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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 801. |
Match the followingColumn-I (Electrolysis)Column-II (Observation) (A) Aqueous solution of NaCl using inert electrodes(p) Metal loss at anode(B) Very dilute aqueous solution of NaCl using mercury cathode (q) Chlorine gas evolved at anode(C) CuSO4 using copper electrodes(r) Oxygen gas evolved at anode(D) 50% H2SO4 solution(s) A compound with peroxide bond is formed |
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Answer» A→(q), B→(r), C→(p), D→(s) |
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| 802. |
Electrode potentials of the given half cells, `Pt(H_(2))|H^(+)(C_(1)),Pt(Cl_(2))|Cl^(-)(C_(2)),Ag|Ag^(+)(C_(3))`A. Will increase on increasing `C_(1),C_(2) and C_(3)`B. Will decrease on increasing `C_(1),C_(2) and C_(3)`C. Will decrease on increasing `C_(1) and C_(3)` and increase on increasing `C_(2)`D. Will remain constant if `C_(1)` or `C_(2)` is doubled and `p_(1)` or `p_(2)` is made four times. |
| Answer» Correct Answer - C::D | |
| 803. |
Assuming that a constant current is delivered, how many kW-h of electricity can be produced by the reacation of `1.0` mole `Zn` with `Cu^(2+)` ion in a Daniel cell in which all the concentration remains `1.00 M` ? `(E_(Zn//Zn^(2+))^(@) = 0.76 V , E_(Cu//Cu^(2+))^(@) = -0.34 V)` |
| Answer» Correct Answer - `0.0059 kW-h ;` | |
| 804. |
In an analytical determination of aresenic , a solution containing aresenious acid, `(H_(3)AsO_(3))`. `KI` and a small amount of starch is electrolysed. The electrolysis produes free `I_(2)` from `I^(-)` ion and the `I_(2)` immediately oxidises the arsenious acid to hydrogen arsenate ion, `(HAsO_(4)^(2-))`. `I_(2(aq.))+H_(3)AsO_(3(aq.))+H_(2)O_((l))rarr2I_((aq.))^(-)+HAsO_(4(aq.))^(2-)+4H_((aq.))^(+)` When the oxidation of arsenic is complete, the free iodine combines with the starch to give a deep blue colour. If during a particular run, it takes `65.3 s` for a current of `10.5 mA` to give an end point (indicated by the blue colour), how many grams of arsenic and `H_(3)AsO_(3)` are present in the solution? `("At. wt. of "As = 75)` |
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Answer» Correct Answer - As `= 2.66 xx 10^(-4)g`, `H_(3)AsO_(3) = 5.0 xx 10^(-4)g ;` |
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| 805. |
The temperature coefficient of a cell whose operation is based on the reaction: `Pb_(s) + HgCl_(2(aq.)) rarr PbCl_(2(aq.)) + Hg_(l)` os `-1.5 xx 10^(-4)VK^(-1)`. The heat of rection at `25^(@)C` is (in `kJ mol^(-1)`) (Given `E = 0.03 V`):A. `8.4`B. `16.4`C. `14.41`D. `4.5` |
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Answer» Correct Answer - C `DeltaH = nF[T[(deltaE)/(deltaT)]_(P)-E]`, `:. DeltaH = 2 xx 96500[-1.5 xx 10^(-4)] xx 298 - 0.03]` `= -14417.1 J` |
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| 806. |
The solubility product of `Pb_(3)(AsO_(4))_(2)` is `4.1xx10^(-36). E^(c-)` for the reaction `:` `Pb_(3)(AsO_(4))_(2)(s)+6e^(-)hArr3Pb(s)+2AsO_(4)^(2-)` `E_((Pb)2^(+)|Pb)^(Θ)=-0.13V`A. `+0.478V`B. `-0.13V`C. `-0.478V`D. `+0.13V` |
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Answer» Correct Answer - c `E^(c-)._(AsO_(4)^(2-)|Pb_(3)(AsO_(4))^(2)|Pb)=E^(c-)._(Pb^(2+)|Pb)+(0.059)/(6)logK_(sp)` `=-0.13+(0.059)/(6)log4.1xx10^(-36)` `=-0.13-0.348=-0.478V` |
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| 807. |
Calculate `E^(c-)` for the reactions `:` `ZnY^(2-)hArrZn(s)+Y^(4-)` where `Y^(4-)` is the completely deprotonated anion of EDTA. The formation constant for`ZnY^(2-)` is `3.2xx10^(16)` and `E^(c-)` for `Zn rarr Zn^(2+)+2e^(-)` is `0.76V`.A. `-1.25V`B. `0.48V`C. `+0.68V`D. `-0.27V` |
