag)=(Zit)/(96500)` `0.42=(108xx3xxt)/(96500)` `:. t=(0.42xx96500)/(108xx3)=(40530)?(324)=125.90s`

Discussion

208.	The value of reaction quotient `Q` for the cell `Zn(s)\|Zn^(2+)(0.01M)\|\|Ag^(o+)(1.25M)\|Ag(s) is `A. 156B. 125C. 1.25xx10^(-2)D. 64xx10^(-3)
Answer» Correct Answer - d Write cell reaction `:` `Zn(s)+2Ag^(o+)(1.25M)rarr Zn^(2+)(0.01M)+2Ag(s)` `A=([Zn^(2+)])/([Ag^(o+)])=(0.1)/((1.25)^(2))=64xx10^(-3)`

Discussion

209.	Electrolysis of molten anphdrous calcium chloride produces .A. calciumB. phosphorusC. sulphurD. sodium
Answer» Correct Answer - A Calcium is produced when molten anydrous cacium chloride is electrolysed.

Discussion

210.	Electrolysis of molten anhydrous calcium chloride produces .A. CalciumB. PhosphorusC. SulphurD. Sodium
Answer» Correct Answer - A Calcium is produced when molten anhydrous calcium chloride is electrolysed.

Discussion

211.	How long has a current of 3 ampere to be applied through a solution of silver nitrate to coat a metal surface of `80 cm^(2)` with 0.005 mm thick layer ? Density of silver is `10.5 g//cm^(3)`.
Answer» Mass of silver to be deposited = volume `xx` density = area `xx` thickness `xx` density Given : Area `=80 cm^(2)`, thickness =0.0005 cm and density `=10.5 g//cm^(3)` Mass of silver to be deposited `=80xx0.0005xx10.5` `=0.42 g` Applying to silver `E=Zxx96500` `Z=108/96500 g` Let the current be passed for t seconds. We know that, `W=ZxxIxxt` So, `0.42=108/96500xx3xxt` or `t=(0.42xx96500)/(108xx3)=125.09` second

Discussion

212.	In acid medium, `MnO_(4)^(c-)` is an oxidizing agent. `MnO_(4)^(c-)+8H^(o+)+5e^(-) rarr Mn^(2+)+4H_(2)O` If `H^(o+)` ion concentration is doubled, electrode potential of the half cell `MnO_(4)^(c-), Mn^(2+)\|Pt` willA. Increase by `28.36mV`B. Decrease by `28.36mV`C. Increase by `14.23mV`D. Decrease by `142.30mV`
Answer» Correct Answer - a `MnO_(4)^(c-)+8H^(o+)+5e^(-)rarr Mn^(2+)+4H_(2)O` `E_(MnO_(4)^(c-)\|Mn^(2+))` `=E^(c-)._(MnO_(4)\|Mn^(2+))-(0.059)/(5)log(([Mn^(2+)])/([MnO_(4)][H^(o+)]^(8)))` `=E^(c-)._(MnO_(4)^(c-)\|Mn^(2+))-(8)/(5)xx0.059pH` `-(0.059)/(5)log(([Mn^(2+)])/([MnO_(4)^(c-)]))` `implies[H^(o+)]` is doubled, `i.e., pH` is reduced by `log 2 -= 0.3,` then `E_(MnO_(4)^(c-)\|Mn^(2+))` will be changed `(` increase `)` by `(8)/(5)xx0.059xx0.3V=28.36mV`

Discussion

213.	`CH_(3)COOH` is titrated with `NaOH` solution. Which of the following statements is true ?A. Conductance decreases upto equivalence point, after which it increases.B. Conductance increases upto equivalence point,after which it decreases.C. Conductance first decreases `(` but not rapidly`)` and then increases upto equibalence point and then increases rapidley after equivalence point.D. None of these
Answer» Correct Answer - c Refer Section.

Discussion

214.	A factory produces 40 kg of calcium in two hours by electrolysis. How much aluminium can be produced by same current in 2 hours if current efficiency is 50%?A. 22 kgB. 18 kgC. 9 kgD. 27 kg
Answer» Correct Answer - B `("Wt. of Ca")/("Wt. of Al") = ("Eq. mass of Ca")/("Eq. mass of Al") , (40)/("Wt. of Al") = (40//2)/(27//3)` or Wt. of Al = 18 kg .

