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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.
| 951. |
Write the relation between molar conductivity and molar ionic conductivities for the following electrolytes : (a) KBr, (b) Na2SO4,(c) AlCl3. |
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Answer» (a) If ∧0 is molar conductivity of an electrolyte at infinite dilution and \(\lambda^0_+\) and \(\lambda^0_-\)are molar ionic conductivities then, ∧0KBr = \(\lambda^0_{K^+}\) + \(\lambda^0_{Br^-}\) (b) \(∧_{0Na_2SO_4}\) = \(2\lambda^0_{Na^+}\) + \(\lambda^0_{SO_4^{2+}}\) (c) \(∧_{0Alcl_3}\) = \(\lambda^0_{Al_{3+}}\)+ \(3\lambda^0_{Cl^-}\) |
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| 952. |
What information is provided by measurement of conductivities of solutions? |
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| 953. |
Which of the following is not correct ? (a) Gibbs energy is an extensive property (b) Electrode potential or cell potential is an intensive property. (c) Electrical work = -ΔG (d) If half reaction is multiplied by a numerical factor, the corresponding E0 value is also multiplied by the same factor. |
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Answer» Correct answer is (d) If half reaction is multiplied by a numerical factor, the corresponding E0 value is also multiplied by the same factor. |
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| 954. |
For the reactionNi(s) + Cu2+ (1 M) → Ni2+ (1 M) + Cu(s), \(E^0_{cell}\)= 0.57 V. Hence ΔG0 of the reaction is(a) 110 kJ (b) -110 kJ (c) 55 kJ (d) -55 kJ |
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Answer» Correct answer is (b) -110 kJ |
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| 955. |
Consider the half reactions with standard potentialsi. \(Ag^+_{(aq)} + e^- \longrightarrow Ag_{(s)} \,E^0 = 0.8\,V\)ii. \(I_{2(s)} + 2e^- \longrightarrow\) \(2I^-_{(aq)}\,E^0 = 0.53\,V\)iii. \(Pb^{2+}_{(aq)} + 2e^- \longrightarrow\) \(Pb_{(s)}\, E^0 = -0.13\, V\)iv. \(Fe^{2+}_{(aq)} + 2e^- \longrightarrow\) \(Fe_{(s)}\, E^0 = -0.44\, V\)The strongest oxidising and reducing agents respectively are(a) Ag and Fe2+ (b) Ag+ and Fe(c) Pb2+ and I-(d) I2 and Fe2+ |
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Answer» Correct answer is (b) Ag+ and Fe |
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| 956. |
For the cell, Pb(s) |Pb2+ (1 M)|| Ag+ (1 M)|Ag(s),if concentration of an ion in the anode compartment is increased by a factor of 10, the emf of the cell will(a) increase by 10 V (b) increase by 0.0296 V (c) decrease by 10 V (d) decrease by 0.0296 V |
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Answer» Correct answer is (d) decrease by 0.0296 V |
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| 957. |
The standard potential of the cell in which the following reaction occursH2+ (g, 1 atm) + Cu2+ (1 M) → 2H (1 M)+ Cu(s), (\(E^0_{Cu} = 0.34\,V\)) is(a) – 0.34 V (b) 0.34 V (c) 0.17 V (d) -0.17 V |
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Answer» Correct answer is (b) 0.34 V |
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| 958. |
1 S m2 mol-1 is equal to(a) 10-4 S m2 mol-1 (b) 104 Ω-1 cm2 mol-1 (c) 10-2 S cm2 mol-1 (d) 102 Ω-1 cm2 mol-1 |
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Answer» Correct answer is (b) 104 Ω-1 cm2 mol-1 |
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| 959. |
The emf `(E^(@))` of the following cels are : `Ag|Ag^(+) (1 M)||Cu^(2+) (1 M)|Cu, E^(@)=-0.46` volt `Zn|Zn^(2+) (1 M)||Cu^(2+) (1 M)|Cu, E^(@)=+1.10` volt Calculate the emf of the cell : `Zn|Zn^(2+) (1 M)||Ag^(+) (1 M)|Ag` |
