94420 + Interview Questions in Chemistry in Class 12 Page 1820 InterviewSolution

90951.	A certain compound has the molecular formula X_(4)O_(6) If 10 gm of X_(4)O_(6) has 5.72 gm X, the atomic mass of X is:
Answer» 32 amu 37 amu 42 amu 98 amu Solution :ATOMIC weight of x be a 10g of `X_(4)O_(6)` …........ 4a GX grams of x, `10 xx 4a=(4a+96)5.72`, a = 32

Discussion

90952.	A certain compound has the molecular formula M_4O_6.If 10.0 g of the compound contains 5.62 g of M, then the atomic mass of M is
Answer» 62.0 u 6.8 u 30.8 u 42 u Solution :If x is the atomic mass of A , then `(4x)/(4x + 96) xx 10G = 5.62 THEREFORE x = 30.8 `amu

Discussion

90953.	A certain compound gives negative test with ninhydrin and positive test with Benedict's solution. The compound is
Answer» a protein a monosaccharide a lipid an AMINO acid Solution :SINCE it gives a negative test with ninhydrin , itcannot be a protein or an amino acid.Since it gives a positivetest with Benedict.ssolution , it MUSTBE a monosaccharidebut NOTA LIQID.

Discussion

90954.	A certain compound gives negative test with ninhydrin and positive test with Benedict's solution, it is
Answer» an AMINO acid a monosaccharide a lipid a protein Solution :MONOSACCHARIDES gives positive test with Benedict's solution because they are reducing sugars but they GIVE negative test with ninhydrin because this test is PERFORMED by PROTEINS, peptides and amino acids.

Discussion

90955.	A certain compound gives negative test with ninhydrin, but positive test with Benedict.s solution. The compound is :
Answer» Protein Monosaccharide Lipid Amino acid Answer :B

Discussion

90956.	A certain compound contain 16% oxgyen and 4% sulphur by mass. Find minimum possible molecular weight of compound.
Answer» `600 GM` `480 gm` `800 gm` `100 gm` Solution :for 16% OXGYEN `(16)/(100) xx M = 16`[minimum 1 atom of oxgyen will be present] `M = 100 gm` For 4% sulphur `(4)/(100) M = 32` `M = 800 gm`

Discussion

90957.	A certain composed(X) is used in laboratoryfor analise. Its aqueous solution gives the following reaction:(i) On addition of copper sulphate solution brown ppt, is oibtained which turns whgite on addition of excess of Na_(2) S_(2)O_(3)solution.(ii) On addition of Ag ion solution , a yellow curdy precipitate is obtained which is insoluble in ammonium hydroxide. Identify(X)and give equation of the reaction at step(i) and(ii)
Answer» SOLUTION :Kl `2CuSO_(4)+4kltoCu_(2)l_(2)darr+l_(2)+2k_(2)SO_(4)` (i)`l_(2)+2Na_(2)S_(2)O_(3)to2Nal+Na_(2)S_(4)O_(6)` (ii)`Ag+ltoAgldarr`

Discussion

90958.	A certain buffer solutions contains equal concentration of X^(-) and HX. The K_(b) for HX is 10^(-10). The pH of the buffer is
Answer» 4 7 10 14 Solution :`X^(-) + H_(2)O hArr OH^(-) + HX` `K_(b) = ([OH^(-)][HX])/([X^(-)])` `HX hArr H^(+) + X^(-)` `K_(a) = ([H^(+)][X^(-)])/([HX]), :. K_(a) xx K_(b) = [H^(+)][OH^(-)] = K_(w) = 10^(-14)` HENCE `K_(a) = 10^(-4)` Now as `[X^(-)] = [HX], pH = pK_(a) = 4`.

Discussion

90959.	A certain colourless liquid Q does not respond to Tollen's test but it reacts with semicarbazide to form crystalline derivative. Q can be
Answer» `CH_(3)(CH_(2))_(2)CHO` `CH_(3)CH_(2)COCH_(3)` `CH_(3)(CH_(2))_(2)OC_(2)H_(5)` `CH_(3)CH_(2)-overset(CH_(3))overset(\|\|)(C)-O-C_(2)H_(5)` Answer :B

Discussion

90960.	A certain buffer solutions contains equal concentration of X^(-) and HX. The K_(a) for HX is 10^(-8). The pH of the buffer is
Answer» 3 8 11 14 Solution :The EQUAL CONC. Of SALT and ACID.

