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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.
| 2451. |
The e.m.f. of the cell `Zn|Zn^(2+)(0.01M)||Fe^(2+)(0.001M)|Fe` at 298K is 0.2905 then the value of equilibrium for the cell reaction isA. `e^((0.32)/(0.0295))`B. `10^((0.32)/(0.0295))`C. `10^((0.26)/(0.0295))`D. `10^((0.32)/(0.0591))` |
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Answer» Correct Answer - B For this cell, reaction is:` Zn+Fe^(2+)toZn^(2+)+Fe` `E=E^(o)-(0.0591)/(n)"log"(c_(1))/(c_(2)),E^(o)+(0.0591)/(n)"log"(c_(1))/(c_(2)) ` `=0.2905+(0.0591)/(2)"log"(10^(-2))/(10^(-3))=0.32V`. `E^(o)=(0.0591)/(2)log" "K_(c),log" "K_(c)=(0.32xx2)/(0.0591)=(0.32)/(0.0295)` `thereforeK_(c)=10^((0.32)/(0.295))`. |
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| 2452. |
`Cu^(+)` ion is not stable in aqueous solution because of disproportionation reaction. `E^(@)` value for disproportionation of `Cu^(+)` is (Given `E_(Cu^(2+)//Cu^(+))^(@)=0.15,E_(Cr^(2+)//Cu)^(@)=0.34V`)A. `-0.49V`B. `0.49V`C. `-0.38V`D. `0.38V` |
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Answer» Correct Answer - D The required reaction `(Cu^(++)+Cuto2Cu^(+))` can be obtained by using the following reactions. `Cu^(++)+e^(-)toCu^(+),E_(Cu^(++)//Cu^(+))^(o)=0.15V` . . .(i) `Cu^(++)+2e^(-)toCu,E_(Cu^(++)//Cu)^(o)=0.34V` . . (ii) Multiplying eq. (i) by 2 we get `2Cu^(++)+2e^(-)to2Cu^(+)` . . . (iii) `DeltaG_(1)=-nFE=-2xxFxx0.15` `Cu^(++)+2e^(-)toCu` . . . (iv) Subtract the eq. (iv) from (iii) `Cu^(++)+Cuto2Cu^(+)` `DeltaG_(3)=-nFE=-1xxFxxE^(o)` Also `DeltaG_(3)=DeltaG_(1)-DeltaG_(2)` `-1FE^(o)=(-2Fxx0.15)-(-2Fxx0.34)` `E^(o)=-0.38` this is the value of the reaction `Cu^(++)+Cuto2Cu^(+)` but the given reaction is just reverse of it `thereforeE_(cell)` for given reaction =+0.38V. |
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| 2453. |
The variation in `Lambda_m` with concentration for a strong electrolyte can be represented by the equation, `Lambda_m=Lambda_m^@ -AC^(1//2)` The value of constant A for a given solvent and temperature depends upon the type of electrolyte i.e., cations and anions produced on dissociation of electrolyte in the solution . Which of the following statements is correct regarding variations of molar conductivity with concentration.A. Molar conductivity decrease with decrease in concentrationB. Variation is molar conductivity of weak and strong electrolytes is same.C. Molar conductivity increases with decrease in concentration.D. When concentration of the solution approaches zero, the molar conductivity is known as conductance. |
| Answer» Correct Answer - C | |
| 2454. |
In an electrolytic cell, the flow of electrons is formA. cathode to anode in solutionB. cathode to anode through external supplyC. cathode to anode through internal supply.D. anode to cathode through internal supply. |
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Answer» Correct Answer - C (c ) is the correct answer. |
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| 2455. |
In the electrolytic cell, flow of electrons is from:A. cathode to anode in solutionB. cathode to anode through external supplyC. cathode to anode through internal supplyD. anode to cathode through internal supply |
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Answer» Correct Answer - C Current flows from anode to cathode in electrolytic cell. Thus electrons flow from cathode to anode through internal supply. |
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| 2456. |