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Answer» `Zn|Zn^(2+) (1 M)||Ag^(+) (1 M)|Ag` `E_(cell)=E_(o x(Zn//Zn^(2+)))+E_(red (Ag^(+)//Ag))` With the help of the following two cells, the above equation can be obtained : `Ag|Ag^(+) (1 M)||Cu^(2+) (1M)|Cu, E^(@)=-0.46" volt"` or `Cu|Cu^(2+) (1 M)||Ag^(+) (1 M) |Ag, E^(@)" will be "+0.46" volt"` or `+0.46=E_("ox"(Cu//Cu^(2+)))+R_("red "(Ag^(+)//Ag))` ...(i) `Zn|Zn^(2+) (1 M)||Cu^(2+)|Cu, E^(@)=+1.10" volt"` `+1.10=E_(o x (Zn//Zn^(2+)))+E_(red (Cu^(2+)//Cu))` ...(ii) Adding eqs. (i) and (ii), `+1.56=E=E_("red "(Ag^(+)//Ag))+E_(o x (Zn//Zn^(2+)))+E_("red "(Cu^(2+)//Cu))` since, `E_(o x(Cu//Cu^(2+)))=-E_("red "(Cu^(2+)//Cu))` So, `+1.56=E_(o x (Zn//Zn^(2+)))+E_("red "(Ag^(+)//Ag))` Thus, the emf of the following cell is `Zn|Zn^(2+)(1 M)||Ag^(+) (1 M)|Ag` is `+ 1.56` volt |
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| 960. |
The standard oxidation potential potential of zinc is 0.76 volt and of silver is - 0.80 volt. Calculate the emf of the cell : `Zn|underset(0.25 M)(Zn(NO_(3))_(2))||underset(0.1 M)(AgNO_(3))|Ag` at `25^(@)C`. |
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Answer» The cell reaction is `Zn+2Ag^(+) rarr 2Ag+Zn^(2+)` `E_(o x)^(@)` of `Zn =0.76` volt `E_("red")^(@)` of `Ag=0.80` volt `E_(cell)^(@)=E_(o x)^(@)` of `Zn+E_("red")^(@)` of `Ag=0.76+0.80=1.56` volt We know that, `E_(cell)=E_(cell)^(@)-0.0591/n"log" (["Products"])/(["Reactants"])` `=E_(cell)^(@)-0.0591/2"log" 0.25/(0.1xx0.1)` `=1.56-0.0591/2xx1.3979` `=(1.56-0.0413)` volt `=1.5187` volt Alternative method : First of all, the single electrode potentials of both the electrodes are determined on the basis of given concentrations. `E_("ox (Zinc)")=E_(o x)^(@)-0.0591/2log 0.25` `=0.76+0.0177=0.7777` volt `E_("red (Silver)")=E_("red")^(@)+0.0591/1 log 0.1` `=0.80-0.0591` `=0.7409` volt `E_("cell")=E_("ox (Zinc)")+E_("red (Silver)")` `0.7777+0.7409` `=1.5186` volt |
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| 961. |
Calculating the Gibbs Energy change from electrode Potentials: Using standard electrode potentials, clacu-late the standard Gibbs energy change at `25^(@)C` for the following cell reaction. Strategy: To calculate `DeltaG_("cell")^(@)`, we use the equation `DeltaG_("cell")^(@) = -nFE_("cell")^(@)`, where `E_("cell")^(@)` is obtained by using a table of standard potentails and `n` can be inferred from the balanced chemical equation. The cell reaction equals the sum of the half-cell reactions after they have been multiplied by factors so that the electrons cancel in the summation. Note that `n` is the number of electrons involved in each half-reaction. |
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Answer» The half-reactions, corresponding half-cell potentials, and their sums are displayed below `{:(Zn(s)hArrZn^(2+)(aq., 1M)+2e^(-)" "E_(Zn//Zn^(2+))^(@)=0.76V),(2Ag^(+)(aq., 1M)=2e^(-)hArr 2Ag(s)" "E_(Ag^(+)//Ag)^(@) = 0.80 V),(bar(Zn(s)+2Ag^(+)(aq., 1M)hArr Zn^(2+)(aq.,1M)+2Ag(s)E_("cell")^(@)=1.56V)):}` Note that each half-reaction involves two electrons, hence, `n = 2` Also, `E_("cell")^(@) = 1.56 V`, and the Faraday constant, `F`, is `9.65 xx 10^(4) C`. Therefore, the standard Gibbs energy change is `DeltaE_("cell")^(@) = -nFE_("cell")^(@)` `-(2 "mole"^(-))(96,500 C//mol e^(-)) (1.56 V) ((1 J)/(1 C. V))` ` = -3.01 xx 10^(5)J`, Recell that (Coulombs) `xx` (volts) = joules `= -301 kJ (1 kJ = 10^(3)J` Short cut : `F` is about `10^(5)C//mol e^(-)` and `E^(@)` is about `1.5 V`, so `DeltaG^(@) = -nFE^(@)` is approximately `- (2 mol e^(-))(10^(5)C//mol e^(-)(1.5 V) = -3.0 xx 10^(5) J`, or `300 kJ`. |
