3195 + Interview Questions in ELECTROCHEMISTRY IN GENERAL AWARENESS Page 19 InterviewSolution

901.	How many hour are required for a current of `3.0` ampere to decompose `18 g` water?
Answer» `H_(2)O rarr H_(2)+1//2O_(2) [(2H^(+)+2e rarr H^(2)),(O^(2-) rarr (1)/(2)O_(2)+2e)]` `:.` Eq. of `H_(2)O = (i.t)/(96500)` Equivalent weight of `H_(2)O = 18//2` as two electrons are used for `1` mole `H_(2)O` to decompose in `H_(2)O` and `O_(2)`. `:. (18)/(18//2) = (3 xx t)/(96500)` `t = 64333.3 sec` `= 1072.2` minute `= 17.87` hr

Discussion

902.	How many hour are required for a current of `3.0` ampere to decompose `18 g` water?A. 9 hrsB. 12 hrsC. 18 hrsD. 24 hrs
Answer» Correct Answer - C `H_(2) O hArr H_(2) + (1)/(2) O_(2)` `therefore` 18 gms (1 mole) liberates 2gms.of `H_(2)` (M.mass of `H_(2) = 2`) `E = ("At.mass")/("valency") = (1)/(1) implies therefore Z = (1)/(96500)` t = ? , t = 3 A , W = 2gm . `t = (W xx 96500)/(E xxi) = (2 xx 96500)/(3) 64332 "sec" = (64322)/(3600) ` hrs = 17.9 hrs = 18 hrs .

Discussion

903.	How long will it take for a current of 3 amperes to decompose 36g of water? (Eq. wt. of hydrogen is 1 and that of oxygen is 8)A. `36` hrs .B. 18 hrsC. 9 hrsD. 4.5 hrs
Answer» Correct Answer - A Eq. mass of `H_(2)O = 1 + 8 = 9` . Hence 9 g is decomposed by 96500 C . So 36 g will be decomposed by Q = `96500 xx 4 C` . Hence , `t = (Q)/(T) = (96500 xx 4 )/(3) s = (96500 xx 4)/(3 xx3600)` hr = 35.74 hr

Discussion

904.	One Faraday of electricity when passed through a solution of copper sulphate deposits .A. 1mole of CuB. 1 gm atom of CuC. 1molecule of CuD. 1 gm equivalent of Cu
Answer» Correct Answer - D One Faraday=1gm equivalent of Cu.

Discussion

905.	Weight of Ag (At.mass = 108) deposited when 560 ml of `O_(2)` (NTP) is evolved isA. 207 gmB. 5.4 gmC. 8.1 gmD. 10.8 gm
Answer» Correct Answer - D 22,400 mL = 32 g of oxygen `therefore` 560 mL = 0.8 g oxygen 0.89 oxygen = 0.1 gm eq. of oxygen 0.1 gm eq. i.e. 10.8 g of silver will be deposited . (Eq.mass of Ag = 108)

Discussion

906.	In the electrolysis of fused salt, the weight of the substance deposited on an electrode will not depend on:A. temperature of the bathB. current intensityC. electrochemical equivalent of ionsD. time for electrolysis
Answer» Correct Answer - A The weight deposited `w prop i`.t and is independent of temperature .

Discussion

907.	The volume of hydrogen at NTP displaced by that amount of current which displaced 1.08 g of Ag (equivalent weight of Ag=108) will beA. 1120ccB. 11.2ccC. 112ccD. 11200cc.
Answer» Correct Answer - A g eq. of hydrogen =g eq. of Ag `(W)/(1)=(1.08)/(108)=0.01g` `2g H_(2)` has volume at N.T.P. =22400cc. 0.01g `H_(2)` has volum e at N.T.P. `=(22400)/(2)xx0.01` =112cc.