Discussion

215.	The time required to coat a metal surface of `80 cm^(2)` with `5 xx 10^(-3) cm` thick layer of silver (density `1.05 g cm^(-3)`) with the passage of `3A` current through a silver nitrate solution is:A. 1150 sB. 1250 sC. 1350 sD. 1450 s
Answer» Correct Answer - B Vol. of Ag = `800 xx 5 xx 10^(-4) cm^(3)` . Mass of Ag = `V xx d = 0.4 xx 10.5 = 4.2` g `Ag^(+) + e^(-) to Ag`, 108 g Ag require , 96500 C , `therefore 4.2` g require = `(96500)/(108) xx 4.2 = 3753` C . Hence , `t = (Q)/(I) = (3753)/(3) = 1251` sec .

Discussion

216.	During the electrolysis of `AgNO_(3)`, the volume of `O_(2)` formed at `STP` due to passage of `2A` of current for `965 s` isA. `0.112L`B. `0.224L`C. `11.2L`D. `22.4L`
Answer» Correct Answer - a Number of Faradays `=(It)/(96500)=(2xx9650)/(96500)=0.02` Anode `:2H_(2)O(l) rarr O_(2)(g)+4H^(o+)(aq)+4e^(-)` `implies4F` electricity `=1 mol O_(2)-=22.4L O_(2)` at `STP` `implies0.2F` electricity `=22.4xx(0.02)/(4)` `=0.112L O_(2)` at `STP`

Discussion

217.	At STP 1.12 litre of `H_(2)` is obtained on flowing a current for 965 seconds in a solution . The value of current isA. 10B. 1C. 1.5D. 2
Answer» Correct Answer - A `2H^(+)+underset(2"mole"2xx96500C)(2e^(-))rarrunderset("1mole"22.4"lt")(H_(2)` To liberate 22.5 lt of `H_(2)` gas, electricity required `=2xx96500C` To liberate 1.12 lt of `H_(2)` gas, electricity required `=(2xx96500xx1.12)/(965"sec")=1 0` ampere

Discussion

218.	The mass of gases evolved in Question isA. `0.06g`B. `0.6g`C. `6.0g`D. `60g`
Answer» Correct Answer - a Total mass of `C_(2)H_(6)+2CO_(2)+H_(2)-=30+88+2=120g` `:. 2xx96500Cimplies120g` `0.1xx965C=(120xx0.1xx965)/(2xx96500)implies0.06g`

Discussion

219.	At STP 1.12 litre of `H_(2)` is obtained on flowing a current for 965 seconds in a solution . The value of current in amperes isA. 10B. `1.0`C. `1.5`D. `2.0`
Answer» Correct Answer - A `1.12 L H_(2) = 0.1 g eq. of H_(2)` Q = 0.1 F = 9650 C . Hence , `I = (Q)/(t) = (9650)/(965) = 10 A` .

Discussion

220.	The volume of gases evolved at `STP` by passing `0.1A` of current for `965g`, through an aqueous solution of potassium acetateA. `22.4mL`B. `11.2mL`C. `89.6mL`D. `44.8mL`
Answer» Correct Answer - d `CH_(3)COOK rarr CH_(3)OO^(c-)+K^(o+)` At cathode, `H_(2)O` undergoes reduction of give `H_(2)`. `H_(2)O+2e^(-) rarr 2overset(c-)(O)H+H_(2)` `2F-=1 mol of H_(2)` At anode, `2CH_(3)COO^(e)rarrunderset(underset(CH_(3)-CH_(3)+2CO_(2))(darr))(2CHOO^(*)+2e^(-))` `2F-=1 mol of C_(2)H_(6)` and `2 mol of CO_(2)` Total volume at cathode and anode `=4 mol =4xx22.4L` `2F=2xx96500Cimplies4xx22.4L` `0.1xx965Cimplies(4xx22.4xx0.1xx965)/(2xx96500)` `implies0.0448L=44.8 mL`

Discussion

221.	What is the electrode potential of a gasous hydrogen electrode dipped in a solution `pH=5.0` relative to the calomel electrode with an electrode potential of `+0.28V` ?A. `0.0125V`B. `+0.575V`C. `+0.015V`D. `-0.575V`
Answer» Correct Answer - d `E_(H_(2))=-0.059xx5=-0.295V` `E_(calomel)=0.28V` `E_(H_(2))` relative to the calomel electrode is `=-0.295-0.28=-0.575V`

Discussion

222.	The specific conductance of saturated solution of AgCl is found to be `1.86xx10^(-6) ohm^(-1) cm^(-1)` and that of water is `6 xx 10^(-8) ohm^(-1) cm^(-1)`. The solubility of agCl is ... . Given, `Lambda_(AgCl)^(@)=137.2 ohm^(-1) cm^(2) eq^(-1)`A. `1.7xx10^(-3) M`B. `1.3xx10^(-5) M`C. `1.3xx10^(-4) M`D. `1.3xx10^(-6) M`
Answer» Correct Answer - B `kappa_(AgCl)=kappa_(AgCl" (Solution)")-kappa_(H_(2)O)` `=1.86xx10^(-6)-6xx10^(-8)=1.8xx10^(-6) ohm^(-1) cm^(-1)` `Lambda_(AgCl)^(@)=kappaxx1000/S` `:. S=(kappaxx1000)/(Lambda_(AgCl)^(@))=(1.8xx10^(-6)xx1000)/137.2=1.31xx10^(-5) M`