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| 962. |
Calculating the cell `emf` from free-energy change: Suppose the reaction of zinc ions and chloride ions are formed in aqueous solution. `Zn(s) + Cl_(2)(g) overset(H_(2)O)rarrZn^(2+)(aq.) + 2Cl^(-)(aq.)` Calculate the standard `emf` for this cell at `25^(@)C` from standard free energies of formation. Strategy: Calculate `Delta_(r)G^(@)` and substitute it along with the value of `n` into the equation `Delta_(r)G^(@) = -nfE_("cell")^(@)`. Solve for `E_("cell")^(@)` |
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Answer» Write the equation with `Delta_(r)G^(@)` beneath `underset(0)(Zn(s))+underset(0)(Cl_(2))(g) overset(rarr)(larr) underset(-147)(Zn)^(2+)(aq.)+ underset(2xx(-131)kJ)(2Cl^(-)(aq.))` Hence `Delta_(r)G^(@) = sumv_(p)Delta_(f)G^(@)("products")-sumv_(R)Delta_(f)G^(@)("reactants")` `= [-147 + 2 xx (-131)] kJ - [0 + 0] kJ` `= -409 kJ` `= -4.09 xx 10^(5)J` We ontain `n` by splitting the cell reacton into half-cell reactions `Zn(s)overset(rarr)(larr)Zn^(2+)(aq.)+2e^(-)` `2e^(-)+Cl_(2)(g) overset(rarr)(larr)2Cl^(-)(aq.)` Each half-cell reaction involves two electrons, so `n = 2`. Now we substitute into `Delta_(r)G^(@) = -nfE_("cell")^(@)` `-4.09 xx 10^(5)J = -(2 "mole" e^(-))(9.65 xx 10^(4)J//V.mol e^(-))` Solving for `E_("cell")^(@)`, we get `E_("cell")^(@) = 2.12 V` |
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| 963. |
Resistance of `0.2 M` solution of an electrolyte is `50 ohm`. The specific conductance of the solution is `1.4 S m^(-1)`. The resistance of `0.5 M` solution of the same electrolyte is `280 Omega`. The molar conductivity of `0.5 M` solution of the electrolyte in `S m^(2) mol^(-1)` idA. `5 xx 10^(3)`B. `5 xx 10^(2)`C. `5 xx 10^(-4)`D. `5 xx 10^(-3)` |
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Answer» Correct Answer - C `K = (1)/(R) xx (l)/(A)` `(K_(1))/(K_(2)) = (R_(2))/(R_(1))` `(1.4)/(K_(2)) = (280)/(50)` `K_(2) = (7)/(28) = (1)/(4) Sm^(-1)` `lambda_(m) =(K)/(M xx 1000) = (1)/(4) xx (1)/(0.5 xx 1000)` `= 5 xx 10^(-4) Sm^(2) mol^(-1)` |
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| 964. |
Calculating half-reaction potential: Using the data in Table 3.1, find the standard potential of the half-reaction for the reduction of iron from the `+3` to the `0` oxidation state. Strategy: Write the desired half-cell reaction: `Fe^(3+)(aq.)+3e^(-)hArrFe(s)` Inspect the table for half-reactions that include these oxidantion states of iron. |
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Answer» The required half reactions are `{:(Fe^(3+)(aq.)+e^(-)hArr Fe^(2+)(aq.)" "E^(@)=0.77V),(Fe^(2+)(aq.)+2e^(-)hArr Fe(s)" "E^(@)=-0.44V):}` We see that the two half-reactions should be added to product the desired half-reaction. The `E^(@)` of the new half-reaction can be calculated by using Equation (3.13): `E_(T)^(@) = (n_1E_1^(@)+n_2E_2^(@))/(n_(T))` `= ((1)(0.77 V) + (2)(-0.044 V))/(3)` `= -0.036` |
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| 965. |
Calculating cell constant: Conductivity of a decimolar solution of potassium chloride at `180^(@)C` is `1.12 Sm^(-1)`. The resistance of a conductivity cell containing the solution at `180^(@)C` was found to be `55 ohm`. What is the cell constant. Strategy: Use Equation (3.18) and solve for cell constant. |
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Answer» Conductance `(G) = 1//R = (1)/(55) S` Conductance `(K) = 1.12 Sm^(-1)` Substituting these results into Equation (3.18) solving for cell constant, we have cell constant `(l//A) = (K)/(G)` `= (1.22 Sm^(-1))/(1//55)S` `61.6 m^(-1)` `= 0.616 cm^(-1)` |