Discussion

908.	A current of 1 amp was passed for t seconds through cells P,Q andR connected in series. These contain respectively silver nitrate, mecruric nitrate and mercurous nitrate. AT the cathode of the cell P,0.21 6 g of Ag was deposited. The weights of mercury deposited in the cathode of Q and R respectively are.A. 0.4012 and 0.8024gB. 0-4012abd 0.2006gC. 0-2006 and 0.4012gD. 0.1003 and 0.2006 g .
Answer» Correct Answer - C In Q cell : g eq. of `H_(2+)=` g eq. Of `Ag^(+)` `(W)/(200.6//@)=(0.216)/(108)therefore W=0.2006g` In R cell: g eq. of `Hg^ (+)=` g eq. of `Ag^(+)` `(W)/(200.6//2)=(0.216)/(108)=therefore W=0.4012`

Discussion

909.	In the electrolysis of fused salt, the weight of the substance deposited on an electrode will not depend on:A. Temperature of the bathB. Current intensityC. Electrochemical equivalent of ionsD. Time of electrolysis
Answer» Correct Answer - A The amount deposited in directly proportional to current intensity, electrochemical equivalent of ions and the time for electrolysis and is independent o of the temperature.

Discussion

910.	An elecric current 0.25 ampere was passed through acidified water for two hours, the volume of `H_(2)` produced at N.T.P isA. 20.16 litresB. 0.2016litresC. 2.016litresD. 0.4032 litres.
Answer» Correct Answer - B `W=(E)/(96500)xxIxxt` `=(1)/(96500)xx0.25xx2xx60=0.01865g` `2gH_(2)` has volume =22.4 litre `0.01865g` of `H_(2)` has volume `=(22.4)/(2)xx0.01865=0.2016` litre.

Discussion

911.	How many coulombs are required for the oxidation of 1 mol of `H_(2)O_(2)` ?A. `93000C `B. `1.93 xx 10^(5) C`C. `9.65 xx 10^(4) C`D. `19.3 xx 10^(3) C`.
Answer» Correct Answer - B `H_(2)O_(2) to H_(2)O + (1)/(2) O_(2)` Oxidation of 1 mole of `H_(2)O_(2)` means both `O^(-)` ions should change to `O_(2)` , i.e,`2 O^(-) to O_(2) + 2e^(-)` Thus , 1 mole of `H_(2)O_(2)` requires = 2 F = `2 xx 96500 C = 1.93 xx 10^(5)` C (Remember that the given reaction is a disproportionation reaction `2H_(2)O_(2) to 2H_(2)O + O_(2)` in which 1 mole is reduced) , `H_(2)O_(2) to H_(2)O + (O)` and 1 mole is oxidized , `H_(2)O_(2) + (O) to H_(2)O + O_(2)`

Discussion

912.	Two electrolytic cells, one containing acidified ferrous sulphate and another acidified ferric chloride, are in series. The ratio of masses of Iron deposited at the cathode in the two cells will beA. `3:1`B. `2:1`C. `1:1`D. `3:2`
Answer» Correct Answer - D Since the wt. of iron deposit is proportional to its equivalent weight `therefore ("Wt. of ferrous ion")/("Wt. of ferric iron")=(at.wt//2)/(at.wt//3)=3//2`

Discussion

913.	Two electrolytic cells, one containing acidified ferrous sulphate and another acidified ferric chloride, are in series. The ratio of masses of Iron deposited at the cathode in the two cells will beA. `3 : 1`B. `2 : 1`C. `1 : 1`D. `3 : 2`
Answer» Correct Answer - D Eq. of mass of `Fe^(2+) ` = At.mass/2 = 56/2 . Eq. mass of `Fe^(3+)` = At.mass /3 = 56/3 . Weights deposited are in the ratio of their equivalents . `therefore` Ratio of wts.deposited is `Fe^(2+) : Fe^(3+) = (56)/(2) : (56)/(3) = (1)/(2): (1)/(3) = 3 : 2`