The variation in `Lambda_m` with concentration for a strong electrolyte can be represented by the equation, `Lambda_m=Lambda_m^@ -AC^(1//2)` The value of constant A for a given solvent and temperature depends upon the type of electrolyte i.e., cations and anions produced on dissociation of electrolyte in the solution . NaCl, `MgCl_2` and `CaSO_4` are known asA. 1-1,2-1, and 2-2 type electrolytes respectivelyB. strong, weak and strong electrolytes respectivelyC. electrolytes with different value of AD. electrolytes with same molar conductivity . |
| Answer» Correct Answer - A | |
| 2457. |
In an electrolytic cell, the flow of electrons is formA. from cathode to anode in the solutionB. from cathode to anode through external supplyC. from cathode to anode through internal supplyD. from anode to cathode through internal supply |
| Answer» Correct Answer - C | |
| 2458. |
Match the column I with column II and mark the appropriate choice. A. A - (i), B-(iii), C-(ii) ,D-(iv)B. A-(iii) , B-(i), C-(iv),D-(ii)C. A-(ii),B-(iv), C-(iii), D-(i)D. A-(iv), B-(ii),C-(i), D-(iii) |
| Answer» Correct Answer - B | |
| 2459. |
Match the column I with column II and mark the appropriate choice. A. A-(i),B-(ii),C-(iii),D-(iv)B. A-(ii),B-(iii),C-(iv),D-(i)C. A-(iii),B-(iv),C-(i),D-(ii)D. A-(iv),B-(i),C-(ii),D-(iii) |
| Answer» Correct Answer - B | |
| 2460. |
Electrolytic conductance `……………………….` with increase of temperature while metallic conductance `………………………….` with increase of temperature. |
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Answer» Correct Answer - Increase,decrease Increase, decrease |
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| 2461. |
Which of the following is not a source of electrical energy ?A. Lead storage batteriesB. A generator setC. Fuel cellsD. Ni-Cd cells . |
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Answer» Correct Answer - C Appolo moon flights used fuel cells. |
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| 2462. |
Choose the option with correct words to fill in the blanks. According to preferential discharge vtheory, out of number of ions the one which requires ____ energy will be liberted ____ at a given electrode.A. least, firstB. least, lastC. highest, firstD. highest, last |
| Answer» Correct Answer - A | |
| 2463. |
The cell potential of dry cell when it is discharged will be :A. Carbon dissolvesB. `NH_(3)` evolvedC. `MnO_(2)` decomposed to MnD. ZnO converted into Zn |
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Answer» Correct Answer - B In dry cell , the net reaction is `Zn + 2 NH_(4)^(+) + 2 MnO_(2) to Zn^(2+) + Mn_(2)O_(3) + 2 NH_(3) + H_(2)O` |
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| 2464. |
According to the preferential discharge theory, out of a number of ions, the one which requires`…………………..` energy will be liberated `………………………..` at a given substance. |
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Answer» Correct Answer - Leaset,first Least, first |
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| 2465. |
During discharge of Dry cell , the gas liberated at the cathode is :A. `H_(2)`B. `O_(2)`C. `Cl_(2)`D. `N_(2)` |
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Answer» Correct Answer - A In dry cell , reduction take place at C cathode as - ` 2NH_(4)^(+) + 2e^(-) to 2 NH_(3) + H_(2)` |
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| 2466. |
Statement : In case of `H^(+)` and `Na^(+)` present in a solution discharge of `H^(+)` is preferred at cathode. Explanation :The higher is discharge potential of ion, lesser is its tendency to get discharged.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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| 2467. |