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Answer» Correct Answer - a `E^(c-)._(Zn^(2+)|ZnY^(2-)|Y^(4-))=-E^(c-)._(Zn^(2+)|Zn)+(0.059)/(2)logK` `K_(f)=([ZnY^(2-)])/([Y^(4-)])impliesK=(1)/(K_(f))` `E^(c-)._(Zn^(2+)|ZnY^(2-)|Y^(4-))` `=-0.76+(0.059)/(2)log.(1)/(3.2xx10^(16))=-1.25V` |
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| 808. |
Show that in a first order reaction, time pequired for fompletion of `99.9%` is 10 times of half-life `(t_(1//2))` of the reaction. |
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Answer» When reaction is completed `99.9%, [R]_(n)=[R]_(0)-0.999[R]_(0)` `k=(2.303)/(t)log ""([R]_(0))/([R])` `=(2.303)/(t )log ""([R]_(0))/([R]_(0)-0.999[R]_(0))=(2.303)/(t)log 10^(3)` `t=6.909//K` ltbr For half-life of the reaction `t_(1//2) =0.693//K` `(t)/(t_(1//2))=(6.909)/(k)xx(k)/(0.693)=10` |
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| 809. |
Which of the following statement is correct in the context of a battery?A. It is an electrochemical cellB. It is used as a source of energyC. The stored energy is released during the redox reactionD. All of these |
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Answer» Correct Answer - D We commonly use voltaic cells as convenient, portable source of energy. Any battery (actually it may have one or more than one cell connected in series) or cell that we use as a source of electrical energy is basically a galvanic cell where cell chemical energy of the redox reaction is converted into electrical energy. However, for a battery to be of practical use it should be reasonably light, compact and its voltage should not very appreciably during its use. There are mainly two types of batteries. |
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| 810. |
`30 mL` of `0.13 M NiSO_(4)` is electrolysed using a current of `360` milliamperes for `35.3` minutes. The mass of the metal that would have been plated out if current efficiency is only `60%` (`Ni = 58.7 u`) isA. `0.9131 g`B. `0.3911 g`C. `0.1391 g`D. `0.2474 g` |
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Answer» Correct Answer - C The current efficiency is define as the ration of the amount of the converted substance to the amount of the substance that would react if the whole current is used in a given reaction. Thus `%` current efficiently `= ("Actuall mass of Ni deposited")/("Expected mass of Ni deposited") xx 100%` Calculate the mass of `Ni` that should be deposited : Charge in coulombs = current in amperes `xx` time in seconds `= (0.36 A)(35.3 "minutes")(60 "seconds/min")` `= 762.48 C` No. of faraday `= (762.48 C)/(96500 C//F)` `= 0.0079 F` Since `1 F` (1 mole of electric charge) produces `1` equivalent of the substance, no of equivalents of `Ni` that should be deposited for `100%` current efficiency is equal to no. of faradays. Thus, mass of `Ni` deposited for `100%` current efficiency will be (no of equivalents)(`gm` equivalent mass) `= (0.0079 eq.)(58.7//2 g eq.^(-1))[Ni^(2+)+2e^(-)rarrNi]` But the current effciency is `60%` only. Thus Actual mass of `Ni` deposited `= 0.2318 xx (60)/(100)` `= 0.1391 g` |
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| 811. |
The decomposition of hydrocarbon follows the equation `K=(4.5xx10^(11) s^(-1))e^(-18000K//T).` Calculate `E_(a).` |
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Answer» According to the available data `k=Ae ^(-E_(a)//RT)-(i)` According to the available data `k=(4.5xx10s^(-1))e^(-28000)k//T-(ii)` on comparing both equations `-(E_(a))/(RT)=(-2800k)/(T)` `E_(a) =(28000k)xxR` `=(28000k)xx(8.314K^(-1)J mol^(-1))` `=232792Jk mol^(-1)` `E_(a) =231.792 kJ^(-1) mol ^(-1)` |