Discussion

914.	Two electrolytic cells, one containing acidified ferrous sulphate and another acidified ferric chloride, are in series. The ratio of masses of Iron deposited at the cathode in the two cells will beA. `3:1`B. `2:`C. `1:1`D. `3:2 `
Answer» Correct Answer - D At cathode: `Fe^(2+)+2e^(-)toFe,Fe^(3+)+3e^(-)toFe` `(E_(Fe))=("Atomic.weight")/(2),(E_(Fe))_(2)=("Atomic.weight")/(3)` Ratio of weight of Fe liberated `=("Atomic weight")/(3):("Atomic weight")/(2)=3:2`

Discussion

915.	Match the following `{:(,"column I" ,, "column II"),((A) ,wedge_(c )"versus" sqrt(c )"(for KCI solution)" , (p),"Positive slope"),((B) , E_(Cu^(2+)//Cu) "versus" log_(10)_(10)[Cu^(2+)](for Cu^(2+)+2e^(-)rarrCu) , (q),"Negative intercept on y axis"),((C ), E_(cell) "versus temperature (in K) (temperature coefficent lt 0)" , (r),"Positive intercept on y axis "),((D) , E_(Au//Au^(3+))"Versus" log_(10)[Au^(3+)] ("for" Au rarr Au^(3+)+3e^(-)) , (s),"Negative slope"),(,,(t),"parameter on y axis varies with temperature"):}`
Answer» A(r,s,t),B(p,r,t),C(r,s,t),D(q,s,t)

Discussion

916.	Unlike dry cell, the mercury cell has a constant cell potential throughout its useful life. Why?
Answer» Ions are not involved in the overall cell reaction of mercury cells.

Discussion

917.	Match the following `{:(,"column I" ,, "column II"),((A) ,"Leclanche cell" , (p),"primary battery" ),((B) , "Car battery" , (q),"Secondary battery"),((C ), "Fuel cell" , (r),"converts energy of combustion of" H_(2)","CH_(4) "etc into eletrical energy"),((D) , "NICAD cell" , (s),"Anode :"Zn rarr Zn^(2+)+2e^(-)),(,,,"Cathode :" MnO_(2)+NH_(4)^(+)+e^(-)rarrMn(OH)+NH_(3)),(,,(t),"Cd(s)+2Ni(OH)_(3)(s)rarrCdO(s)+2Ni(OH)_(2)(s)+H_(2)O(I)):}`
Answer» A(p,s)B(q),C(r ),D(q,t)

Discussion

918.	What is Battery ? Give one example each of primary battery and secondary battery.
Answer» For Answer, Consult Section 22.

Discussion

919.	Why does the cell potential of mercury cell remain constant throughout its life ?
Answer» This is due to the reason that the overall reaction does not involve any ions in the solution whose concentration changes during the reaction. `Zn(s)+H_(2)O(s) to ZnO(s)+H_(2)O(l)` For more details, consult Section 22.

Discussion

920.	(a) Explain electrochemical series. (b) Can we store 1M `CuSO_(4)` solution in zinc vessel or not, why ?
Answer» (a) For Answer, Consult Section 7. (b) No, it is not possible. Zinc `(E^(@)=-0.76V)` will reduce `Ca^(2+)` ions `(E^(@)=+0.34V)` by taking part in the displacement reaction. `Zn(s)+Cu^(2+)(aq) to Zn^(2+)(aq)+Cu(s)`

Discussion

921.	Standard Electrode Potential
Answer» For Answer, Consult Section 5.

Discussion

922.	What do you understand by standard e.m.f. of a cell ? Derive a relationship between standard emf of a cell and equilibrium constant.
Answer» Standard emf of a cell is the difference in the standard reduction potentials of cathode and anode when no current flows through the cell. For the relationship between standard emf `(E_(cell)^(@))` are equilibrium constant`(K_(c))`, consult Section 9.11.