Discussion

90961.	A certain buffer solution contains equal concentration of X^- and HX. The K_b for X^- is 10^-10. What is the pH of the buffer?
Answer» 4 7 10 14 Answer :A

Discussion

90962.	A certain buffer solution contains equal concentration of X^- and HX. The K_a for HX is 10^-8. What is the pH of the buffer solution?
Answer» 3 8 11 14 Answer :B

Discussion

90963.	A certain buffer solution contains equal concentration of X^- and HX. The K_afor HX is 10^(-8) The pH of the buffer is
Answer» 5 8 6 14 Solution :Foracidicbuffer `pH =pK_a+log("[SALT]")/(["ACID"])` `:.["salt"]=["Acid"]` ` thereforepH = pK_a=- log 10^(-8) =8`

Discussion

90964.	A certain buffer solution contains equal concentration of X^(-) and HX. K_(b) for X^(-) is 10^(-10). Calculate pH of buffer
Answer» Solution :Given that `becauseK_(b(X^(-)) = 10^(-10)` and for conjugate acid-base pair `K_(a(HX)) XX K_(b(X^(-))) = 10^(-14) RARR K_(a(HX)) = 10^(-4)` But `underset("(acid) = (salt)")([HX]=[X^(-)])` Now USING Handerson's equation `:. pH = -log K_(a) + log.(["Salt"])/(["Acid"]) rArr pH = -log 10^(-4) = 4`

Discussion

90965.	A certain aqueous solution of FeCl_(3) (formula mass = 162) has a density of 1.1 g/ml and contains 20.0% FeCl_(3). Molar concentration of this solution is
Answer» `0.028` `0.163` `1.27` `1.47` ANSWER :C

Discussion

90966.	A certain amount of methyl acetate was hydrolysed in the presence of excess of "0.05 M HCl at "25^(@)C. 20 mL of reaction mixture were removed and titrated with NaOH solution, the volume V of alkali required for neutralisation after time ‘t’ were as follows : {:("t(min)",0,20,40,60,oo),("v(mL)",20.2,25.6,29.5,32.8,50.5):} Show that the reaction is the first order reaction.
Answer» Solution :`K=(2.303)/(t)log""((V_(oo)-V_(o))/((V_(oo)-V_(t))))` `V_(oo)-V_(o)=50.4-20.2=30.2` When `t="20 MTS. "k=(2.303)/(20)log""(50.4-20.2)/(50.4-25.6)` `=0.1151 log""(30.2)/(24.8)` `=0.1151 log 1.2479` `=01151 xx 0.0959 =11.03xx10^(-3)" MIN"^(-1)` when t=40 mts `k=(2.303)/(40)log""(50.4-20.2)/(50.2-29.5)` `=0.0576xxlog""(30.2)/(20.9)=0.0576xx0.1596` `=9.19xx10^(-3)" min"^(-1)` `" when"t="60 mts"` `k=(2.303)/(60)log""(50.4-20.2)/(50.2-32.8)=0.03838xxlog""(30.2)/(17.6)` `=0.03838xx0.2343=8.99xx10^(-3)" min"^(-1)` The constant values of k show that the reaction is of FIRST order.

Discussion

90967.	A certain acid - base indicator is red in acid and blue in basic solution. At pH = 5, 75% of the indicator is present in the solution in its blue form. Calculate dissociation constant (K_a) for the indicator and pH range over which the indicator changes from 90% red- 10% blue to 90% blue-10% red.
Answer» Solution :Since `K_a` is asked, the indicator must be an acid. Let the acid be represented by HIN. HI n `hArrH^+ + "In"^(-)` In acid solution, the indicator will be predominatly present in the form of HIn (due to common ion effect.) Since in acid solutin the the COLOUR is red, this would be due to HIn. In basic solution, the indicator will be predominatly in the form of `In^–`. Since the indicator is blue in basic solution, so `In^–` must be blue in colour. At PH = 5, the indicator is 75% blue. This also means it is 25% red. `therefore K_a=(10^(-5)xx0.75)/0.25=3xx10^(-5)`pH when it is 90% red & 10% blue `: {H^+ ] =(K_axx[HIn])/([In^-])=(3xx10^(-5)xx0.9)/0.1 = 2.7xx10^(-4)` `therefore` pH=3.56 pH when it is 90% blue and 10% red : `[H^+]=(K_axx[HIn])/([In^-])=(3xx10^(-5) xx0.1)/0.9=3.3xx10^(-6)`