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| 812. |
Sucrose decompose in acid solution into glucose and fructose according to the first order rate law, with `t_(1/2)=3.00` hours. What fraction of sample of sucrose remains after 8 hours ? |
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Answer» For `1^(st)` order reactions `k=(0.693)/(t_(1//2))=(0.693)/((3.h))` ` t=(2.303)/(k)log""([A]_(0))/([A])` `log ""([A]_(0))/([A])=(kxxt)/(2.303)` `log ""([A]_(0))/([A])=(0.693)/(3) xx((8h))/(2.303)=0.8024` `([A]_(0))/([A])="Antilog"0.8024=6.345` `[A]_(0)=1M,` `[A]=([A]_(0))/(6.345)=(1M)/(6.345)=0.1576M` After 8 hours sucrose left `=0.1576M` |
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| 813. |
In the electrolysis of aqueous `NaCl` ,what volume of `Cl_2`(g) is produced in the time that it takes to liberate 5.0 liter of `H_2`(g) ? Assume that both gases are measured at STP.A. 5B. 2.5C. 7.5D. 10 |
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Answer» Correct Answer - A |
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| 814. |
The quantity of charge required to obtain one mole of aluminium from `Al_(2)0_(3)` isA. 1 FB. 6 FC. 3 FD. 2 F |
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Answer» Correct Answer - C `Al_2O_3 to 2Al` `2Al^(3+) + 6e^(-) to 2Al ` `therefore Al^(3+) + 3e^(-) to Al` Thus, 3 F of charge is required to obtain 1 mole of Al from `Al_2O_3` |
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| 815. |
`Lambda_((m)(NH_(4)OH))^(@)` is equal toA. `Lambda_(m(NH_4OH))^@ + Lambda_(m(NH_4Cl))^@-Lambda_(m(HCl))^@`B. `Lambda_(m(NH_4Cl))^@+Lambda_(m(NaOH))^@-Lambda_(m(NaCl))^@`C. `Lambda_(m(NH_4Cl))^@+Lambda_(m(NaCl))^@-Lambda_(m(NaOH))^@`D. `Lambda_(m(NaOH))^@+Lambda_(m(NaCl))^@-Lambda_(m(NH_4Cl))^@` |
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Answer» Correct Answer - B `Lambda_(m(NH_4Cl))^@+Lambda_(m(NaOH))^@-Lambda _(m(NaCl))^@` `Lambda_(m(NH_4^+))^@+Lambda_(m(Cl^-))^@+Lambda_(m(Na^+))^@+Lambda_(m(OH^-))^@-Lambda_(m(Na^+))^@-Lambda_(m(Cl^-))^@` `=Lambda_(m(NH_4^+))^@+Lambda_(m(OH^-))^@=Lambda_(m(NH_4OH))^@` |
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| 816. |
While charging the lead storage battery:A. `PbSO_4` anode is reduced to PbB. `PbSO_4` cathode is reduced to PbC. `PbSO_4` cathode is oxidised to PbD. `PbSO_4` anode is oxidised to `PbO_2` |
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Answer» Correct Answer - A On Charging the battery the reaction is reversed and `PbSO_(4(s))` is the converted into Pb at anode and `PbO_2` at cathode . |
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| 817. |
The quantity of charge required to obtain one mole of aluminium from Al2O3 is ___________.(i) 1F(ii) 6F(iii) 3F(iv) 2F |
| Answer» The correct answer is (iii) 3F | |
| 818. |
In the electrolysis of aqueous sodium chloride solution which of the half cell reaction will occur at anode?A. `Na^(+)(aq)+e^(-)toNa(s),E_(Cell)^(Theta)=-2.71V`B. `2H_(2)O(l)toO_(2)(g)+4H^(+)(aq)+4e^(-),E_(Cell)^(Theta)=1.23V`C. `H^(+)(aq)+e^(-)to(1)/(2)H_(2)(g),E_(Cell)^(Theta)=0.00V`D. `Cl^(-)(aq)to(1)/(2)Cl_(2)(g)+e^(-),E_(Cell)^(Theta)=1.36V` |
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Answer» Correct Answer - B oxidatio occurs at anode. As (b) has higher oxidation potential than (a), (b) occurs. |
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| 819. |
In the electrolysis of aqueous sodium chloride solution which of the hall cell reaction will occur at anode?A. `Na^(+)(aq)+e^(-) to Na(s),E_(cell)^(Ï´)=2.71" V "`B. `2H_(2)O(l) to O_(2)(g)+4H^(+)(aq)+4e^(-),E_(cell)^(Ï´)=1.23" V "`C. `H^(+)(aq)+e^(-)to(1)/(2)H_(2)(g),E_(cell)^(Ï´)=0.00" V "`D. `Cl^(-)(aq)to(1)/(2)Cl_(2)(g)+e^(-),E_(cell)^(@)=1.36" V "` |