Discussion

923.	`I_(2)(s)\|I^(-)`(0.1M) half cell is connected to a `H^(+)` (aq)\|`H_(2)`(1 bar)\|Pt half celland e.m.f. is found to be 0.7714 V. If `E_(I_(2)\|I^(-))^(@)`=0.535 V, find the pH of `H^(+)\|H_(2)` half cell.A. 1B. 3C. 5D. 7
Answer» Correct Answer - B

Discussion

924.	The time required for `10%` completion of a first order reaction at 298 K is equal to that required for its `25%` completiion at `308K.` If the value of A is `4xx10^(10)s^(-1).` calculate k at 318K and `E_(a),`
Answer» Calculation of activation energy `(E_(a))` For `1^(st)` order reaction : `k=(2.303)/(t)log ""([A]_(0))/([A])` `At 298K` `K_(1)=(2.303)/(t) log """"(100)/(90)-(i)` At 308 K `K_(2)=(2.303)/(t)log ""(100)/(75) -(ii)` Dividing eq (ii) by (i) `(k_(2))/(k_(1))=(log ""(100)/(75))/(log ""(100)/(90))=(0.1249)/(0.0458)=2.73` According to Arrhenius theory `log ""(k_(2))/(k_(1))=(E_(a))/(2.303 R)xx(T_(2)-T_(1))/(T_(1)T_(2))` `log 2.73 =(E_(a))/(2.303R)((308-298)/(298xx308))` `E_(a)=(0.4361xx2.303xx(8.314J mol ^(-1))xx298xx308)/(10)` `E_(a) k =log A (-E_(a))/(2.303 xx(8.314J mol^(-1) K^(-1))xx(318K))` `log k =10.6021 -12.5870 =-1.9849` k = Antilog `(-19849)=` Antilog `(bar2.0151)=1.035xx10^(-2)s^(-1)` `E_(a)=76.640kJ mol ^(-1)` `k=1.035 xx10^(-2)s^(-1)`

Discussion

925.	The decomposition of A into product has value of k as `4.5 xx10^(3) s^(-1)at 10^(@)C` and energy of activation `60 kJ mol ^(-1).` At what temperature would k be `1.5 xx10^(4) s^(-1)?`
Answer» According to Arrhenius equation `log ""(k_(2))/(k_(1))=(E_(a))/(2.303R)xx(T_(2)-T_(1))/(T_(1)T_(2))` `k_(1)=4.5xx10^(3)s^(-1)` `k_(2) =1.5 xx10^(4)s^(-1)` `T_(1)=10^(@)C=283K` `log ""(1.5 xx10^(4))/(4.5xx10^(3))=((6000Jmol ^(-1)))/(2.303 xx(8.314J mol ^(-1)))=((T_(2)-283)/(283T_(2)))` `log 3.333 =3133.62 (T_(2)-283)/(283T_(2))` `1-(283)/(T_(2))=0.04776` `or T_(2)=(283)/(1-0.04776)=(283)/(0.9524)` ` T_(2) =297.19 K =(297.19-273.0)=24.19^(@)C` Temperature `=24.19^(@)C`

Discussion

926.	The rate constant for the first order decomposition of `H_(2)O_(2)` is given by the following equation `: log k =14.34-1.25 K//T.` Calculate `E_(a)` for this reactin and at what temperature will its half-life period be 256 minutes ?
Answer» a) Calculation of activation energy `E_(a)` According to Arrhenius equation: `k=Ae ^(-E_(a)//RT)` `log k= log A-(E_(a))/(2.303RT)-(i)` `log K =14.34-(1.25 xx10^(4)K)/(T)-(ii)` on comparing both equations. `(E_(a))/(2.303RT) =(1.25xx10^(4)K)/(T)` `E_(a) =1.25 xx10^(4) K xx 2.303 xx8.314 (JK^(-1) mol ^(-1))` `= 23.93 xx10^(4) J mol ^(-1) =239.3 kJ mol ^(-1)` b) Calculation of required temperature If `t_(1//2)=256 min. for 1^(st)` order reaction, `k=(0.693)/(t_(1//2))=(0.693)/((256min))=(0.693)/(256xx60S)` According to Arrhenius theory `log k = 14.34 -(1.25 xx10^(4)k)/(T)` `log (4.51xx10^(-5))=14.34 -(1.25xx10^(4)k)/(T)` `(1.25xx10^(4))/(18.69)=669K` `E_(a)=239 . 3 kJ mol^(-1)` `T=669K`