Discussion

90968.	A certain acid-base indicator is red in acid solution and blue in basic medium. At pH=5 75% of the indictor is present in the solution in its blue form. Calculate the pH at which indicator shows 90% red form ?
Answer» Solution :`K_(sp) = [Mg^(+)][OH^(-2)]` `[oH^(-)]=SQRT((2.5xx10^(-11))/0.01)=5xx10^(-5)` `K_(B)=([NH_(4)^(+)][OH^(-)])/([NH_(3)])` So `[NH_(4)^(+)]=0.04M`

Discussion

90969.	(A): Central atom usually forms three single and one double bonds in an oxyacid of phosphorus (R): Phosphorus atom usually undergoes sp^(3) hybridisation in its oxyacids
Answer» Both (A) and (R ) are TRUE and (R ) is the CORRECT explanation of (A) Both (A) and (R ) are true and (R ) is not the correct explanation of (A) (A) is true but (R ) is false Both (A) and (R ) are false ANSWER :1

Discussion

90970.	A centi- molar solution is diluted 10 times. The molaruity would become
Answer» `1/10` 1/100 1/1000 `1/1` SOLUTION :Centimolar=1/100, DILUTED 10 TIMES, `M=1/1000`

Discussion

90971.	(A) Cellulose is not digested by human beings. (R) Human beings are not having cellulose digestable enzymes.
Answer» Both A & R are TRUE and R is the correct EXPLANATION of A Both A & R are true, but R is not the correct explanation of A A is true, R is FALSE A is false, R is true Answer :A

Discussion

90972.	A cellwith two electrodes, one of grey tin and the other white tin, both dipping in solution of (NH_4)_2 SNCI_6 showed zero emf at 18^@C. What conclusion may be draw from this "
Answer» The emf DEVELOPED at the electrode -SOLUTION PHASE boundary cancels the normal emf Grey tin being non-medallic ceases ti orivude a reversible electrode reaction Electrode surface develops a ortective layer and the cell develops a very large internal resistance The free energy CHANGO of the cell becomes zero Answer :D

Discussion

90973.	A cell with N/50 KCl solution showed a ressitance of 550 ohms at 25^(@)C. The specific conductivity of N/50 KCl at 25^(@)C is 0.002768 ohm^(-1)cm^(-1). The cell filled with N/10 ZnSO_(4) solution at 25^(@)C shows a resistance of 72.18 ohms. find the cell constant and molar conductivity of ZnSO_(4) solution.
Answer» SOLUTION :`(N)/(10)ZnSO_(4)" SOL"=(M)/(20)ZnSO_(4)" sol. And "wedge_(m)=kappaxx(1000)/("Molarity")`.

Discussion

90974.	A cell utilises the following reaction: 2CO^(3+) (aq)+Zn(s) to 2CO^(2+)(aq)+Zn^(2+) (aq) What is the effect on cell emf of each of the following changes? Additional water is added to the anode compartment
Answer» Solution :On DILUTING the solution in ANODIC compartment, `[ZN^(2+)]` shall DECREASE and so `E_(cell)` will increase.

Discussion

90975.	A cell utilises the following reaction: 2CO^(3+) (aq)+Zn(s) to 2CO^(2+)(aq)+Zn^(2+) (aq) What is the effect on cell emf of each of the following changes? Co(II) nitrate is dissociated in the cathode compartment.
Answer» Solution :For the GIVEN cell, we have `E_(cell)=E_(cell)^@-0.0591/2log""([Co^(2+)]^2[ZN^(2+)])/[Co^(3+)]^2` USING the above QUESTION, we know that if `[Co^(2+)]` is increased `E_(cell)` will decrease.

Discussion

90976.	A cell utilises the following reaction: 2CO^(3+) (aq)+Zn(s) to 2CO^(2+)(aq)+Zn^(2+) (aq) What is the effect on cell emf of each of the following changes? The size or area of Zn (s) electrode is doubled
Answer» SOLUTION :The emf is an INTENSIVE property and so the size of the ELECTRODE shall not AFFECT the emf of the cell.