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| 966. |
Resistance of `0.2 M` solution of an electrolyte is `50 ohm`. The specific conductance of the solution is `1.4 S m^(-1)`. The resistance of `0.5 M` solution of the same electrolyte is `280 Omega`. The molar conductivity of `0.5 M` solution of the electrolyte in `S m^(2) mol^(-1)` idA. `5xx10^(2)`B. `5xx10^(-4)`C. `5xx10^(-3)`D. `5xx10^(3)` |
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Answer» Correct Answer - B (b) In first case : `k=(1)/(R )xx(1)/(a)` `(l)/(a)=kxxR=(1.40" ohm"^(-1)m^(-1))xx(50" ohm")` `=70 m^(-1)` In second case : `k=(1)/(R )xx(l)/(a)` `=(1)/((282" ohm"))xx(70 m^(-1))=0.25" ohm"^(-1)m^(-1)` `^^_(m)=(k)/(Mx1000)` `C=0.5 M =(0.5 mol)/(1L)=(0.5 mol)/(m^(3))=0.5 mol m^(-3)` `^^_(m)=((0.25" ohm"^(-1)m^(-1)))/((10^(3)xx0.5 mol m^(-3)))=5xx10^(-4)" ohm"^(-1) m^(2) mol^(-1)` |
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| 967. |
Calculating conductivity and molar conductivity: Resistance of a conductivity cell filled with `0.1 M KCl` solution is `100 Omega`. If the resistance of the same cell when filled with `0.02 M KCl` solutions `520 Omega` and the conductivity of `0.1 KCl` solution is `1.29 S m`, calculate the conductivity and moalr conductivity of `0.02 M KCl` solution. Strategy : Calculate the cell constant with the help of `0.01 M KCl` solution (both `R` and `kappa` are Known). Use the cell constant to determine the conductivity of `0.02 M KCl` solution and finally find its molar conductivity using the molarity |
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Answer» Consider `0.01 M KCl` solution Cell constant `(G^(**))` = conductivity `xx` resistance `= (1.29 S m^(-1)) xx (100 Omega)` `= 1.29 Sm^(-1)` Now consider `0.02 MKCl` solution : Conductivity `(kappa)` = Cell constant `(G^(**))`/resistance `(R)` `= 129 m^(-1)//520Omega` `= 0.248 S m^(-1)` Molar concentration `= 0.02 mol L^(-1)` `= ((1000 L)/(m^(3))) xx (0.02 mol L^(-1))` `= 20 mol m^(-3)` Therefore, molar conductivity is given as `Lambda_(m) = (kappa)/(C)` `= (0.248 S m^(-1))/(20 mol m^(-3))` `= (248 xx 10^(-3) S m^(-1))/(2 xx 10^(1) mol m^(-3))` `= 124 xx 10^(-4) s m^(2)mol^(-1)` |
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| 968. |
The equivalent conctance at infinite dilution of the salt MX is 160.84 `ohm^(-1)cm^(2)eq^(-1)`. If the transport number of `M^(+)` is 0.40, calculate the ionic mobility of the ion. |
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Answer» Ionic conductance of ion=Transport no. of that ion`xx(AA)_(eq)^(@)` of strong electrolyte containing that ion `=(1-0.40)xx160.84=96.504` Ionic mobility`=("Ionic conductance")/(96,500)=(96.504)/(96,500)=10^(-3)`. |
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| 969. |
What current is to be passed for 0.25 s for deposition of a certain weight of metal which is equal to its electrochemical equivalent?A. `4A`B. `100A`C. `200A`D. `2A` |
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Answer» Correct Answer - A Electrochemical equivalent is the weight deposited by 1 coulomb. `Q=Ixxttherefore1=Ixx0.25` or `I=4A`. |
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| 970. |
STATEMENT-1: If `((dE_(cell))/(dT))_Pgt0,` For a cell reaction then `/_` S is positive. STATEMENT-2: `/_SnFT((dE)/(dT))_p`A. If both the statements are TRUE and STATEMENTS-2 is the correct explantion of STATEMENTS-11B. If both the statements are TRUE but STATEMENTS-2 is NOT the correct explanation of STATEMENTS-13C. If STATEMENTS-1 is TRUE and STATEMENTS-2 is FALSED. If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE |
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Answer» Correct Answer - C |
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| 971. |
Electrolysis of dilute aqueous `NaCl` solution was carried out by passing `10mA` current. The time required to liberate `0.01mol` of `H_(2)` gas at the cathode is `(1F=96500C mol ^(-1))`A. `9.65 xx 10^(4)`secB. `19.3 xx 10^(4)` secC. `28.95 xx 10^(4)`secD. `38.6 xx 10^(4)`sec |
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Answer» Correct Answer - B No. of Faraday of electricity ot liberate `0.01` Mole `= 0.01 xx 2F = 0.02F` `=2 xx 10^(-2) xx 9600 = 1930` coulomb `A = 10` miliampere `= 0.01` ampere `Q = 1 xx t` `t = (Q)/(I) = (1930)/(0.01) = 1.93 xx 10^(5) = 19.3 xx 10^(4)` sec |
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| 972. |
The compound exhibiting maximum conductance in a fused state isA. `SrCl_(2)`B. `CaCl_(2)`C. `MgCl_(2)`D. `BeCl_(2)`. |
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Answer» Correct Answer - D In molten state, smaller the cation more is the conductance. |
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| 973. |
If a cell reaction is spontaneous, then :A. `E_(cell)^(@)` in negativeB. `E_(cell)^(@)` is positiveC. `DeltaG` is negativeD. `DeltaG` is positive |
| Answer» Correct Answer - B::C | |
| 974. |
Which of the following statement(s) is (are) incorrect?A. Reduction occurs at the cathode in both galvanic and electrolytic cellsB. Oxidation takes place at the cathode in both galvanic and electrolytic cellsC. The anode is the negative terminal and the cathode is the positive terminal in a galvanic cellD. The anode is the negative terminal and the cathode is the positive terminal for an electrolytic cell |
| Answer» Correct Answer - B::D | |
| 975. |
In the electrolysis of which solution `OH^(-)` ions are discharged in preference to `Cl^(-)` ions?A. Dilute NaClB. Very dilute NaClC. Fused NaClD. Solid NaCl. |
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Answer» Correct Answer - B In very dilute solution of NaCl, the following reaction takes place on electrolysis. Anode: `2OH^(-)rarrH_(2)O+1//2O_(2)+2e^(-)` Cathode: `2H^(+)+2e^(-)rarrH_(2)` |
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| 976. |
In the electrolysis of which solution, `OH^(-)` ion are discharged in preference to `Cl^(-)` ions?A. Dilute NaClB. very dilute NaClC. fused NaClD. solid NaCl |
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Answer» Correct Answer - B In very dilute solution of `NaCl`, the following reactions take place on electrolysis: Anode: `2OH^(-)toH_(2)O+(1)/(2)O_(2)+2e^(-)` Cathode: `2H^(+)+2e^(-)toH_(2)` |
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| 977. |
Suppose the equilibrium constant for the reaction, `3M^(3+) rarr 2M^(2+)(aq) +M^(5+) (aq)` |
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Answer» Correct Answer - `010` In aqueous medium is `x x 10^(y)`. Find the value of y from given information. `E_(M^(5+)//M^(2+))^(@) = 0.6V` `E_(M^(3+)//M^(2+))^(@) = 0.8V` `(2.303RT)/(F) = 0.06V` |
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| 978. |
At 298K the standard free energy of formation of `H_(2)O(l)` is `-256.5 kJ..mol`, while that of its ionisation to `H^(+)` & `OH^(-)` is `80 kJ//mol`. What will be emf at 298 K of the cell. `H_(2)(g,1 bar) |H^(+)(1M) || O_(2)(g,1bar)` Fill your answer by multiplying it with 10. |
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Answer» Correct Answer - `005` `H_(2) +H_(2)O +(1)/(2)O_(2) rarr 2H^(+) +2OH^(-)` `DeltaG^(@) =- 256.5 +2 xx 80 = -96.5 kJ` `-DeltaG^(@) = nFE^(@)` `+96.5 xx 1000 = 2 xx 96500 xx E^(@)` `E^(@) = 0.5` volt |