Discussion

927.	Consider the following `E^o` values : `E^o _Fe^(3+)//FE^(2+)o = + 0.77 V` `E_(Sn^(2+)//Sn) =- 0.14 V` Under standard conditions the potential for reaction `Sn(s) +2Fe^(3+) (aq) rarr 2Fe^(2+) (sq) + Sn^(2+) (aq)` is.A. `0.91V`B. `1.40V`C. `1.68V`D. `0.63V`
Answer» Correct Answer - A `Sn -2e rarr Sn^(2+)` `(2Fe^(3+)+2e rarr 2Fe^(2+))/(Sn+2Fe^(3+) rarr Sn^(2+) +2Fe^(2+))` `E^(@)cell = (E_(C)^(@) -E_(A)^(@))_(R)` `=0.77V +0.14V = 0.91V`

Discussion

928.	Assertion: When acidified zinc sulphate solution is electrolysed between zinc electrodes, it is zinc that is deposited at the cathode and hydrogen evolution does not take place. Reason: The electrode potential of zinc is more negative than hydrogen as the overvoltage for the hydrogen as the evolution on zinc is quite large.A. if both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is falseD. If the assertion and reason both are false.
Answer» Correct Answer - A

Discussion

929.	Assertion: A large dry cell has high e.m.f. Reason: The e.m.fof a dry cell is proportional to its size.A. if both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is falseD. If the assertion and reason both are false.
Answer» Correct Answer - D

Discussion

930.	Assertion `(A):` For four half`-` cell reactions involving different number of electrons, `E_(4)=E_(1)+E_(2)+E_(3)` Reaction `(R): DeltaG_(4)=DeltaG_(1)+DeltaG_(2)+DeltaG_(3)`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 `(A)` is incorrect, `(R)` is correct.
Answer» Correct Answer - D `(A)` is incorrect since `EMFs` are neither additive nor subtractice. `(R)` is correct but not the explanation of `(A)`. `(R)` is an independent statement and `DeltaG` are additive or subtrative.

Discussion

931.	Statement : If two half reaction with electrode potential `E_(1)^(@)` and `E_(2)^(@)` gives a third reaction then, `DeltaG_(3)^(@) = DeltaG_(1)^(@) + DeltaG_(2)^(@)` Explanation : `E_(3)^(@) = E_(1)^(@) + E_(2)^(@)`
Answer» Correct Answer - A In such case `E^(@)` are not additive.

Discussion

932.	Solutions of two electrolytes ‘A’ and ‘B’ are diluted. The Λm of ‘B’ increases 1.5 times while that of A increases 25 times. Which of the two is a strong electrolyte? Justify your answer.
Answer» Electrolyte ‘B’ is strong as on dilution the number of ions remains the same, only interionic attraction decreases therefore increase in ∧_m is small.