Discussion

223.	The specific conductivity of N/10 KCl solution at `20^(@)C` is `0.0212 ohm^(-1) cm^(-1)` and the resistance of the cell containing this solution at `20^(@)C` is 55 ohm. The cell constant is :A. `4.616 cm^(-1)`B. `1.166 cm^(-1)`C. `2.173 cm^(-1)`D. `3.324 cm^(-1)`
Answer» Correct Answer - B `kappa=Cxxl/A` `l/A=kappaxx1/C=kappaxxR=0.0212xx55=1.166 cm^(-1)`

Discussion

224.	Out of Cu, Ag, Fe and Zn, the metal which can displace all others from their salt solutions is :A. AgB. CuC. FeD. Zn
Answer» Correct Answer - D

Discussion

225.	Out of `Cu, Ag, Fe` and `Zn`, the metal which can displace all others from their salt solution isA. `Ag`B. `Cu`C. `Zn`D. `Fe`
Answer» Correct Answer - C The metal placed above in electromical series, can displace the metals placed below, from their salt solution. Hence, `Zn` can displace `Fe, Ag, Cu`

Discussion

226.	Fuel cell is an electrical cell which converts chemical energy into electrical energy. The most successful fuel cell is `H_(2)-O_(2)` fuel cell, which is known as Bacon cell. It had been used to fulfil the electric power supply required in Appolo mission. This fuel cell is pollution free. Fuel cells are preferred to other energy producing devices in space because ofA. high efficiencyB. pollution free natureC. less weightD. all of these
Answer» Correct Answer - D (d) All are correct.

Discussion

227.	Fuel cell is an electrical cell which converts chemical energy into electrical energy. The most successful fuel cell is `H_(2)-O_(2)` fuel cell, which is known as Bacon cell. It had been used to fulfil the electric power supply required in Appolo mission. This fuel cell is pollution free. The cell used in Appolo mission wasA. Leclanche cellB. Daniell cellC. Voltaic cellD. Bacon cell
Answer» Correct Answer - D (d) Bacon cell was used in Appolo mission. It is named after the British Scientist who developed it.

Discussion

228.	Fuel cell is an electrical cell which converts chemical energy into electrical energy. The most successful fuel cell is `H_(2)-O_(2)` fuel cell, which is known as Bacon cell. It had been used to fulfil the electric power supply required in Appolo mission. This fuel cell is pollution free. The fuel used in the cell used in Appolo mission wasA. `H_(2)`B. `H_(2)-O_(2)`C. `CH_(4)`D. `O_(2)`
Answer» Correct Answer - B (b) It is a hydrogen-oxygen fuel cell.

Discussion

229.	The molar conductances of `NaCI, HCI` and `CH_3 COONa` at infinite dilution are `12.45, 426. 16` and ` 91 "ohm"^(-1) cm^2 "hol"^(-1)` respectively. The molar conductance of `CH_3COOH` at infinite dilution is .A. ` 201 .28 ohm^(-1) cm^2 "mol^"(-1)`B. ` 39071 "ohm"^(-1) cm^2 "mol"^(-1)`C. ` 698.28 "ohm"^(-1) cm^2 "mol"^(-1)`D. ` 540.48 ohm^(-1) cm^2 mol^(-1)`
Answer» Correct Answer - B ` ^^_m^@ (CH_3COOH)` `= (CH_3COONa) + ^^@(HCl) -^^@ (NaCl)` `= 91 + 426 . 16 - 12 . 45` ` = 390. 71 "ohm"^(-1) cm^@ "mol"^(-1)`.

Discussion

230.	Calculte molar conductance at infinite dilution for acetic acid, given `A_(m)^(oo)HCI=425 ohm^(-1)cm^(-1),A_(m)^(oo)NaCI=188 ohm^(-1)cm^(-1),A_(m)^(oo)CH_(3)COOHNa=96 ohm^(-1)cm^(-2)mol^(-1)`
Answer» Correct Answer - `333 ohm^(-1)cm^(2)mol^(-1)` `Lambda_(m(CH_(3)COOH))^(oo)=Lambda_(m(CH_(3)COONa))^(oo)+Lambda_(m(HCl))^(oo)-Lambda_(m(NaCl))^(oo)` `=(96.0+425.0-188.0)" ohm"^(-1)cm^(2)mol^(-1)=333" ohm"^(-1)cm^(2)mol^(-1)`.