Statement : In concentration cell neither electronation occurs at cathode nor e-electronation at anode. Explanation : The electrical energy is produced due to decrease in frree energy during the transfer of concentration from high to low region.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 - B In concentration cells no doubt oxidation occurs at anode and reduction at cathode but net redox change is zero. |
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| 2468. |
The electrochemical cell shown below is a concentration cell `M//M^(2+)` (saturated solution of a sparingly soluble salt, `MX_(2))||M^(2+)(0.001 mol dm^(-3))|M` The emf of the cell depends on the difference in concentrations of `Mn^(2+)` ions at the two electrodes. The emf of the cell at `298 K` is `0.059 V`. The solublity product `(K_(SP) , "mol"^(3)"dm"^(-9))` of `MX_(2)` at `298 K` based on the information available for the given concentration cell is : (take `2.303 xx R xx 298//F = 0.059V`)A. `1 xx 10^(-15)`B. `4 xx 10^(-15)`C. `1 xx 10^(-12)`D. `4 xx 10^(-12)` |
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Answer» Correct Answer - B For concentration cell, `E_(cell) = (0.059)/(n)"log"(C_(2(RHS)))/(C_(1(LHS)))` `E_(cell) = 0.059V, C_(2(RHS)) = 0.001` `0.059 = (0.0591)/(2)"log"(0.001)/(C_(1))` or `(2 xx 0.59)/(0.0591) = "log"(0.001)/(C_(1))` or antilog `2 = (0.001)/(C_(1))` `:. C_(1) = (0.001)/(100) = 10^(-5)` `C_(1) =` concentration or solubility of `M^(2+) = 10^(-5)` `MX_(2) hArr underset(S)(M^(2+))+underset(2S)(2X^(-))` `K_(SP) = S(2S)^(2) = 4S^(3)` `K_(SP) = 4 xx (10^(-5))^(3) = 4 xx 10^(-15) "mol"^(3) "dm"^(-9)` |
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| 2469. |
What happens Δ0m for weak electrolytes obtained by using Kohlrausch law if the migration of ions is increased to three. |
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Answer» Δ0m for weak electrolytes obtained by using Kohlrausch law is independent of the migration of ions. |
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| 2470. |
Why is the rusting of iron faster in saline water than in pure water? |
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Answer» Strong electrolytes such as sodium chloride are present in saline water. The ions produced from NaCl help in the reduction of oxygen to form water. Hence, the rusting of iron is faster in saline water than in pure water. |
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| 2471. |
Statement : Electrolysis involves electronaltion or de-electronation as a result of passage of current. Explanation : The species undergoes electronation at anode and show show de-electronation at cathode.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 - A Electronation (reduction) occurs at cathode and de-electronation (oxidation) occurs at anode. |
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| 2472. |
Statement : Very pure form of iron does not show rusting. Explanation : Rusting is catalysed by impurities present in iron and `H^(+)` ions.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 - D Both are facts and true. |
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| 2473. |
If 9 gm `H_(2)O` is electrolysed completely with the currect of 50 % efficiency then :A. 96500 charge is requiredB. `2 xx 96500` C charge is requiredC. 5.6 L of `O_(2)` at STP will be formedD. 11.2 L of `O_(2)` at STP will be formed |
| Answer» Correct Answer - B::C | |
| 2474. |
Calculate the equilibrium constant for the reaction `Cu(s)+2Ag^(+)(aq)rarrCu^(2+)(aq)+2Ag(s), E_(cell)^(@)=0.46 V`.A. `579jK^(-1)`B. `386jK^(-1)`C. `193jK^(-1)`D. None of these. |