Discussion

90977.	A cell utilises the following reaction: 2CO^(3+) (aq)+Zn(s) to 2CO^(2+)(aq)+Zn^(2+) (aq) What is the effect on cell emf of each of the following changes? Co(III) nitrate is dissolved in the cathode compartment
Answer» Solution :Similarly, if `[CO^(3+)]` is increased, `E_(cell)` will INCREASE.

Discussion

90978.	A cell reaction is represented as Pt\|H_(2)Q,Q,H^(+)\|\|KCl\|AgCl\|Ag. E_(H_(2)Q\|Q)^(@) = +0.4V, E_(Ag^(+)\|Ag)^(@)=0.8 V, K_(sp)=10^(-10), then calcualte a four digit number 'abcd' where: (a) K_(eq) of the reaction H_(2)Q + 2AgCl rightarrow 2Ag+2Cl^(-)+2H^(+) + Q=10^(ab) (b) Milli"mole"s of Ag formed if 1 litre of saturated solution of AgCl reacts with excess of hydroquinone (H_(2)Q) = 10^(-cd). If initially negligible AgCl(s) is present.
Answer» ANSWER :2002

Discussion

90979.	A cell necessarily does not contain:
Answer» An anode A cathode An ELECTOLYTE or a fuel A POROUS diaphagm Answer :D

Discussion

90980.	A cell is set up between copper and silver electrodes as :Cu\|Cu^(2+)(aq)\|\|Ag^(+)(aq)\|Ag. If its two half cells work under standard conditions, calculate the e.m.f. of the cell. [Given E_(Cu^(2+)//Cu)^(@)(E_(red)^(@))=+0.34"volt",E_(Ag^(+))//Ag)^(@)(E_(red)^(@))=+0.80"volt"]
Answer» ANSWER :0. 46 VOLT

Discussion

90981.	A cell necessarily does not contain
Answer» an ANODE a cathode an electrolyte or a fuel a porous diaphragm SOLUTION :Porous diaphragm is not NECESSARY for cell .

Discussion

90982.	A cell is represented by Zn\| Zn_((aq))^(+ +) \|\|Cu_((aq))^(+ +) \| CugivenCu^(+ +)+2 e^(-) toCu , E^(@)= + 0.35Vand Zn^(+ +) +2e^(- )to Zn ,E^@=- 0.763V Calculate emf of the cell and state whether the cell reaction will be spontaneous or non-spontaneous ?
Answer» 1.113, spontaneous `-0.567`, non-spontaneous `-1.113`, non-spontaneous 5.678 spontaneous Solution :Halfcellreactionof aodei.e,L.H.Selectrons ` Zn-2e^(-) hArrZn^( + +)` halfcellreactionof cathodei.e.,R.H.Selectrode ` Cu^( + +) hArrZn^( ++)+cu` `E_("cell")^@=E_("RED ")^(@)` ofR.H.S-` E_("Red ")^@ `ofL.H.S `= + 0.35- ( - 0.763)= 1.113` VOLT since `E_("cell")^@ `is POSITIVEAND hence`DELTAG `is negativeandthereforeteh cellreactionis spontaneous

Discussion

90983.	A cell is represented by the equation Cu(s)+2Ag^(+)(aq) to Cu^(2+)(aq)+2Ag(s). The voltage of the cell will increase with the increase in
Answer» size of the COPPER rod size of the silver rod CONCENTRATION of `Ag^(+)` ions concentration of `Cu^(2+)` ions Solution :Increase in concentration of cathodic COMPARTMENT solution will increase EMF of cell.