Discussion

231.	Why `wedge_(m)^(@)` for `CH_(3)COOH` cannot be determined experimentally?
Answer» Molar conductivity of weak electrolytes keeps on increasing with dilution and does not become constant even at very large dilutions.

Discussion

232.	Molar conductance at infinite dilution of acetic acid cannot be determined experimentally.1. How can it be determined?2. State the law.3. Give two sets of electrolytes which can be used to nd out the molar conductance of acetic acid at infinite dilution.
Answer» 1. It can be determined using Kohlrausch’s law of independent migration of ions. 2. Molar conductivity at infinite dilution of an electrolyte is the sum of molar ionic conductivities of the cation and anion at infinite dilution. 3. CH₃COONa, HCl, NaCl & CH₃COOK, HCl, KCl.

232.

Molar conductance at infinite dilution of acetic acid cannot be determined experimentally.1. How can it be determined?2. State the law.3. Give two sets of electrolytes which can be used to nd out the molar conductance of acetic acid at infinite dilution.

Answer»

1. It can be determined using Kohlrausch’s law of independent migration of ions.

2. Molar conductivity at infinite dilution of an electrolyte is the sum of molar ionic conductivities of the cation and anion at infinite dilution.

3. CH₃COONa, HCl, NaCl & CH₃COOK, HCl, KCl.

Discussion

233.	Copper sulphate solution is electrolysed between two platinum electrodes. A current is passed unit 1.6 g of oxygen is liberated at anode. The amount of copper deposited at the cathode during the same period is:A. `6.36 g`B. `63.6g`C. `12.7 g`D. `3.2g`
Answer» Correct Answer - A

Discussion

234.	Exactly 0.2 mole electrons are passed through two electrolytic cells in series containing `CuSO_(4)` and `ZnSO_(4)` respectively. How many grams of each metal will be deposited on the respective cathodes in the two cells ?
Answer» Gram equivalent mass of copper `=(63.5)/(2)=31.75 g` Gram equivalent mass of zinc `=(65.0)/(2)=32.50 g` Now, `" " 1.0" mole"` of electrons deposit copper=31.75 g 0.2 mole of electrons deposit copper `=((31.75 g)xx(0.2 mol))/((1.0 mol))=6.35 g` Similarly, 1.0 mole of electrons deposit zinc=32.50 g 0.2 mole of electrons deposit zinc`=((32.50g)xx(0.2 mol))/((1.0 mol))=6.50 g`

Discussion

235.	A solution of copper (II) sulphate us electrolysed between copper electrodes by a current of 10 amperes exactly for 1 hour. What changes occur at the electrodes and in the solution ?
Answer» Quantity of charge passed (Q) `=Ixxt=(10 amp)xx(60xx60s)`=3600 As =3600 C Gram equivalent mass of copper `=("Gram atomic mass")/("Valency")=(63.5)/(2)=31.75` 96500 C of charge deposit copper =31.75 g 36000 C of charge deposit copper `=((31.75g))xx((36000 C))/((96500 C))=11.84 g` Thus, 11.84 g of copper will dissolve from the anode and the same amount of copper will be deposited on the cathode. The concentration of the solution will remain unchanged.

Discussion

236.	An electrochemical cell is used to convertA. potential energy changes into kinetic energyB. kinetic energy changes into potntial energyC. chemical energy changes into electrical energyD. electrical energy changes into chemical energy
Answer» Correct Answer - C In the electrochemical cell chemical energy chanes into electrocal energy.

Discussion

237.	Which of the following statements for the electrochemical daniel cellA. Electrons flow from copper electrode to zinc electrodeB. Current flows from zinc electrode to copper electrodeC. Cations move towards copper electrode which is cathodeD. Cations move towards zinc electrode
Answer» Correct Answer - C In Daniel cell copper rod acts as cathode so there cations move towards copper electrode and reduction take place on copper rod.

Discussion

238.	`E^(@)` for `Fe//Fe^(2+)` is `+0.44 V` and `E^(@)` for `Cu//Cu^(2+)` is `-0.32 V`. Then, in the cell,A. `Cu` oxidises `Fe^(2+)` ionB. `Cu^(2+)` oxidises ironC. `Cu` reduces `Fe^(2+)` ionD. `Cu^(2+)` reduces `Fe`
Answer» Correct Answer - B `underset("Oxidation")(Fe)+underset("Reduction")(Cu^(2+)) rarr Fe^(2+) +Cu` `E_("cell")^(@)=E_(Fe//Fe^(2+))^(@)+E_(Cu^(2+)//Cu)^(@)` `=0.44+0.32` `=0.76 V`

Discussion

239.	Which of the following statement is true for the electrochemical Daniell cell ?A. Electrons flow from copper electrode to zinc electrodeB. Current flows from zinc electrode to copper electrodeC. Cations move toward copper electrode which is cathodeD. Cations move toward zinc electrode
Answer» Correct Answer - C In Daniel cell copper rod acts as cathode so there cations move towarrds copper electrode and reduction take place on copper rod.