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| 979. |
An equeous solution of `Na_(2)SO_(4)` was electrolysed for 10 min. 82 ml of a gas was produced at anode and collected over water at `27^(@)C` at a total pressure of 580 torr. Determine the current that was used in amp. Given: Vapour pressure of `H_(2)O` at `27^(@)C = 10` torr `R = 0.082 atm L//mol-K` Write your answer exclusing decimal places. |
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Answer» Correct Answer - 1 An anode reaction will be `2H_(2)O rarr O_(2) +4H^(+) +4e^(-)` `Vo_(2)` collected `= 82 ml` By `PV = nRT ((580 -10))/(760) xx (82)/(1000) = (n_(o_(2))) xx 0.0821 xx 300` `(n_(o_(2)) =(1)/(400))` By Faraday law `(W)/(E) = (i xx t)/(96500)` `((W)/(M)) xx n = (ixxt)/(96500) ((1)/(400)xx4) = (ixx 10 xx 60)/(96500)` `i= 1.6` |
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| 980. |
Same amount of electric current is passed through solutions of `AgNO_(3)` and HCl. If 1.08 g of silver is obtained in the first case, the amount of hydrogen liberated at S.T .P. in the second case is:A. `112cm^(3)`B. `22400cm^(3)`C. `224cm^(3)`D. `1.008g`. |
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Answer» Correct Answer - A `("Mass of Ag")/("Mass of " H_(2))=("Eq. mass of Ag")/("Eq. mass of " H_(2))` `(1.08)/("mass of" H_(2))=(108)/(1)` Mass of `H_(2)=( 1.08)/(108)=10^(-2)g` 2g of `H_(2)` at S.T.P. occupy 22400 `cm^(3)` `therefore10^(-2)g` of `H_(2)` at STP occupy `=(22 400)/(2)xx10^(-2)10^(-2)cm^(3)=112cm^(3)` |
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| 981. |
A solution contains `A^(+)` and `B^(+)` in such a concentration that both deposit simultaneously. If current of `9.65` amp was passed through `100ml` solution for 55 seconds then find the final concentration of `A^(+)` ion if initial concentration of `B^(+)` is `0.1M`. [Fill your answer by multiplying it wity `10^(3)]`. Given: `{:(A^(+)+e^(-)rarrA,,E^(@) =- 0.5 "volt"),(B^(+)+e^(-)rarrB,,E^(@) =- 0.56 "volt"),((2.303RT)/(F) =0.06,,):}` |
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Answer» Correct Answer - 5 `-0.5 -(0.06)/(1)log.(1)/([A^(+)]) =- 0.56 -(0.06)/(1)log.(1)/([B^(+)])` `0.06 = (0.06)/(1)log.([B^(+)])/([A^(+)])` `([B^(+)])/([A^(+)]) = 10` `[A^(+)]_("final") = 0.01` Total equivalent of charge `= (9.65 xx 55)/(96500) = 5.5` meq Final `[A^(+)] = (0.5)/(100) = 0.005` |
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| 982. |
A solution of sodium chloride in water is electrolysed using inert electrodes . The solution is formed in vessel isA. `NaOH`B. `H_(2)O`C. `NaCl`D. `HCl` |
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Answer» Correct Answer - A `Na^(+) + OH^(-) to NaOH` |
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| 983. |
Time required to deposit one milli"mole" of aluminium metal by the passage of 9.65 amp through aqueous solution of aluminium ion is:A. 30sB. 10sC. 30,000sD. 10,000s. |
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Answer» Correct Answer - A 1 mol of Al requires =3xx96500C `10^(-3)` mol of Al requires `=3xx96500xx10^(-3)C` `=3xx96.5C` `therefore` Time (s)`=(3xx96.5As)/(9.65A)=30s` |
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| 984. |
When an electrolytic solution conducts electricity , current is carried out byA. electrons B. cations and anionsC. neutral atomsD. none |
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Answer» Correct Answer - B Passage of current in electrolytic solution is due to migration of ions towards opposite electrodes . |
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| 985. |