Discussion

90984.	A cell is prepared by dipping a copper rod in 1 M CuSO_(4) solution and a nickel rod in 1 M NiSO_(4) solution. The standard reduction potentials of copper electrode and nickel electrode are 0.34 volt and -0.25 volt respectively. (a) What will be the cell reaction? (b) What will be the standard EMF of the cell? (c) Which electrode will be positive? (d) How will the cell be represented?
Answer» Solution :(a) The cell reaction can be: either `Ni+CuSO_(4)toNiSO_(4)+Cu,""i.e.,""Ni+Cu^(2+)toNi^(2+)+Cu` or `Cu+NiSO_(4)toCu^(2+)+Ni,"i.e.,"Cu+Ni^(2+)toCu^(2+)+Ni` The correct reaction will be the one which gives positive E.M.F. if the first reaction is the cell reaction, we willhave `NitoNi^(2+)+2e^(-), "std. oxdn. POT"=-(-0.25)" volt"` `underline(""Cu^(2+)+2e^(-)toCu,"Std. redn. pot."=+0.34" volt")` Overall reaction: `Ni+Cu^(2+)toNi^(2+)+Cu,"Std. EMF=+0.59 volt"` Thus, EMF is positive for this reaction. hence, this is the correct cell reaction. (if the other reaction is tried, EMF will come out to be -0.59 volt) (b) Standard EMF of the cell as calculated above `=0.59` volt. (c) Since oxidation takes place at the nickel electrode, therefore, it acts as anode or since electrons are produced at this electrode, it acts as a negative pole. obviously, COPPER electrode will act as cathode or a positive pole. (d) By CONVENTION, the cell will be represented as : `Ni\|NiSO_(4)(1M)\|\|CuSO_(4)(1M)\|Cu`.

Discussion

90985.	A cell is constructed using Pb^(2+), Pb and Ni^(2+), Ni electrodes. If E^@ values of Pb and Ni electrode are respectively -0.13 and -0.25 V, write (a) the cell reaction and (b) cell notation.
Answer» Solution :Reduction POTENTIAL of Ni electrode is less and hence it reduces `PB^(2+)` ions. The CELL REACTIONS is written as, `Ni + Pb^(2+) to Pb + Ni^(2+)` The oxidation half reaction is , `Ni to Ni^(2+) + 2e^(-)` The reduction half reaction is , `Pb^(2+) + 2e^(-) to Pb`. The cell is written as, `Ni , Ni^(2+) (aq)//Pb^(2+)(aq),Pb.`

Discussion

90986.	A cell is constructed Pb^(2+) Pb and Ni^(2+), Ni electrodes. If E^(@) values of Pb and Ni electrodes are respectively -0.13 and -0.25V, write (a) the cell reaction and (b) cell notation.
Answer» Solution :Reduction potential of NI electrode is less and hence it REDUCES `Pb^(2+)`ions. The CELL reaction is written as, `Ni+Pb^(2+) RARR Pb+Ni^(2+)` The oxidation half reaction is,`Ni rarr Ni^(2+) +2e^(-)` The reduction half reaction is , `Pb^(2+)+2e^(-) rarr Pb` The cell is written as, `Ni, Ni_((aq))^(2+) //Pb_((aq))^(2+), Pb`.

Discussion

90987.	A cell involves the reaction Sn(S)+2Ag^(+)(aq) to Sn^(2+)(aq)+2Ag(s) The EMF of cell will invrease by
Answer» increase in concentration of `AG^(+)` ions decrease in concentration of `Ag^(+)` ions increase in concentration of `SN^(2+)` ions increase in size of Ag electrode Solution :Sn is undergoing oxidation and in solid form but `Ag^(+)` is solution with limited concentration . To continue working of CELL `Ag^(+)` must be increased time to time.

Discussion

90988.	A cell from the following which converts electrical energy into chemical energy?
Answer» dry cell ELECTRO chemical cell Electrolytic cell Lithium ION battery ANSWER :C

Discussion

90989.	A cell Cu\|Cu^(++)\|\|Ag^(+)\|Ag initially contains 2MAg^(+) and 2MCu^(++) ions in 1 L electrolyte. The change in cell potential after the passage of 10 amp. Curret for 4825 sec is:
Answer» `-0.00738V` `-1.00738V` `-0.0038V` none Solution :`Q=10xx4825=48250C` no. of FARADAY `=(48250)/(96500)=0.5` `AG+undersetunderset(2-0.25)(2.00)((1)/(2)Cu^(++))toundersetunderset(2+0.50)(2.00)(Ag^(+))+(1)/(2)Cu` `E_(cell)=E_(cell)^(@)-(0.0591)/(1)log(([Ag^(+)])/([Cu^(++)]^(1//2)))` `E_(1)=E_(cell)^(@)-(0.0591)/(1)log((2.00)/((2.00)^(1/2)))` `E_(2)=E_(cell)^(@)-(0.0591)/(1)log((2.50)/((1.75)^(1//2)))` `triangleE=E_(2)-E_(1)=(0.0591)/(1)[logsqrt(2)-log((2.50)/(sqrt(1.75)))]=(0.0591)/(1)[log1.41-log1.88]` `=(0.0591)/(1)[0.1492-0.2742]=-(0.0591)/(1)xx0.125=-0.00738V`.