Discussion

240.	Which one of the following metals cannot veolve `H_2` from acids or `H_2O` from its componds ?A. ` Hg`B. ` Al`C. `Pb`D. `Fe`
Answer» Correct Answer - A Hg has greater reduction potntial than that of `H^+` and hence cannot displace hudrogen from acid .

Discussion

241.	The chemical reaction, `2AgCl_((s))+H_(2(g))to2HCl_((aq))+2Ag_((s))` taking place in a galvanic cell is represented by the notationA. `Pt\|H_(2(g)),1"bar"\|1M" "KCl_((aq))//AgCl_((s))\|Ag_((s))`B. `Pt_((s))\|H_(2(g)),1"bar"\|1M" "HCl_((aq))\|\|1MAg_((aq))^(+)\|Ag_((s))`C. `Pt_((s))\|H_(2(g)),1"bar"\|1M" "HCl_((aq))\|AgCl_((s))\|Ag_((s))`D. `Pt_((s))\|H_(2(g)),1"bar"\|1M" "HCl_((aq))\|Ag_((s))\|AgCl_((s))`
Answer» Correct Answer - B `2AgCl_((s))+H_(2(g))to2HCl_((aq))+2Ag_((s))` the activities of solids and liquid are takes as unity and at low concentrations, the activity of a solute is approximated to its molarity. The cell reaction will be `Pt_((s))\|H_(2(g)),1"bar"\|H_((aq))^(+)1M\|AgCl_((aq))1M\|Ag_((s))`.

Discussion

242.	The chemical reaction `2AgCl_("(fused)")+H_(2(g))rarr2HCl_((aq))+2Ag_((s))` taking place in a galvanic cell is represented by the notationA. `Pt(s)\|H_(2)(g),1 "bar"\|\|1M KCl(aq)\|AgCl(s)\|Ag(s)`B. `Pt(s)\|H_(2)(g),1"bar"\|1M HCl(aq(\|\|1MAg^(+)(aq)\|Ag(s)`C. `Pt(s)\|H_(2)(g),1"bar"\|1M HCl(aq)\|\|AgCl(s)\|Ag(s)`D. `Pt(s)\|H_(2)(g),1"bar"\|1MHCl(aq)\|\|Ag(s)\|AgCl(s)`.
Answer» Correct Answer - C See A-Level information.

Discussion

243.	Deduce from the following `E^(c-)` values of half cells, what combination of two half cells would results in a cell with the largest potential? `{:(I. , A+e(-) rarr A^(c-),,,,E^(c-)=-0.24V),(II., B^(c-)+e^(-)rarr B^(2-),,,,E^(c-)=+1.25V),(III.,C^(c-)+2e^(-) rarr C^(3-),,,,E^(c-)=-1.25V),(IV. , D+2e^(c-)rarr D^(2-),,,, E^(c-)=+0.68V):}`A. (ii) and (iii)B. (ii) and (iv)C. (i) and (iii)D. (i) and (iv)
Answer» Correct Answer - A Calculate `E_(cell)^(@)` for each using , `E_(Cell)^(@) = E_(RHS)^(@) - E_(LHS)^(@)`

Discussion

244.	`E^(@)` for `Fe//Fe^(2+)` is `+0.44 V` and `E^(@)` for `Cu//Cu^(2+)` is `-0.32 V`. Then, in the cell,A. Cu oxidises `Fe^(2+)` ionB. `Cu^(2+)` oxidises FeC. Cu reduces `Fe^(2+)` ionD. `Cu^(2+)` reduces Fe
Answer» Correct Answer - B `E_(OP)^(@)` of `Fe gt E_(OP)^(@)` of Cu [ `because E_(RP (Fe))^(@) lt E_(RP (Cu))^(@)]` Thus Fe gets oxidised of `Fe to Fe^(2+) + 2e^(-)`, `Cu^(2+) + 2e^(-) to Cu`