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Answer» Correct Answer - B `DeltaS^(@)=nF((dE^(@))/(dT))_(p)` `=2xx96500((0.48-0.46)/(310-300))` `=(2xx96500xx0.02)/(10)=386JK^(-1)` |
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| 2475. |
Calculate the equilibrium constant for the reaction `Cu(s)+2Ag^(+)(aq)rarrCu^(2+)(aq)+2Ag(s), E_(cell)^(@)=0.46 V`.A. `4.0 xx 10^(10)`B. `4.0 xx 10^(15)`C. `2.4 xx 10^(10)`D. `2.0 xx 10^(10)` |
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Answer» Correct Answer - B According to Nernst equation `E_("cell")^(@) = (2.303 RT)/(nF) log K_(eq.)` or `log K_(eq.) = (nFE_("cell")^(@))/(2.303 RT)` `= ((2 mol)(96500 C mol^(-1))(0.46 V))/((2.303)(8.314 JK^(-1))(298 K))` `= 15.56` Thus `K_(eq.) = 4.0 xx 10^(15)` |
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| 2476. |
Which of the following solutions has larger molar conductance ? (a) 0.08 M solution having conductivity equal to `2.0xx10^(-2)ohm^(-1)cm^(-1)`. (b) 0.10 M solution having resitivity equal to 58 ohm cm. |
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Answer» Correct Answer - solution (a) (a)`" " Lambda_(M)=(1000xxk)/(M)=((1000" cm"^(3))xx(2.0xx10^(-2)" ohm"^(-1)cm^(-1)))/((0.08" mol"))=250" ohm"^(-1)cm^(2)"mol"^(-1)`. (b)`" " Lambda_(M)=(1000xxk)/(M)=(1000)/(pxxM)=((1000cm^(3)))/((58" ohm cm")xx(0.1 "mol"))=172.41" ohm"^(-1)cm^(2)"mol"^(-1)`. Solution (a) has larger molar conductance than solution (b). |
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| 2477. |
Which of the following is incorrect in a galvanic cell ?A. Oxidation occurs at anodeB. Reduction occurs at cathodeC. The electrode at which electrons are gained is called cathodeD. The electrode at which electrons are lost is called cathode |
| Answer» Correct Answer - D | |
| 2478. |
The specific conductance of a solution is 0.2 `ohm^(-1)" "cm^(-1)` and conductivity is 0.04 `ohm^(-1)`. The cell constant would beA. 1 `cm^(-1)`B. 0 `cm^(-1)`C. 5 `cm^(-1)`D. 0.2 `cm^(-1)` |
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Answer» Correct Answer - C `K=Cxx`cell constant`=(K)/(C)=(0.2)/(0.04)=5cm^(-1)`. |
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| 2479. |
Which colourlss gas evolves, when `NH_(4)Cl` reacts with zinc in a dry cell batteryA. `NH_(4)`B. `N_(2)`C. `H_(2)`D. `Cl_(2)` |
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Answer» Correct Answer - C `2NH_(4)Cl+Znto2NH_(3)+ZnCl_(2)+H_(2)uarr`. |
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| 2480. |
Which of the following statement about galvanic cell is incorrectA. Anode is positiveB. Oxidation occurs at the electrode with lower reduction potentialC. Cathode is positiveD. Reduction occurs at cathode |
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Answer» Correct Answer - A Anode has negative polarity. |
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| 2481. |
The relationship between standard reduction potential of a cell and equilibrium constant is shown byA. `E_(cell)^(o)=(n)/(0.059)log" "K_(c)`B. `E_(cell)^(o)=(0.059)/(n)log" "K_(c)`C. `E_(Cell)^(o)=0.059n" log "K_(c)`D. `E_(cell)^(o)=(log" "K_(c))/(n)` |
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Answer» Correct Answer - B `E_(cell)^(@)=(2.303RT)/(nF)"log K"=(0.0591)/(n)log" "K_(C)` at 298K . |
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| 2482. |
The relationship between standard reduction potential of a cell and equilibrium constant is shown byA. `E_("cell")^(@)=(n)/(0.059)"log"k_(c)`B. `E_("cell")^(@)=(0.059)/(n)"log"k_(c)`C. `E_("cell")^(@)=0.059n "log"k_(c)`D. `E_("cell")^(@)=("log"k_(c))/(n)` |
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Answer» Correct Answer - B `E_("cell")^(@)=(0.0591)/(n)"log" k_(c)`. |