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Answer» Correct Answer - B Is the correct answer. |
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| 820. |
EΘcell for some half cell reactions are given below. On the basis of these mark the correct answer.(a) H+ (aq) + e- →1/2H2(g) ; EΘCell = 0.00V(i) In dilute sulphuric acid solution, hydrogen will be reduced at cathode.(ii) In concentrated sulphuric acid solution, water will be oxidised at anode.(iii) In dilute sulphuric acid solution, water will be oxidised at anode.(iv) In dilute sulphuric acid solution, SO42– ion will be oxidised to tetrathionate ion at anode. |
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Answer» (i) In dilute sulphuric acid solution, hydrogen will be reduced at cathode. (iii) In dilute sulphuric acid solution, water will be oxidised at anode. |
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| 821. |
The cell constant of a conductivity cellA. changes with change of electrolyteB. changes with change of concentration of electrolyteC. changes with temperature of electrolyteD. remains constant for a cell. |
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Answer» Correct Answer - D The cell constanet (I/a) remains constant for a cell. |
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| 822. |
The positive value of the standard electrode potential of Cu2+/Cu indicates that ____________.(i) this redox couple is a stronger reducing agent than the H+/H2 couple.(ii) this redox couple is a stronger oxidising agent than H+/H2.(iii) Cu can displace H2 from acid.(iv) Cu cannot displace H2 from acid. |
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Answer» (ii) this redox couple is a stronger oxidising agent than H+/H2. (iv) Cu cannot displace H2 from acid. |
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| 823. |
Electrode potential for `Mg` electrode varies according to the equation `E_(Mg^(2+)|Mg)=E_(Mg^(2+)|Mg)^(ϴ) -(0.059)/2 "log" 1/([Mg^(2+)])` The graph of `E_(Mg^(2+)|Mg) vs log [Mg^(2+)]` is |
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Answer» Correct Answer - B (b) `E=E^(@)+(0.059)/(2)"log"[Mg^(2+)]`.Hence the plot E vs log `[Mg^(2+)]` is linear with positive slope intercept=`E^(@)`. |
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| 824. |
Which cell will measure standard electrode potential of copper electrode?A. `Pt(s) | H_(2)(g, 0.1 " bar")|H^(+)(aq., 1 M)||Cu^(2+)(aq., 1 M)|Cu`B. `Pt(s) | H_(2)(g, 1 " bar")|H^(+)(aq., 1 M)||Cu^(2+)(aq., 2 M)|Cu`C. `Pt(s) | H_(2)(g, 1 " bar")|H^(+)(aq., 1 M)||Cu^(2+)(aq., 1 M)|Cu`D. `Pt(s) | H_(2)(g, 1 " bar")|H^(+)(aq., 0.1" M")||Cu^(2+)(aq., 1 M)|Cu` |
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Answer» Correct Answer - C (c ) For standard electrode potential `(E^(@))`. `[Cu^(2+)]=1M` |
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| 825. |
The metal which can be used to obtain metallic copper from aqueous `CuSO_(4)` is :A. NaB. AgC. HgD. Fe |
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Answer» Correct Answer - D (d) Sodium reacts with water violently. Both silver and mercury are weaker reducing agents than copper. Therefore, metal iron can reduce `Cu^(2+)` ions to Cu. `Fe(s)+CuSO_(4)(aq) to FeSO_(4)(aq)+Cu(s)` |
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| 826. |
`Zn|Zn^(2+)(a=0.1M)||Fe^(2+)(a=0.01M)Fe.` The `EMF` of the above cell is `0.2905`. The equilibrium constant for the cell reaction isA. `10^(0.32//0.0591)`B. `10^(0.32//0.0295)`C. `10^(0.26//0.0295)`D. `e^(0.32//0.295)` |