Which one of the following material conducts electricity?A. diamondB. crystalline sodium chlorideC. barium sulphateD. fused potassium chloride |
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Answer» Correct Answer - D Out of the given five choices only fisued KCl conducts electricity. |
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| 986. |
Which is correct for cell reaction?A. `Zn+2Ag^(+)rarrZn^(2+)+2Ag`B. `2Ag+Zn^(2+)rarr2Ag^(+)+Zn`C. BothD. None |
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Answer» Correct Answer - A R.P. of `AggtR.P`. Of Zn `therefore Zn` must oxidine and `Ag^(+)` will be reduced. |
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| 987. |
During electrolysis of NaOHA. `H_(2)` is liberarted at cathodeB. `O_(2)` is liberated at cathodeC. `H_(2)` is liberated at anodeD. `O_(2)` is liberated at anode. |
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Answer» Correct Answer - D During electrolysis of NaOH (molten), `O_(2)` is liberated at anode. `4OH^(-)rarr2H_(2)O+O_(2)+4e^(-)` |
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| 988. |
Which of the following (1M) conducts more electricity?A. sulphuric acidB. boric acidC. nitric acidD. aluminium. |
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Answer» Correct Answer - A More the number of ions especially smaller cations (hydrated `H^(+)` ions) more the conductivity At 1 M concentration weak acids like `H_(3)PO_(3),H_(3)BO_(3)` will be weakly dissociated resulting in poor conductivity of `HNO_(3)` and `H_(2)SO_(4)` both strong acis `H_(2)SO_(4)` solution will be more conducting due to larger number of `H^(+)` ions in solution. |
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| 989. |
Which of the following conducts electricity ?A. Fused NaClB. `CO_2`C. `Br_2`D. `Si` |
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Answer» Correct Answer - A `NaClunderset(harr)(rarr) Na^(+) +Cl^(-)`. So. It conducts electricity . |
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| 990. |
One faraday is equal toA. 96.5 coulomb `"equivalent"^(-1)`B. `96.5 xx 10^(3)` coulomb `"equivalent"^(-1)`C. `96.5 xx 10^(10)` coulomb `"mol"^(-1)`D. `96.5 xx 10^(23)` coulomb `"mol"^(-1)` |
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Answer» Correct Answer - B 1 F = N `xx` e , where N is Av. No and e is charge on electron . |
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| 991. |
`H_2`(g) and `O_2`(g), can be produced by the electrolysis of water. What total volume (in L) of `O_2` and `H_2` are produced at 1 atm and 273K when a current of 30 A is passed through a `K_2SO_4` (aq) solution for 193 min?A. 20.16B. 40.32C. 60.48D. 80.64 |
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Answer» Correct Answer - C |
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| 992. |
Galvanic cells are classified into primary and secondary cells.(a) Write any two differences between primary cell and secondary cell.(b) (i) What is a fuel cell?(ii) Write the overall cell reaction in H2 – O2 fuel cell. |
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Answer» (a) Primary cell Cell reaction cannot be reversed and hence can not be recharged, cannot be reused again. e.g. Dry cell, Mercury cell Secondary cell Cell reaction can be reversed and hence can be recharged, can be resued again. e.g. Lead storage battery, nickel-cadmium cell (b) (i) Fuel cell is a galvanic cell that is designed to convert the energy of combustion of fuels directly into electncal energy. (ii) 2H2(g) + O2(g) → 2H2O(l) |
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| 993. |