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| 2483. |
On electrolysis, which of the following does not give out oxygen ?A. Acidic water using Pt electrodeB. Fused NaOH using Pt electrodeC. Dilute `H_(2)SO_(4)` using Pt electrodeD. Dilute `H_(2)SO_(4)` using Cu electrode |
| Answer» Correct Answer - D | |
| 2484. |
A smuggler could not carry gold by chemicaly depositing iron on the gold surface sinceA. gold is denserB. iron rustsC. gold has higher reduction potential than ironD. gold has lower reduction potential thatn iron. |
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Answer» Correct Answer - C Gold has higher electrode potential than iron. |
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| 2485. |
A smuggler could not carry gold by depositing iron on the gold surface since :A. gold is denserB. iron rustsC. gold has higher reduction potential than ironD. gold has lower reduction potential than iron |
| Answer» Correct Answer - C | |
| 2486. |
The hydrogen electrode is prepared electrochemically by depositingA. Platinum on copper metalB. Silver on platinum metalC. Platinum on platinum metalD. Palladium on palladium metal |
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Answer» Correct Answer - C The normal hydrogen electrode or standard hydrogen electrode, which is used as a primary reference, consists of a piece of platinum foil, which is cpateed electrolytically with a very finely-devided platinum known as platinum black to give it a large surface area, suspended in a molar solution of hydrogen ions. Pure hydrogen gas at `1` bar pressure is continually bubbed into the solution to keep it saturated and to expose the soil alternately to gas and solution. The platinum act as an electrical conductor and also facilitates the attainment of equilibrium between the gas and its ions in solution. In practice a 1.18 `M` solution of hydrochloric acid is used so that the activity (i.e. the effectivity concentration) of the hydrogen ions in solution is exactly unity. |
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| 2487. |
The emf of a cell is the ratio of the difference inA. charge in coulombs to the electrical potential energy in joulesB. electrical potential energy in joules to the charge in coulombsC. electrical potential energy in joules to the resistanceD. electrical potential energy in joules to the time |
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Answer» Correct Answer - B The fact that electrons flow from one elctrode to the other indicates that there is a potential (or voltage) difference between the two electrodes. This potential difference between the electrodes, called the electromotive froce, or `emf(E)`, is most conveniently described as potential, which is the potential energy per unit charge. We measure this quantity in volts. The volt `(V)` is the `SI` unit of potential difference. The electrical work expended in moving a charge through a conductor is Electrical work = charge `xx` potential difference Corresponding `SI` units for the terms in the terms in this equationare Joules = coulombs `xx` volts |
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| 2488. |
`lamda_(ClCH_(2)COONa)=224ohm^(-1)cm^(2)gmeq^(-1)` `lamda_(NaCl)=38.2ohm^(-1)cm^(2)gmeq^(-1)` `lamda_(HCl)=203ohm^(-1)cm^(2)gmeq^(-1)` What is the value of `lamda_(ClCH_(2)COOH)`A. 288.5 `ohm^(-1)cm^(2)gmeq^(-1)`B. 289.5 `ohm^(-1)cm^(2)gmeq^(-1)`C. `388.8ohm^(-1)cm^(2)gmeq^(-1)`D. 59.5 `ohm^(-1)cm^(2)gmeq^(-1)` |