Discussion

245.	The SOP , `E^(@)` for the half reactions are `Zn to Zn^(2+) + 2e , E^(@) = + 0.76 V` `Ag to Ag^(+) + e , E^(@) = -0.77 V` `E^(@)` of the cell , `Ag^(+)+ Zn to Zn^(2+) + Ag` is :A. `+ 1.53`B. `-1.53 `C. `-0.01`D. `+0.01`
Answer» Correct Answer - A `E_("cell")^(@) = E_(OP_(Zn))^(@) + E_(RP_(Ag))^(@) = 0.76 + 0.77 = 1.53` V

Discussion

246.	The chemical reaction , `2 Ag Cl (s) rarr 2HCl(aq) + 2Ag(s)` taking place in a galvanic cell is represented by the notation .A. `Pt \|H_2(g), 1 "bar" \|1 M KCl(aq) \|AgCl (s) \|Ag(s)`B. `Pt(s) \|H_2(g), 1 "bar" \|1 "bar" \|a M HCl(aq) \|\| 1 Mag^+ (aq) Ag(s)`C. `Pt (s) +H_2(g).1 \|"bar" \|MHCl (aq) \|AgCl(s) \|Ag(s)`D. `Pt (s) +H_2(g).1 \|"bar" \|MHCl (aq) \|AgCl(s) \|Ag(s)`
Answer» Correct Answer - C ` 2 AgCl(s) + H_2(g) rarr HCl (aq) + 2 Ag (s)` The activites of solids and liquids are taken as unity and at low concentratios, the activity of a solute is approximated to its molarity . The cell reaction will be `Pt (s) \| H_2 (g) , 1 "bar "\|H^+ (aq) 1M \| Ag Cl (aq) 1M \| Ag(s)`.

Discussion

247.	Following reactions are taking place in a Galvanic cell , `Zn to Zn^(2+) + 2e^(-) , Ag^(+) + e^(-) to Ag` Which of the given representations is the correct method of depicting the cell ?A. `Zn_((s)) \| Zn_((aq))^(2+)"\|\|"Ag_((aq))^+ \| Ag_((s))`B. `Zn^(2+)\| Zn\|\|Ag\|Ag^+`C. `Zn_((aq))\|Zn_((s))^(2+) "\|\|"Ag_((s))^+ \| Ag_((aq))`D. `Zn_((s)) \| Ag_((aq))^+ "\|\|" Zn_((aq))^(2+) \| Ag_((s))`
Answer» Correct Answer - A `Zn+2Ag^(+) to Zn^(2+) + 2Ag ` can be represented as `Zn_((s))\|Zn_((aq))^(2+)\|\|Ag_((aq))^+ \|Ag_((s))`

Discussion

248.	Which of the following expressions represent molar conductivity of Al2(SO4)3 ?(a) \(3 \lambda^0_{Al^{3+}} + 2\lambda^0_{SO^{2-}_4}\)(b) \(2 \lambda^0_{Al^{3+}} + 3\lambda^0_{SO^{2-}_4}\)(c) \(\frac{1}{3} \lambda^0_{Al^{3+}} + \frac{1}{2}\lambda^0_{SO^{2-}_4}\)(d) \(\lambda^0_{Al^{3+}} + \lambda^0_{SO^{2-}_4}\)
Answer» Correct answer is (b) \(2 \lambda^0_{Al^{3+}} + 3\lambda^0_{SO^{2-}_4}\)

248.

Which of the following expressions represent molar conductivity of Al2(SO4)3 ?(a) \(3 \lambda^0_{Al^{3+}} + 2\lambda^0_{SO^{2-}_4}\)(b) \(2 \lambda^0_{Al^{3+}} + 3\lambda^0_{SO^{2-}_4}\)(c) \(\frac{1}{3} \lambda^0_{Al^{3+}} + \frac{1}{2}\lambda^0_{SO^{2-}_4}\)(d) \(\lambda^0_{Al^{3+}} + \lambda^0_{SO^{2-}_4}\)

Answer»

Correct answer is

(b) \(2 \lambda^0_{Al^{3+}} + 3\lambda^0_{SO^{2-}_4}\)

Discussion

249.	The cell potential of the following cell is (\(E^0_{Al^{3+}/Al}\) = – 1.66 V)Al \| Al2(SO4)3(aq) \|\| HCl(aq) \| H2(g) \| Pt 0.5M 1M 1 atm(a) 1.66 V (b) -1.66 V (c) 0.5533 V (d) 2.14 V
Answer» Option : (b) -1.66 V

Discussion

250.	The cell reaction `Hg_2Cl_2(s)+cu(s)to cu^(2+)(aq)+2Cl^-"(aq)+2Hg(l)`,best represented by :A. `cu(s)\|cu^(2+)(aq)"\|\|"Hg_2Cl_2(s)\|Hg(l)`B. `cu(s)\|cu^(2+)(aq)"\|\|"Hg(l)\|HgCl_2(s)`C. `cu(s)\|cu^(2+)(aq)"\|\|"Cl^-"(aq)\|Hg_2Cl_2(s)\|Hg(l)\|Pt(s)`D. `Hg_2Cl_2(s)\|Cl^-"(aq)\|\|Cu^(2+)(aq)\|Cu(s)`
Answer» Correct Answer - C

Discussion

Explore topic-wise InterviewSolutions in .