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Answer» Correct Answer - B `Zn|Zn^(+2) (0.01M)||Fe^(+2) (0.001M)|Fe` `E_(cell)^(@) = E_(cell)^(@) - (RT)/(nF) In K` `E^(@) = E_(Cell)^(@) - (RT)/(2F) In K` `E^(@) = 0.2905 - (0.0591)/(2) log.(10^(-3))/(10^(-2))` `=0.2905 - (0.0591)/(2) log 10^(-1)` `= 0.2905 +(0.0591)/(2) = 0.2905 +0.02950` `= 0.32V` `E^(@) = (RT)/(2F) In K` `E^(@) = (0.0591)/(2) log K` `log K = (0.32)/(0.0295) rArr K = 10^(0.32//0.0295)` |
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| 827. |
`Zn|Zn^(2+)(a=0.1M)||Fe^(2+)(a=0.01M)Fe.` The `EMF` of the above cell is `0.2905`. The equilibrium constant for the cell reaction isA. `10^(0.32//0.0591)`B. `10^(0.32//0.0295)`C. `10^(0.26//0.0295)`D. `e^(0.32//0.2995)` |
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Answer» Correct Answer - b `Zn|Zn^(2+)(a=0.1M)||Fe^(2+)(a=0.01M)|Fe` the cell reaction `i. Zn(s)rarr Zn^(2+)(aq)+2e^(-)` `ii. Fe^(2+)(aq)+2e^(-) rarr Fe(s)` `ulbar(Zn(s)+Fe^(2+)(aq) rarr Zn^(2+)+Fe(s))` On applying Nernst equation, `E_(cell)=E^(c-)._(cell)-(0.0591)/(n)log.([Zn^(2+)])/([Fe^(2+)])` `0.2905=E^(c-)._(cell)-(0.0591)/(2)log.(0.1)/(0.01)` `0.2905=E^(c-)._(cell)-0.02905xxlog10` `0.295=E^(c-)._(cell0-0.2905xx1` `:. E^(c-)._(cell)=0.2905+0.0295=0.32V` At equilibrium `(E_(cell)=0)`, `E_(cell)=E^(c-)._(cell)-(0.0591)/(n)log K_(c)` `:. 0=E^(c-)._(cell)-(0.0591)/(n)logK_(c)` or `E^(c-)._(cell)=(0.0591)/(2)logK_(c)` `0.32=(0.0591)/(2)logK_(c)` or `K_(c)=10^(0.32//0.295)` |
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| 828. |
which of the following plots will obtained for a conductometric titration of strong acid against a weak base?A. B. C. D. |
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Answer» Correct Answer - C |
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| 829. |
The cell potential becomes half if the cell reaction is divided by 2 througout. |
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Answer» Correct Answer - F Cel potential is an intensive property |
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| 830. |
Using the date in the preceding problem, calculate the equilibrium constant of the reaction at `25^(@)C` `Zn +Cu^(++) hArr Zn^(++) +Cu, K ([Zn^(2+)])/([Cu^(2+)])`A. `8.314 xx 10^(24)`B. `4.831 xx 10^(31)`C. `8.314 xx 10^(36)`D. `4.831 xx 10^(44)` |
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Answer» Correct Answer - C `{:(E^(@)=(0.0591)/(2)logK,1.1 xx 2 = 0.06 log K),(36.22 = logK, 10^(36.22) = K_(f)),(8.314 xx 10^(36) =K,):}` |
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| 831. |
Name the electrodes used in a fuel cell. |
| Answer» These are porous carbon electrodes through which hydrogen and oxygen are bubbled under high pressure. | |
| 832. |
The standard electrode potential (reduction) of `Ag^(+)|Ag` is `0.800 V` at `25^(@)C`. Its electrode potential in a solution containing `10^(-3)M` ion of `Ag^(+)` ions is:-A. `0.623V`B. `-0.977V`C. `0.892V`D. `1.246V` |
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Answer» Correct Answer - A `E_(cell) = 0.80 -(0.0591)/(1)log[(1)/(10^(-3))] = 0.623 V` |
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| 833. |
Given that : (R)CO3+ + e- → CO2+ E˚ = 1.82 V2H2O → O2 + 4H+ + 4 e- E˚ = -1.23 V Explain why CO3+ is not stable in aqueous solution? |
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Answer» The E˚ cell can be calculated as : 4[CO3+ + e- → CO2+] E˚= 1.82 V 2H2O → O2 + 4H+ + 4e- E˚= -1.23 V Cell reaction : 4CO3+ + 2H2O → CO2+ O2 + 4H+ E˚ cell = 1.82 V-(-1.23 V) = 3.05 V Since E˚cell is positive, the cell reaction is spontaneous. CO3+ iron will take part in the reaction and hence |