Discussion

90990.	A cell in which electric current is produced by net oxidation and reduction process is called:
Answer» VOLTAIC cell Electrolytic cell Concentration cell None Answer :A

Discussion

90991.	A cell Cu\| Cu^(2+) \|\| Ag^(+) \| Ag initially contains 2 M Ag^(+) and 2 M Cu^(2+) ions in 1 L electrolyte. The magnitude of change in potential after charging the cell by passage of 10 amp current for 4825 sec at 298 K is:
Answer» 0.0074 V 1.0074 V 0.0038 V 0.0089 V Answer :A

Discussion

90992.	A cell contains two hydrogen electrodes. The negative electrode is in contact with a solution of 10^-6 Mhydrogen ions. The emf of the cell is 0.118V at 25^@C. Calculate the concentration of hydrogen ions at the positive electrode.
Answer» SOLUTION :POSITIVE ELECTRODE is CATHODE. `10^-4 M`

Discussion

90993.	A cell constructed by coupling a standard copper electrode and a standard magnesium electrode has emf of 2.7 volts. If the standard reduction potential of copper electrode is +0.34 volt, that of magnesium electrode is
Answer» +3.04 VOLTS -3.04 volts +2.36 volts -2.36 volts Answer :D

Discussion

90994.	A cell constant is generally found by measuring the conductivity of aqueous solution of
Answer» `BaCl_(2)` `KCl` `NACL` `MgCl_(2)` Solution :`KCl` is USED since its conductivity is known ACCURATELY at various CONCENTRATIONS and at different temperatures.

Discussion

90995.	A cell consists of two hydrogen electrodes. The negative electrode is in contact with a solution having 10^(-6)MH^(+) ion concentration. Calculate the concentration of H^(+) ions at the positive electrode, if the emf of the cell is found to be 0.118 V at 298 K.
Answer» Solution :Here, `C_(1)=10^(-6)M,""C_(2)=`tobe CALCULATED. For given concentration CELL, `E_(cell)=(0.0591)/(n)"log"(C_(2))/(C_(1))` `0.118=(0.0591)/(1)"log"(C_(2))/(10^(-6))" or ""log"(C_(2))/(10^(-6))=2" or "(C_(2))/(10^(-6))="Antilog 2"=10^(2) ""thereforeC_(2)=10^(2)xx10^(-6)=10^(-4)M`.

Discussion

90996.	A cell consists of two hydrogen electrode . The negative electrode is in contact with a solution having pH = 6 . The positive electrode is in contact with a solution of pH = x . Calculate the value of x if the e.m.f. of the cell is found to be 0.118 V at 298 K .
Answer» Solution :The cell may be represented as `H_2 \|H^+ (10^(-6) M) \|\|M^+ (M_1) \|H_2` `E = E^(@) - (0.059)/(1) "log" ([H^(+)]_("anode"))/([H^(+)]_("cathode"))` `0.118 = 0 - (0.059)/(1)"log" ((10^(-6)))/(M_(1))` log `((10^(-6)))/(M_(1))= - (0.118)/(0.059) = -2` `(10^(-6))/(M_(1)) = 10^(-2)` `therefore M_(1) = (10^(-6))/(10^(-2)) = 10^(-4)` M `therefore X = 4`

Discussion

90997.	A cell constructed by coupling a standard copper electrode and a standard magnesium electrode has EMF to 2.7 volts. If the standard reduction potential of copper electrode is +0.34 volt that of magnesium electrode is:-
Answer» `+3.04`VOLTS `-3.04` volts `+2.36` volts `-2.36` volts Solution :Mg lies above Cu in electrochemical series and hence Cu ELECTRODE acts as CATHODE. `E_(CELL)^(o)=E_(Cu^(++)//Cu)^(o)-E_(Mg^(++)//Mg)^(o)=-2.36V`

Discussion

90998.	A cell, Ag \| Ag^(+) \|\| Cu^(2+) \| Cu , initially contains 1 M Ag^+ and 1 M Cu^(2+)ions. Calculate the change in the cell potential after the passage of 9.65 A of current for 1 h.
Answer» SOLUTION :0.1355 V