Discussion

933.	In the following process of disproportionation: `underset(underset("ion")("Chlorate"))(2 ClO_(3)^(-))hArr underset(underset("ion")("Per chlorate"))(ClO_(4)^(-))` `{:(E_(ClO_(4)^(-)//ClO_(3)^(-))^(@),= + 0.36 V),(E_(ClO_(3)^(-)//ClO_(2)^(-))^(@),= + 0.33 V):}` Initial concentration of chlorate ion was 0.1 M. The equilibrium concentration of per chlorate ion will be :A. 0.19 VB. 0.1 MC. 0.024 MD. 0.019 M
Answer» Correct Answer - D `{:(ClO_(3)^(-) (aq.) +H_(2)O (l) rarr ClO_(4)^(-) (aq.)+2H^(+) (aq.) +2e^(-)),(2H^(+)+ClO_(3)^(-) (aq.) +2e^(-) rarr ClO_(2)^(-) (aq.)+H_(2)O(l)),(ulbar(2ClO_(3)^(-) (aq.) hArr ClO_(2)^(-) (aq.) +ClO_(4)^(-) (aq.))):}` `E_(cell)^(@)=0.33-0.36=-0.03 V` `E=E^(@)-0.059/n log Q` At equilibrium, `E=0, n=2, Q=K` `0=-0.03-0.059/2 log K` `log K=-1` `K=1/10` ...(1) `{:(,2ClO_(3)^(-) (aq.),hArr,ClO_(2)^(-) (aq.),+,ClO_(4)^(-) (aq.)),(t_(0),0.1,,0,,0),(t_(eq.),0.1-2x,,x,,x):}` `K=(x xx x)/((0.1-2x)^(2))=1/10` `x=0.019`

Discussion

934.	which of the following statements are correct regarding to galvanic cell?A. A reaction is spontaneous from right to left if `E_(cell)gt0`B. A reaction occurs from right to left if `E_(cell)lt0`C. If the system is at equilibrium no net reaction occursD. `E_(cell)` is temperature-independent
Answer» Correct Answer - A::B::C

Discussion

935.	Stronger the oxidising agent , greater is the :A. reduction potentialB. oxidation potentialC. ionic behaviourD. none
Answer» Correct Answer - A More as SRP values `F_(2) gt Cl_(2) gt Br_(2) gt I_(2) is E_(RP)^(@)` , more is the tendency to get itself oxidised , more is reducing nature .

Discussion

936.	Which one will liberate `Br_(2)` from `KBr`?A. `HI`B. `I_(2)`C. `Cl_(2)`D. `SO_(2)`
Answer» Correct Answer - C Cl is placed above Br in electrochemical series , the non- metal placed above in series , replaces other from its solution , which is placed below of them .

Discussion

937.	The e.m.f. of the cell in which the following reaction `Zn_((s))+Ni^(2+)(a=1.0)hArrZn^(2+)(a=10)+Ni_((s))` occurs, is found to be 0.5105V at 298K. The standard e.m.fof the cell isA. 0.54B. 0.4810VC. 0.5696VD. `-0.5105V`
Answer» Correct Answer - B

Discussion

938.	Equivalent weight of `FeS_(2)` in the half reaction `FeS rarr Fe_(2)O_(3) + SO_(2)` is :A. `M//10`B. `M//11`C. `M//8`D. `M//7`
Answer» Correct Answer - D `S^(2-) rarr S^(4+) + 6e` `2Fe^(2+) rarr (Fe^(3+))_(2) + 2e` `FeS Fe_(2)O_(3) + SO_(2)` involves a change of `7` electrons per mole.

Discussion

939.	Why molbilities of `H^(o+)` ions in ice is greater as compared to liquid water.
Answer» The density of liquid water is more as compared to that if ice. Therefore, ionic mobility is expected to be less in water as compared to ice. Moreover in water, the `H^(+)` ions get hydrated i.e., `H^(+)`(aq). Since the hydrated ions are heavier than the `H^(+)` ions, the ionic mobility of the hydrated ions is less.

Discussion

940.	The equivalent conductance of NaCl at concentration C and at infinite dilution are `lamda_(C) and lamda_(oo)`, respectively. The correct (where the constant B is positive)A. `lamda_(C)=lamda_(oo)+(B)C`B. `lamda_(C)=lamda_(oo)-(B)C`C. `lamda_(C)=lamda_(oo)-(B)sqrt(C)`D. `lamda_(C)=lamda_(oo)+(B)sqrt(C)`
Answer» Correct Answer - C `lamda_(C)=lamda_(oo)-(B)sqrt(C)` (debye Huckel onsagn equation).