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| 834. |
The number of faradays required to produce one mole of water from hydrogen-oxygen fuel cell containing aqueous alkali as electrolyte is:A. 1B. 3C. 2D. 4 |
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Answer» Correct Answer - C `overset(0)H_(2) +(1)/(2)O_(2) rarr overset(+1)H_(2) O^(-2) = 2F` |
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| 835. |
State two advantages of `H_(2)-O_(2)` fuel cell over ordinary cell. |
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Answer» (i) They do not cause any pollution. (ii) They have high efficiency of 60-70%. |
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| 836. |
What is the overall electrochemical reaction taking place in rusting? |
| Answer» `2Fe(s)+O_(2)(g)+4H^(+)(aq)to2Fe^(2+)(aq)+2H_(2)O(l)` | |
| 837. |
The standard Gibbs free energy change `(DeltaG^(@)" in kJ mol"^(-1))`, in a Daniel cell `(E_("cell")^(@)=1.1 V)`, when 2 moles of `Zn (s)` is oxidized at 298 K, is closest toA. 212 . 30B. `-212.30`C. `106.15`D. `-106.15` |
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Answer» Correct Answer - B `Delta G = - n F E_(cell) = - 2 xx 96500 xx 1.1 J = - 212.3 kJ` . |
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| 838. |
The electrolysis of molten sodium hydride liberates `………………` gas it the `………………`. |
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Answer» Correct Answer - `H_(2)` gas, anode The electrolysis of molten sodium hydride liberates hydrogen gas at the anode. |
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| 839. |
TRENDS IN THE `E_(M^(+3)//M^(+2))` VALUES OF 3d SERIES |
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Answer» Correct Answer - T TRUE `E=E^(c-)-(0.0591)/(n)log.(1)/([M^(n+)])` `=E^(c-)+(0.0591)/(n)[M^(+n)]` |
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| 840. |
One faraday of electricicy is passed through aqueous solution of sodium chloride. It producesA. one mole of oxygen at anodeB. 1 gm of hydrogen at cathodeC. neither hydrogen nor oxygen is producedD. sodium is deposited at cathode in equivalent proportion. |
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Answer» Correct Answer - B `NaCl + aq to Na^(+) (aq) + Cl^(-) (aq)` `H_(2)O to H^(+) + OH^(-)` At cathode : `H^(+) + e^(-) to (1)/(2) H_(2)` 1 F liberates 1 g eq.of `H_(2)` gas , i.e., 1 g of `H_(2)` at cathode . |
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| 841. |
Which of the following involves electrochemnical phenomenon?A. Manufacture of aluminium from bauxiteB. Manufacture of caustic soda by Castner- Kellner methodC. Sensory signals sent to the brain through cellsD. All the three above |
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Answer» Correct Answer - D All the given processes involve electrochemical phenomena. |
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| 842. |
A reaction `:` `(1)/(2)H_(2)(g)+AgCl(s)hArrH^(o+)(aq)+Cl^(Θ)(aq)+Ag(s)` occurs in a galvanic cell. The structure of the cell will beA. `Ag,AgCl(s)|KCl(sol)|AgNO_(3)(sol),Ag`B. `Pt,H_(2)(g)|HCL(sol)|AgNO_(3)(sol),Ag`C. `Pt,H_(2)(g)|HCl(sol)|AgCl(s),Ag`D. `Pt,H_(2)(g)|KCl(sol)|AgCl(s),Ag` |
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Answer» Correct Answer - c Factual Statement |
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| 843. |
The number of atoms of `Ca` that will be deposited from a solution of `CaCl_(2)` by a current of `25mA` for `60s` will beA. `4.68xx10^(18)`B. `4.68xx10^(15)`C. `4.68xx10^(10)`D. `2.34xx10^(15)` |