For gold plating, the electrolyte used isA. `AuCl_(3)`B. `HAuCl_(4)`C. `K[Au(CN)_(2)]`D. None of these |
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Answer» Correct Answer - C For gold plating, the used electrolyte is `K[Au(CN)_(2)]`. |
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| 994. |
The equivalent constant of the reaction: `Cu(s)+2Ag^(+)(aq.) rarr Cu^(2+)(aq.)+2Ag(s)` `E^(@)=0.46 V` at `298 K`,is:A. `2.0 xx 10^(10)`B. `4.0xx10^(10)`C. `4.0xx10^(15)`D. `2.4xx10^(10)` |
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Answer» Correct Answer - C K= antilog `[(nE^(@))/0.059]=` antilog `[(2xx0.46)/0.059]` = antilog 15.593 `=3.9xx10^(15)` `~~ 4xx10^(15)` |
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| 995. |
The electrolytic bath used in gold plating of copper articles containsA. molten goldB. `CuSO_(4)`C. `AuCl_(3)`D. `AuCl_(3) + NaCN` |
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Answer» Correct Answer - D The complex formation enhances the electrolytic deposition of Au |
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| 996. |
Which of the following statements is true for fuel cells ?A. They ar more efficientB. They are fee from pollutionC. The run till reactants are activeD. All of these |
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Answer» Correct Answer - D Fuel cells are more efficient , free from pollution and they run till reactants are active. |
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| 997. |
Consider the cell reaction `:` `Mg(s)+Cu^(2+)(aq) rarr Cu(s) +Mg^(2+)(aq)` If `E^(c-)._(Mg^(2+)|Mg(s))` and `E^(c-)._(Cu^(2+)|Cu(s))` are `-2.37` and `0.34V`, respectively. `E^(c-)._(cell)` isA. 2.03 VB. `-2.03 V`C. `+ 2.71 V`D. `-2.71 V` |
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Answer» Correct Answer - C `E_(cell)^(@)=E_("Cathode")^(@)-E_("Anode")^(@)` `=E_("Reduced species")^(@)-E_("Oxidised species")^(@)` `=0.34-(-2.37)=+2.71 V` |
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| 998. |
Consider the cell reaction `:` `Mg(s)+Cu^(2+)(aq) rarr Cu(s) +Mg^(2+)(aq)` If `E^(c-)._(Mg^(2+)|Mg(s))` and `E^(c-)._(Cu^(2+)|Cu(s))` are `-2.37` and `0.34V`, respectively. `E^(c-)._(cell)` isA. ` 2.03 V`B. ` -20.3 V`C. ` +2. 71 V`D. ` -2 . 71 V` |
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Answer» Correct Answer - C ` E_(cell)^@ -E_("cathode")^@ -E_("anode")^@` ` =0.34 - (-2.37) = + 271 V`. |
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| 999. |
The standard reduction potentials of 4 elements are given below . Which of the following will be the most suitable oxidising agent ? ` {:( I = -3.04 V "," ,, II = -1.90 V) , (III = 0 V "," ,, IV = 1.98 V):}`A. IB. IIC. IIID. IV |
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Answer» Correct Answer - D more is the reduction potential , stronger will be the oxidising agent . |
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| 1000. |
Consider the cell reaction `:` `Mg(s)+Cu^(2+)(aq) rarr Cu(s) +Mg^(2+)(aq)` If `E^(c-)._(Mg^(2+)|Mg(s))` and `E^(c-)._(Cu^(2+)|Cu(s))` are `-2.37` and `0.34V`, respectively. `E^(c-)._(cell)` isA. `-2.71V`B. `2.71V`C. `-2.03V`D. `2.03V` |
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Answer» Correct Answer - B Cell reaction is `Mg(s)+Cu^(2)+(aq)rarrCu(s)+Mg^(2+)(aq)` Half reaction at LHE , i.e., oxidation `MghArrMg^(2+)+2e^(-)` Half reaction at LHE, i.e., Reduction `Cu^(2+)(aq)+2e^(-)hArrCu(s)` Now, `E_(LHE)^(@)=E_(Mg^(2+//Mg))^(@)=-237V` `E_(RHE)^(@)=E_(Cu^(2+)//Cu)^(@)=+0.34V` `therefore E_( "cell")^(@)=E_(RHE)^(@)-E_(LHE)^(@)` `=0.34-(-2.37)=2.71V` |
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