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Answer» Correct Answer - C `ClCH_(3)COONa+HCltoClCH_(3)COOH+NaCl` `lamda_(ClCH_(3)COONa)+lamda_(HCl)=lamda_(CiCH_(2)COOH)+lamda_(NaCl)` ltBrgt `224+203=lamda_(ClCH_(2)COOH)+38.2` `lamda_(ClCH_(3)COOH)=427-38.2=388.8ohm^(-1)cm^(2)gmeq^(-1)`. |
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| 2489. |
Which of the following relations is incorrect ( `l =` length, a = area) ?A. `Lambda = kV` (V = dilution of the solution mL/equiv. or mL/mol)B. `G = k (a)/(l) ohm`C. `G = k (a)/(l) ohm^(-1)`D. `R = (1)/(k)(a)/(l) ohm` |
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Answer» Correct Answer - B Conductance `(G)`, the reciprocel of the resistance `(R)` of the condluctor is expressed in the unit celled reciprocel `ohm` (`ohm^(-1)` or `Omega^(-1)`). Thus, the correct relation is `G = k (a)/(l) ohm^(-1)` |
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| 2490. |
The unit of equivalent conductivity `(Lambda_(eq.))` areA. `ohm^(-1)cm^(2)eq^(-1)`B. `ohm^(-2)cm^(2)`C. `ohm^(-1)cm^(-1)`D. `ohm^(-1)cm^(-2)` |
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Answer» Correct Answer - A Imagine `1 c.c.` of a solution of `1` equivalent of an electrolyte placed between two large electrodes `1 cm` apart. In this case Conductance `(G)` = specific conductivity `(k)` = Equivalent conductivity `(Lambda_(eq.))` Suppose the solution is now diluted to `1000 cm^(3)`. We will be having now `1000 cm` cubes of the solution. Therefore, the conductance of the resulting solution will be `1000` times its specific conductance. But even now, as the solution contain only `1 grams` equivalent of the electrolyte between the electrodes, the conductance measured will be equivalent conductance. Thus, in this case, `Lambda_(eq.) = 1000 xx k` If the solution is further diluted to say, `5000 cm^(3)`, there will be `5000 cm` cubes of the solution and hence the equivalent conductance of the resulting solution will be `5000` times its specific conductance. In general, `Lambda_(eq) = kV` where `V` (called dilution) is the volume of the solution in `c.cs`, containing `1 gram` equivalent of the electrolyte. Since the units of conductivity are `Omega^(-1) cm^(-1)` and the units of volume are `cm^(3)`, units of equivalent conductance are : `(Omega^(-1) cm^(-1)) (cm^(3) = Omega^(-1) cm^(2)` In `SI` system, the units of equivalent conductance are `S m^(2)`. Truly speaking the unit of equivalent conductance should be `S m^(2) "equiv"^(-1)` or `ohm^(-1) cm^(2) "equiv"^(-1)` |
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| 2491. |
The value of `Lambda_(eq)^(oo)` for `NH_(4)Cl, NaOH` and `NaCl` are `149.74, 248.1` and `126.4 Omega^(-1) cm^(2)" equiv"^(-1)`. The value of `Lambda_(eq)^(oo)` of `NH_(4)OH` isA. `371,44 Omega cm^(2)" equiv"^(-1)`B. `271,44 Omega cm^(2)" equiv"^(-1)`C. `71,44 Omega cm^(2)" equiv"^(-1)`D. data is insufficient to calculate it |
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Answer» Correct Answer - B `Lambda_(eq)^(oo) (NH_(4)OH)=Lambda_(eq)^(oo)(NH_(4)Cl)+ Lambda_(eq)^(oo) NaOH-Lambda_(eq)^(oo) (NaCl)` `=(149.74+248.1)-126.4` `=271.44 Omega cm^(2)" equiv"^(-1)` |
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| 2492. |
The standard reduction potnetials of 4 elements are given below. Which of the following will be the most suitable reducing agent? `I=-3.04V,II+-1.90V,III=0V,IV=1.90V`A. IB. IIC. IIID. IV |
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Answer» Correct Answer - A Elements with least reduction potential is most suitable reducing agent. |
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| 2493. |
Reduction potential of four elements P,Q,R,S is -2.90,+0.34,+1.20 and -0.76. reactivity decreases in the orderA. PgtQgtRgtSB. QgtPgtRgtSC. RgtQgtSgtPD. PgtSgtQgtR |