State the products of electrolysis obtained on the cathode and the anode in the following cases: (i) A dilute solutoin of `H_(2)SO_(4)` with platinum electrodes (ii) An aqueous solution of `AgNO_(3)` with silver electrodes.

In electrolysis of dilute `H_(2)SO_(4)` what is liberated at anode?A. `H_(2)`B. `SO_(4)^(2-)`C. `SO_(2)`D. `O_(2)`.

When an electric cell is charged thenA. voltage of cell increasesB. electrolyte of cell dilutes.C. resistance of cell increaseD. None of these.

The half cell reaction for rusting of iron are: `2H^(+)+2e^(-)+(1)/(2)O_(2)rarrH_(2)O(l), E^(@)=+1.23V` `Fe^(2+)+2e^(-)rarrFe(s), E^(@)=-0.44V` `DeltaG^(@)` (in KJ) for the reaction isA. `-76`B. `-322`C. `-161`D. `-152`.

The half cell reaction for rusting of iron are: `2H^(+)+2e^(-)+(1)/(2)O_(2)rarrH_(2)O(l), E^(@)=+1.23V` `Fe^(2+)+2e^(-)rarrFe(s), E^(@)=-0.44V` `DeltaG^(@)` (in KJ) for the reaction isA. `-76`B. `-322`C. `-122`D. `-176`

How long a current of `3A` has to be passed through a solution of silver nitrate to coat a metal surface of `80cm^(2)` with a `0.005-mm-` thick layer ? The density of silver is `10.5g cm^(-3)`.

The value of reaction quotient `Q` for the cell `Zn(s)|Zn^(2+)(0.01M)||Ag^(o+)(1.25M)|Ag(s) is `A. 156B. 125C. 1.25xx10^(-2)D. 64xx10^(-3)

Electrolysis of molten anphdrous calcium chloride produces .A. calciumB. phosphorusC. sulphurD. sodium

Electrolysis of molten anhydrous calcium chloride produces .A. CalciumB. PhosphorusC. SulphurD. Sodium

How long has a current of 3 ampere to be applied through a solution of silver nitrate to coat a metal surface of `80 cm^(2)` with 0.005 mm thick layer ? Density of silver is `10.5 g//cm^(3)`.

A factory produces 40 kg of calcium in two hours by electrolysis. How much aluminium can be produced by same current in 2 hours if current efficiency is 50%?A. 22 kgB. 18 kgC. 9 kgD. 27 kg

The time required to coat a metal surface of `80 cm^(2)` with `5 xx 10^(-3) cm` thick layer of silver (density `1.05 g cm^(-3)`) with the passage of `3A` current through a silver nitrate solution is:A. 1150 sB. 1250 sC. 1350 sD. 1450 s

During the electrolysis of `AgNO_(3)`, the volume of `O_(2)` formed at `STP` due to passage of `2A` of current for `965 s` isA. `0.112L`B. `0.224L`C. `11.2L`D. `22.4L`

At STP 1.12 litre of `H_(2)` is obtained on flowing a current for 965 seconds in a solution . The value of current isA. 10B. 1C. 1.5D. 2

The mass of gases evolved in Question isA. `0.06g`B. `0.6g`C. `6.0g`D. `60g`

At STP 1.12 litre of `H_(2)` is obtained on flowing a current for 965 seconds in a solution . The value of current in amperes isA. 10B. `1.0`C. `1.5`D. `2.0`

The volume of gases evolved at `STP` by passing `0.1A` of current for `965g`, through an aqueous solution of potassium acetateA. `22.4mL`B. `11.2mL`C. `89.6mL`D. `44.8mL`

What is the electrode potential of a gasous hydrogen electrode dipped in a solution `pH=5.0` relative to the calomel electrode with an electrode potential of `+0.28V` ?A. `0.0125V`B. `+0.575V`C. `+0.015V`D. `-0.575V`

The specific conductivity of N/10 KCl solution at `20^(@)C` is `0.0212 ohm^(-1) cm^(-1)` and the resistance of the cell containing this solution at `20^(@)C` is 55 ohm. The cell constant is :A. `4.616 cm^(-1)`B. `1.166 cm^(-1)`C. `2.173 cm^(-1)`D. `3.324 cm^(-1)`