Discussion

941.	Resistance of 0.2M solution of an electrolyte is `50Omega`. The specific conductance of the solution of 0.5M solution of the same electrolyte is `280Omega`. The molar conductivity of 0.5M solution of the electrolyte is S `m^(2)mol^(-1)` isA. `5xx10^(-4)`B. `5xx10^(-3)`C. `5xx10^(3)`D. `5xx10^(2)`
Answer» Correct Answer - A `x=1.4S//m` `R=50Omega` `M=0.2 ` `K=(1)/(R)xx(l)/(A)` `implies(l)/(A)=1.4xx50m^(-1)`. Now, new solution has M=0.5, `R=280Omega` `impliesK=(1)/(R)xx(l)/(A)=(1)/(280)xx1.4xx50=(1)/(4)` `implies lamda_(M)=(K)/(1000xxM)=((1)/(4))/(1000xx0.5)=(1)/(2000)=5xx10^(-4)`

Discussion

942.	Statement -1 : In alkaline version of dry cell, `NH_(4)Cl` is replaced by `KOH`. Statement -2 : Zinc container does not undergo corrosion in alkaline medium.A. Statement-1 is true, Statement-2 is true, statement-2 is a correct explanation for statement-1.B. Statement-1 is true , statement-2 is true, statement-2 is not a correct explanation for statement-1.C. Statement-1 is true, statement-2 is false.D. statement-1 is false, statement-2 is true.
Answer» Correct Answer - A

Discussion

943.	How does aqueous and acidified `CuSO_(4)` dissociate on the passage of current ? Also write the electrode reactions at the respective electrodes
Answer» Aqueous acidified copper sulphate dissociates as shown below. `CuSO_(4) rarr Cu^(2+) + SO_(4)^(-2) " , " H_(2)O rarr H^(+) + OH^(-) " , " H_(2) SO_(4) rarr 2H^(+) + SO_(4)^(2-)` Reaction at cathode: Copper gets discharged at the cathode because of lower discharge potential than that of hydrogen. `Cu^(2+) + 2e^(-) rarr Cu` Reaction at anode (i) If anode is platinum or graphite (inactive anode) `OH^(-)` is discharged at the anode because its discharge potential is lower than that of `SO_(4)^(2-)` ion. `4OH^(-) rarr 2H_(2)O + O_(2) + 4e^(-)` (ii) If the anode is copper itself (active anode) `Cu - 2e^(-) rarr Cu^(2+)` No product is formed at anode.

Discussion

944.	When concentrated solution of cupric chloride is taken, chlorine gets liberated at anode, in contrast to a dilute solution where oxygen is liberated. However, when aqueous solution of `CuSO_(4)` is taken, only oxygen is liberated at anode irrespective of concentration. How do yoy account for this ?
Answer» An aqueous solution of cupric chloride contrains `Cl^(-)` ions, `OH^(-)` ions, `Cu^(+2)` ions and `H^(+)` ions. Due to lower discharge potential of `OH^(-) " than " Cl^(-)`, oxygen gas is liberated at anode in preference to chlorine. However, in the concetrated solution, as the concetration of `OH^(-)` ions is very low, only chlorine gets liberated at anode. This is possible because the difference between dischange potentials of `Cl^(-) and OH^(-)` is only marginal. An aqueous solution of `CuSO_(4)` contains `Cu^(+2), H^(+), OH^(-) and SO_(4)^(-2)` ions. Due to large difference in discharge potentials between `SO_(4)^(-2) and OH^(-), " only " OH^(-)` ions get discharged at anode irrespective of the concetration of the electrolyte

Discussion

945.	Assertion: One coulomb of electric charge deposits wegiht equal to the electrochemical equivalent of the substance. Reason: One faraday deposits one mole of the substance.A. if both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is falseD. If the assertion and reason both are false.
Answer» Correct Answer - C One Faraday deposits one gram equivalent of the substance.