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Answer» Correct Answer - a `25xx10^(-3)Axx60 C` of charge carried by `=(25xx10^(-3)xx60s)/(96500C)=1.55xx10^(-5) mol e^(-)` `Ca^(2+)+2e^(-) rarr Ca` `2 mol ` of `e^(-)` will produce `=6xx10^(23)` atoms of `Ca` `1.55xx10^(-5) mol e^(-)` will produce `=(1.55xx10^(-5)xx6xx10^(23)/(2)=4.68xx10^(18)` atoms of `Ca` |
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| 844. |
Which of the following involves electrochemnical phenomenon?A. The transmission of sensory singnals through cell to brain and vice cersa and communication between the cellsB. Manufacture of fluorineC. Refining of metal copperD. All of these |
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Answer» Correct Answer - D Many reactions that take place in living cells are redox reactions. These reaction can be studied with mini-ture electrodes. Element fluorine is manufacured by the electrolysis of a solution of `KHF_(2)` in anhydrous `HF`: `K[HF_(2)]+Hfoverset("electrolysis") rarr H_(2)+F_(2)` Metal copper is refined by electrolysis using `Cu` electrodes with an electrolyte of dilute `H_(2)SO_(4)` and `CuSO_(4)`. |
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| 845. |
The number of faradays required to liberate 1 mole of any element indicatesA. weight of elementB. conductance of electrolyteC. charge on the ion of that elementD. none |
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Answer» Correct Answer - C 1 Faraday deposits E g or (M/charge) g of species . |
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| 846. |
Statement : Poggendroff compensation method is used for the measurement of emf of voltaic cells. Explanation : This method has the advantage of giving the emf of an open circuit when it produces no current and thus determines emf of cells under reversible condition.A. `S` is correct but `E` is wrongB. `S` is wrong but `E` is correct.C. Both `S` and `E` are correct and `E` is correct explanation of `S`.D. Both `S` and `E` are correct but `E` is not correct explanation of `S`. |
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Answer» Correct Answer - C Explanation is correct reason for statement. |
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| 847. |
Find the charge in coulombs on 1 g-ion of `N^(3-)` |
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Answer» Charge on one ion of `N^(3-)=3xx1.6xx10^(-19)`coulombs One g-ion`=6.02xx10^(23)`ions ltBrgt `therefore`charge on 1 g-ion of `N^(3-)=3xx1.6xx10^(-19)xx6.02xx10^(23)` `=2.89xx10^(5)`coulombs |
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| 848. |
1 faraday = _____ coulombsA. 10000B. 95000C. `96.5`D. 96500 |
| Answer» Charge of one mole of electrons is known as one faraday that is equal to 96,500 C. | |
| 849. |
How many coulombs of electricity are required for the oxidation of one mole of water to dioxygen ?A. `1.93xx10^(4)C`B. `19.3xx10^(5)C`C. `9.65xx10^(4)" C"`D. `1.93xx10^(5)" C"` |
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Answer» Correct Answer - D (d) `H_(2)O to 2H^(+)+1//2 O_(2)+underset(2 mol)(2e^(-))` No. of coulombs electricity required `=2xx96500" C"` `=1.93xx10^(5)" C"` |
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| 850. |
How many coulombs of electricity is required to reduce 1 mole of `Cr_2O_7^(2-)` in acidic medium?A. 4 x 96500 CB. 6 x 96500 CC. 2 x 96500 CD. 1 x 96500 C |
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Answer» Correct Answer - B In acidic medium , dichromate ions are reduced according to the equation, `Cr_2O_7^(2-) +14H^+ + 6e^(-) to 2Cr^(3+) + 7H_2O` 1 mole of `Cr_2O_7^(2-)` ion requires 6 moles of electrons or 6 x 96500 C of electricity |
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