Out of Cu, Ag, Fe and Zn, the metal which can displace all others from their salt solutions is :A. AgB. CuC. FeD. Zn

Out of `Cu, Ag, Fe` and `Zn`, the metal which can displace all others from their salt solution isA. `Ag`B. `Cu`C. `Zn`D. `Fe`

Calculte molar conductance at infinite dilution for acetic acid, given `A_(m)^(oo)HCI=425 ohm^(-1)cm^(-1),A_(m)^(oo)NaCI=188 ohm^(-1)cm^(-1),A_(m)^(oo)CH_(3)COOHNa=96 ohm^(-1)cm^(-2)mol^(-1)`

Why `wedge_(m)^(@)` for `CH_(3)COOH` cannot be determined experimentally?

Molar conductance at infinite dilution of acetic acid cannot be determined experimentally.1. How can it be determined?2. State the law.3. Give two sets of electrolytes which can be used to nd out the molar conductance of acetic acid at infinite dilution.

Copper sulphate solution is electrolysed between two platinum electrodes. A current is passed unit 1.6 g of oxygen is liberated at anode. The amount of copper deposited at the cathode during the same period is:A. `6.36 g`B. `63.6g`C. `12.7 g`D. `3.2g`

Exactly 0.2 mole electrons are passed through two electrolytic cells in series containing `CuSO_(4)` and `ZnSO_(4)` respectively. How many grams of each metal will be deposited on the respective cathodes in the two cells ?

A solution of copper (II) sulphate us electrolysed between copper electrodes by a current of 10 amperes exactly for 1 hour. What changes occur at the electrodes and in the solution ?

An electrochemical cell is used to convertA. potential energy changes into kinetic energyB. kinetic energy changes into potntial energyC. chemical energy changes into electrical energyD. electrical energy changes into chemical energy

Which of the following statements for the electrochemical daniel cellA. Electrons flow from copper electrode to zinc electrodeB. Current flows from zinc electrode to copper electrodeC. Cations move towards copper electrode which is cathodeD. Cations move towards zinc electrode

`E^(@)` for `Fe//Fe^(2+)` is `+0.44 V` and `E^(@)` for `Cu//Cu^(2+)` is `-0.32 V`. Then, in the cell,A. `Cu` oxidises `Fe^(2+)` ionB. `Cu^(2+)` oxidises ironC. `Cu` reduces `Fe^(2+)` ionD. `Cu^(2+)` reduces `Fe`

Which of the following statement is true for the electrochemical Daniell cell ?A. Electrons flow from copper electrode to zinc electrodeB. Current flows from zinc electrode to copper electrodeC. Cations move toward copper electrode which is cathodeD. Cations move toward zinc electrode

Which one of the following metals cannot veolve `H_2` from acids or `H_2O` from its componds ?A. ` Hg`B. ` Al`C. `Pb`D. `Fe`

`E^(@)` for `Fe//Fe^(2+)` is `+0.44 V` and `E^(@)` for `Cu//Cu^(2+)` is `-0.32 V`. Then, in the cell,A. Cu oxidises `Fe^(2+)` ionB. `Cu^(2+)` oxidises FeC. Cu reduces `Fe^(2+)` ionD. `Cu^(2+)` reduces Fe

The SOP , `E^(@)` for the half reactions are `Zn to Zn^(2+) + 2e , E^(@) = + 0.76 V` `Ag to Ag^(+) + e , E^(@) = -0.77 V` `E^(@)` of the cell , `Ag^(+)+ Zn to Zn^(2+) + Ag` is :A. `+ 1.53`B. `-1.53 `C. `-0.01`D. `+0.01`

The cell potential of the following cell is (\(E^0_{Al^{3+}/Al}\) = – 1.66 V)Al | Al2(SO4)3(aq) || HCl(aq) | H2(g) | Pt 0.5M 1M 1 atm(a) 1.66 V (b) -1.66 V (c) 0.5533 V (d) 2.14 V

The cell reaction `Hg_2Cl_2(s)+cu(s)to cu^(2+)(aq)+2Cl^-"(aq)+2Hg(l)`,best represented by :A. `cu(s)|cu^(2+)(aq)"||"Hg_2Cl_2(s)|Hg(l)`B. `cu(s)|cu^(2+)(aq)"||"Hg(l)|HgCl_2(s)`C. `cu(s)|cu^(2+)(aq)"||"Cl^-"(aq)|Hg_2Cl_2(s)|Hg(l)|Pt(s)`D. `Hg_2Cl_2(s)|Cl^-"(aq)||Cu^(2+)(aq)|Cu(s)`