Discussion

946.	In a conductivity cell the two platinum electrodes each of area 10 sq. cm are fixed 1.5 cm apart. The cell. Contained `0.05N` solution of a salt. If the two electrodes are just half dipped into the solution which has a resistance of 50 ohms, find equivalent conductance of the salt solution.
Answer» Correct Answer - `120 mho cm^(2) eq^(-1)` Since the electrodes of the cell are just half depped the effective area will be 5 eq cm. Cell constant `= (1)/(a) =(1.5)/(5) = 0.3 cm^(-1)` Specific conductance = conductance `x` cell constant `= (1)/("resistance") xx` cell constant `=(1)/(50)xx0.3 =(3)/(500)"mho cm"^(-1)` Equivalent conductance = specific conductance `x` volume....(8) `(3)/(500)xx 20000 = 120 "mho cm"^(2)` `(0.05 N =N//20 :. V =20,000 "cc")`.

Discussion

947.	A current of 1 ampere is paased for one hour between nickel electrodes in 0.5 L of 2 M Ni `(NO_(3))_(2)` solution. What will be the molarity of the solution at the end of the electrolysis?
Answer» The changes taking place at the electrodes in the electrolysis of `Ni(NO_(3))_(2)` solution using nickel electrodes will be as follows: At cathode: `Ni^(2+)+2etoNi` At anode: `NitoNi^(2+)+2e^(-)` thus, amount of nickel deposited on the cathode from the solution=nickel dissolved from the anode. Hence, molarity of the solution will remain unchanged at the end of the electrolysis.

Discussion

948.	A conductivity cell has two electrodes 20 mm apart and of cross section area 1.8 cm2. Find the cell constant.
Answer» Given : Distance between two electrodes = l = 20 mm = 2 cm Cross section area = a = 1.8 cm Cell constant = b = ? b = \(\frac{l}{a}\) = \(\frac{2}{1.8}\) = 1.111 cm^-1 ∴ Cell constant = 1.111 cm^-1

948.

A conductivity cell has two electrodes 20 mm apart and of cross section area 1.8 cm2. Find the cell constant.

Answer»

Given :

Distance between two electrodes = l

= 20 mm

= 2 cm

Cross section area = a = 1.8 cm

Cell constant = b = ?

b = \(\frac{l}{a}\)

= \(\frac{2}{1.8}\)

= 1.111 cm^-1

∴ Cell constant = 1.111 cm^-1

Discussion

949.	The resistance of a solution is 2.5 × 103 ohm. Find the conductance of the solution.
Answer» Given : Resistance of solution = R = 2.5 × 10³Ω Conductance of solution = G = ? G = \(\frac{1}{R}\) = \(\frac{1}{2.5\times 10^3}\) ohm^-1 (Ω^-1 or S) = 4 × 10^-3 Ω^-1 (or S) ∴ Conductance = G = 4 × 10^-3 Ω^-1

949.

The resistance of a solution is 2.5 × 103 ohm. Find the conductance of the solution.

Answer»

Given :

Resistance of solution = R = 2.5 × 10³Ω

Conductance of solution = G = ?

G = \(\frac{1}{R}\)

= \(\frac{1}{2.5\times 10^3}\) ohm^-1 (Ω^-1 or S)

= 4 × 10^-3 Ω^-1 (or S)

∴ Conductance = G = 4 × 10^-3 Ω^-1

Discussion

950.	Which of the following processes does not involve oxidation of iron ?A. Formation of `Fe(CO)_(5)` from FeB. Liberation of `H_(2)` from steam by iron at high temperatureC. Rusting of iron sheetsD. Decolourisation of blue `CuSO_(4)` solution by iron.
Answer» Correct Answer - A (a) Oxidation state does not change in `Fe(CO)_(5)` formed from Fe (It is zero). Therefore, there is no oxidation in this case.

Discussion

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A current of 1 ampere is paased for one hour between nickel electrodes in 0.5 L of 2 M Ni `(NO_(3))_(2)` solution. What will be the molarity of the solution at the end of the electrolysis?

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