Discussion

90999.	A cell, Ag\|Ag^(+)\|\|Cu^(2+)\|Cu, initially contains 1 M Ag^(+) and 1M" "Cu^(2+) ions. Calculate the change in cell potential after the passage of 9.65 A of a current of 1 h.
Answer» SOLUTION :Quantity of electricity passed passed=`9.65xx60xx60C=34740C` `thereforeAg^(+)` ions DEPOSITED`=34740//96500" mole"=0.36` mole`(Ag^(+)+e^(-)toAg)` `Cu^(2+)` ions deposited`=34740//(2xx96500)=0.18` mole`(Cu^(2+)+2e^(-)toCu)` `therefore[Ag^(+)]"left"=1-0.36=0.64M` `[Cu^(2+)]" left"=1-0.18=0.82M` Cell reaction is: `Cu+2AG^(+)toCu^(2+)+2Ag` `DeltaE=E_(cell)^(@)=(0.0591)/(2)"log"([Cu^(2+)])/([Ag^(+)]^(2))=(0.0591)/(2)"log"(0.82)/((0.64)^(2))=8.89xx10^(-3)V`

Discussion

91000.	A cell Ag\|Ag^+\|\|Cu^(++)\|Cu initially contains 2M Ag^+ and 2M Cu^(++) ions. The change in cell potential after the passage of 10 amp current for 4825 sec is:
Answer» `-0.0074 V` `-1.00738 V` `-0.0038 V` none Solution :Q=10x4825 =48250 C no. of mole of `Ag^+ =48250/96500=0.5` `Ag+1/2 Cu^(++)TOAG^+ +1/2 Cu` `{:(2.00,2.00),(2-0.25,2+0.50):}` `E_"CELL"=E_"cell"^@-0.0591/1 "LOG" ([Ag^+])/([Cu^(++)]^(1//2)) IMPLIES E_1=E_"cell"^@-0.0591/1"log"2.00/((2.00)^1//2)` `E_2=E_"cell"^@-0.0591/1 "log" (2.50)/(1.75)^((1//2)) implies DeltaE=E_2-E_1=0.0591/1["log"SQRT2-"log"2.50/sqrt(1.75)]` `=0.0591/1["log"1.41-"log"1.88]=0.0591/1[0.1492-0.2742]=-0.0591/2xx0.125=-0074 V`

Discussion

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A certain buffer solution contains equal concentration of X^- and HX. The K_afor HX is 10^(-8) The pH of the buffer is

A certain buffer solution contains equal concentration of X^(-) and HX. K_(b) for X^(-) is 10^(-10). Calculate pH of buffer

A certain aqueous solution of FeCl_(3) (formula mass = 162) has a density of 1.1 g/ml and contains 20.0% FeCl_(3). Molar concentration of this solution is

A certain acid - base indicator is red in acid and blue in basic solution. At pH = 5, 75% of the indicator is present in the solution in its blue form. Calculate dissociation constant (K_a) for the indicator and pH range over which the indicator changes from 90% red- 10% blue to 90% blue-10% red.

A certain acid-base indicator is red in acid solution and blue in basic medium. At pH=5 75% of the indictor is present in the solution in its blue form. Calculate the pH at which indicator shows 90% red form ?

(A): Central atom usually forms three single and one double bonds in an oxyacid of phosphorus (R): Phosphorus atom usually undergoes sp^(3) hybridisation in its oxyacids

A centi- molar solution is diluted 10 times. The molaruity would become

(A) Cellulose is not digested by human beings. (R) Human beings are not having cellulose digestable enzymes.

A cellwith two electrodes, one of grey tin and the other white tin, both dipping in solution of (NH_4)_2 SNCI_6 showed zero emf at 18^@C. What conclusion may be draw from this "

A cell utilises the following reaction: 2CO^(3+) (aq)+Zn(s) to 2CO^(2+)(aq)+Zn^(2+) (aq) What is the effect on cell emf of each of the following changes? Additional water is added to the anode compartment

A cell utilises the following reaction: 2CO^(3+) (aq)+Zn(s) to 2CO^(2+)(aq)+Zn^(2+) (aq) What is the effect on cell emf of each of the following changes? Co(II) nitrate is dissociated in the cathode compartment.