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Answer» Correct Answer - D Reducing power i.e. the tendency to loose electrons increases as the reduction potential decreases. |
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| 2494. |
The standard reduction potentials of 4 elements are given below. Which of the following will be the most suitable reducing agent. `I=-3.04V,II=-1.90V,III=0V,IV=1.90V`A. IB. IIC. IIID. IV |
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Answer» Correct Answer - A Less is the reduction potential stronger is the reducing agent. |
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| 2495. |
Name the two half-cell reactions that are taking place in the Daniell cell. |
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Answer» The two half cell reactions that are takes place in the Daniell cell are oxidation and reduction. `Zn to Zn^(+2) +2 e^(-)` (Oxidation) `Cu^(+2)+2e^(-) to Cu` (Reduction) |
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| 2496. |
Write the chemical reaction used in the construction of the Daniell cll together with the half-cell reactions. |
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Answer» The chemical reactions used in the construction of the Daniell cell. `Zn to Zn^(+2) + 2e^(-)`(Oxidation) `Cu^(+2)+2e^(-) to CU` (Reduction) Overall reaction `underline overline(Zn + Cu^(+2)to ZN ^(+2)+Cu)` |
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| 2497. |
The cell reaction `Cr_2O_7^(2-)(aq)+14H^+(aq)+6Fe^(2+)(aq)to6Fe^(3+)(aq)+2Cr^(3+)(aq)+7H_2O(l)` is best represented by:A. `Pt(s)|Fe^(2+)(aq),Fe^(3+)(aq)"||"Cr__2O_7^(2-)(aq),Cr^(3+)(aq)|Pt(s)`B. `Pt(s)|Cr_2O_7^(2-)(aq),Cr^(+3)(aq)"||"Fe^(3+)(aq),Fe^(+2)(aq)|Pt(s)`C. `Fe^(2+)(aq)|Fe^(3+)(aq)"||"Cr_2O_7^(2-)(aq)|Cr^(3+)(aq)`D. `Cr_2O_7^(2-)(aq)|Cr^(3+)(aq)"||"Fe^(3+)(aq)|Fe^(2+)(aq)` |
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Answer» Correct Answer - A |
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| 2498. |
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. `9.65xx10^(4)s`B. `19.3xx10^(4)s`C. `28.95xx10^(4)s`D. `38.6xx10^(4)s` |
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Answer» Correct Answer - b `2H_(2)+e^(-)rarr H_(2)+2overset(c-)(O)H` For `0.01 mol H_(2),0.02mol ` of electrons are consumed. Charge required `-0.02xx96500C=Ixxt` Time required `=(0.02xx96500)/(10xx10^(-3))=19.3xx10^(4)s` |
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| 2499. |
The following reaction takes place during discharge of lead storage battery: `Pb(s)+PbO_(2)(s)+2H_(2)SO_(4)(aq)to2PbSO_(4)(s)+2H_(2)O(l)` If 2.5 ampere of current is drawn for 16 hours 5 min, how much `H_(2)SO_(4)` is consumed? |
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Answer» Oxidation: `Pb(s)+SO_(4)^(2-)(aq)toPbSO_(4)(s)+2e^(-)` `underline("Reduction:"PbO_(2)(s)+4H^(+)(aq)+SO_(4)^(2-)(aq)+2e^(-)toPbSO_(4)(s)+2H_(2)O(l))` Overall reaction: `Pb(s)+PbO_(2)(s)+2H_(2)SO_(4)(aq)to2PbSO_(4)(s)+2H_(2)O(l)` Thus, for drawing 2F of electricity, `H_(2)SO_(4)` consumed=2 mole Actually quantity of electricity drawn`=Ixxt=2.5xx965xx60C=(2.5xx965xx60)/(96500)F=1.5F` `thereforeH_(2)SO_(4)` consumed`=1.5`mole. |
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| 2500. |
01. ampere current is passed for 10 second through copper and silver voltameters . The metal that is deposited more isA. CuB. AgC. both (a) and (b)D. no effect |
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Answer» Correct Answer - B `w alpha E` : provided Q is constant `, E_(Ag)^(@) gt E_(Cu)^(@)` |
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