A cell utilises the following reaction: 2CO^(3+) (aq)+Zn(s) to 2CO^(2+)(aq)+Zn^(2+) (aq) What is the effect on cell emf of each of the following changes? The size or area of Zn (s) electrode is doubled

A cell utilises the following reaction: 2CO^(3+) (aq)+Zn(s) to 2CO^(2+)(aq)+Zn^(2+) (aq) What is the effect on cell emf of each of the following changes? Co(III) nitrate is dissolved in the cathode compartment

A cell necessarily does not contain:

A cell is set up between copper and silver electrodes as :Cu|Cu^(2+)(aq)||Ag^(+)(aq)|Ag. If its two half cells work under standard conditions, calculate the e.m.f. of the cell. [Given E_(Cu^(2+)//Cu)^(@)(E_(red)^(@))=+0.34"volt",E_(Ag^(+))//Ag)^(@)(E_(red)^(@))=+0.80"volt"]

A cell necessarily does not contain

A cell is represented by Zn| Zn_((aq))^(+ +) ||Cu_((aq))^(+ +) | CugivenCu^(+ +)+2 e^(-) toCu , E^(@)= + 0.35Vand Zn^(+ +) +2e^(- )to Zn ,E^@=- 0.763V Calculate emf of the cell and state whether the cell reaction will be spontaneous or non-spontaneous ?

A cell is represented by the equation Cu(s)+2Ag^(+)(aq) to Cu^(2+)(aq)+2Ag(s). The voltage of the cell will increase with the increase in

A cell is constructed using Pb^(2+), Pb and Ni^(2+), Ni electrodes. If E^@ values of Pb and Ni electrode are respectively -0.13 and -0.25 V, write (a) the cell reaction and (b) cell notation.

A cell is constructed Pb^(2+) Pb and Ni^(2+), Ni electrodes. If E^(@) values of Pb and Ni electrodes are respectively -0.13 and -0.25V, write (a) the cell reaction and (b) cell notation.

A cell involves the reaction Sn(S)+2Ag^(+)(aq) to Sn^(2+)(aq)+2Ag(s) The EMF of cell will invrease by

A cell from the following which converts electrical energy into chemical energy?

A cell Cu|Cu^(++)||Ag^(+)|Ag initially contains 2MAg^(+) and 2MCu^(++) ions in 1 L electrolyte. The change in cell potential after the passage of 10 amp. Curret for 4825 sec is:

A cell in which electric current is produced by net oxidation and reduction process is called:

A cell Cu| Cu^(2+) || Ag^(+) | Ag initially contains 2 M Ag^(+) and 2 M Cu^(2+) ions in 1 L electrolyte. The magnitude of change in potential after charging the cell by passage of 10 amp current for 4825 sec at 298 K is:

A cell contains two hydrogen electrodes. The negative electrode is in contact with a solution of 10^-6 Mhydrogen ions. The emf of the cell is 0.118V at 25^@C. Calculate the concentration of hydrogen ions at the positive electrode.

A cell constructed by coupling a standard copper electrode and a standard magnesium electrode has emf of 2.7 volts. If the standard reduction potential of copper electrode is +0.34 volt, that of magnesium electrode is

A cell constant is generally found by measuring the conductivity of aqueous solution of

A cell consists of two hydrogen electrodes. The negative electrode is in contact with a solution having 10^(-6)MH^(+) ion concentration. Calculate the concentration of H^(+) ions at the positive electrode, if the emf of the cell is found to be 0.118 V at 298 K.

A cell consists of two hydrogen electrode . The negative electrode is in contact with a solution having pH = 6 . The positive electrode is in contact with a solution of pH = x . Calculate the value of x if the e.m.f. of the cell is found to be 0.118 V at 298 K .

A cell constructed by coupling a standard copper electrode and a standard magnesium electrode has EMF to 2.7 volts. If the standard reduction potential of copper electrode is +0.34 volt that of magnesium electrode is:-

A cell, Ag | Ag^(+) || Cu^(2+) | Cu , initially contains 1 M Ag^+ and 1 M Cu^(2+)ions. Calculate the change in the cell potential after the passage of 9.65 A of current for 1 h.

A cell, Ag|Ag^(+)||Cu^(2+)|Cu, initially contains 1 M Ag^(+) and 1M" "Cu^(2+) ions. Calculate the change in cell potential after the passage of 9.65 A of a current of 1 h.

A cell Ag|Ag^+||Cu^(++)|Cu initially contains 2M Ag^+ and 2M Cu^(++) ions. The change in cell potential after the passage of 10 amp current for 4825 sec is: