InterviewSolution
Saved Bookmarks
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.
| 2601. |
The conductance of 0.1 M HCl solution is greater than that of 0.1 M NaCL This is because (A. HCl is more ionized than NaClB. HCl is an acid where as NaCl solution is neutralC. `H^(+)` ions have greater mobility than `Na^(+)` ionsD. Interionic forces in HCl are weaker than those of NaCl. |
|
Answer» Correct Answer - C Greater conductance by HCl is due to greater mobility of `H^(+)` ions. |
|
| 2602. |
The molar conductivity of acetic acid at infinite dilution is `387omega^(-1)cm^(2)mol^(-1)`. At the same temperature, but at a concentration of 1 mole in 1000 litres, it is 55 `Omega^(-1)cm^(2)mol^(-1)`. What is the % age dissociation of 0.001 M acetic acid? |
|
Answer» Correct Answer - 0.1421 1 mol in 1000 litres `=0.001` mol `L^(-1)=0.001M`. |
|
| 2603. |
The increase in the molar conductivity of acetic acid with dilution is due toA. decrease in interionic forcesB. increase in degree of ionisationC. increase in self ionisation of waterD. none of these |
|
Answer» Correct Answer - B In a weak electrolyte like acetic acid , degree of ionization increases with dilution . |
|
| 2604. |
Point out the correct statement.A. Equivalent conductance decreases with dilutionB. Specific conductance increases with dilutionC. Specific conductance decreases with dilutionD. Equivalent conductance increases with increase in concentration |
| Answer» Correct Answer - C | |
| 2605. |
The increase in the molar conductivity of acetic acid with dilution is due toA. decrease in interionic forcesB. increase in degree of ionisationC. increase in self ionisation of waterD. None of the above |
| Answer» Correct Answer - B | |
| 2606. |
The increase in the molar conductivity of `HCl` with dilution is due toA. increase in the self ionisation of waterB. hydrolysis of HClC. decrease in the self ionisation of waterD. decrease in the interionic forces . |
|
Answer» Correct Answer - D The increase in molar conductivity of any strong electrolyte with dilution is due to decrease in the interionic forces . |
|
| 2607. |
Given that `I_(2)+2e^(-) rarr 2I^(c-)," "E^(c-)=0.54V` `Br_(2)+2e^(-) rarr 2Br^(-)," "E^(c-)=1.69V` Predict which of the following is true.A. `I^(c-)` ions will be able to reduce bromine.B. `Br^(c-)` ions will be able to reduce iodine.C. Iodine will be able to reduce broming.D. Bromine will be able to reduce iodide ions. |
|
Answer» Correct Answer - a Since the reduction potential of `Br_(2)|2Br^(c-)` is greater than the reduction potential of `I_(2)|2I^(c-)`, so `I^(c-)` ions will reduce `Br_(2)` |
|
| 2608. |
Write the cell reaction that occurs when the following half-cells are combined. `I_(2)+2e^(-)to 2l^(-)(IM) , " "E^(@)=0.54 V` `Br_(2)+2e^(-)to 2Br^(-)(IM) , " "E^(@)=1.08 V` |
|
Answer» The half cell reaction with lesser `E^(@)` value takes place at the anode while the other takes place at the cathode. Thus `{:("At anode: " 2I^(-) to I_(2)+2e^(-) " (oxidation)"),("At cathode : " Br_(2)+2e^(-) to 2Br^(-) " (reduction )"),("cell reaction: " bar(" "2I^(-) +Br_(2)to I_(2) +2Br^(-))):}` |
|
| 2609. |
Which of the following is the cell reaction that occurs when the following half-cells are combined? `I_2 + 2e^(-) to 2I^(-) (1M) , E^@=+0.54 V` `Br_2+2e^(-) to 2Br (1 M) , E^@=+1.09 V`A. `2Br^(-) + I_2 to Br_2 + 2I^-`B. `I_2 + Br_2 to 2I^(-) + 2Br^-`C. `2I^(-) + Br_2 to I_2 + 2Br^-`D. `2I^(-) + 2Br^(-) to I_2 + Br_2` |
|
Answer» Correct Answer - C `{:(2I^(-) to I_2 + 2e^(-),"(Oxidation)"),(Br_2 + 2e^(-) to 2Br^(-),"(Reduction)"),(2I^(-)+Br_2 to I_2 + 2Br^(-),"is net cell reaction"):}` |
|
| 2610. |
Calculate the standard cell potential of galvanic cell in which the following reaction takes place `2Cr_(s)+3Cd_(aq)^(+2)rarr2cr_(aq)^(+3)+3Cd_(s)` Given `E_(Cr^(+3)//Cr)=-0.74(V)E^(@)_(Cd^(+2)//Cd)=-0.04(V)`A. 0.74 VB. 1.14 VC. 0.34 VD. `-0.34 V` |
|
Answer» Correct Answer - C `E_"cell"^@=E_"cathode"^@-E_"anode"^@` `E_"cell"^@=E_(Cd^(2+)//Cd)^@-E_(Cr^(3+)//Cr)^@` =-0.40-(-0.74)=+0.34 V |
|
| 2611. |
The resistance of `0. 01 N NaCl` solution at `25^@C` is `200 Omega` . Cell constant of conductivity cell is ` 1 cm^(-1)` . The equivalent conductnance is .A. ` 5 xx10^2 Omega cm^2 eq^(-1)`B. ` 6 xx 10^3 Omega cm^2 eq^(-1)`C. ` 7 xx 10^4 Omega cm^2 eq^(-1)`D. ` 8 xx 10^5 Omega cm^2 eq^(-1)` |
|
Answer» Correct Answer - A ` lambda=k xx V 1/R xx 1/a xx V = 1/( 200) xx 1 xx 10.000` ` =5 xx 10^2 Omega^(-1) cm^2 eq^(-1)`. |
|
| 2612. |
The standard reduction potential for the half-cell reaction,`Cl_2 + 2e^(-) to 2Cl^(-)` will be `(Pt^(2+)+2Cl^(-)to Pt + Cl_2 , E_"cell"^@=-0.15 V , Pt^(2+) + 2e^(-) to Pt, E^@=1.20 V)`A. `-1.35 V`B. `+1.35 V`C. `1.05 V`D. `+1.05 V` |
|
Answer» Correct Answer - B `{:(Pt+Cl_2toPt^(2+)+2Cl^(-),,E_"cell"^@=0.15 V),(Pt^(2+) + 2e^(-) to Pt,,E^@=1.20 V ),(Cl_2+2e^(-)to 2Cl^(-),,E^(@)=1.35 V):}` |
|
| 2613. |
At `25^(@)C`, the resistance of `0.01 N NaCl` solution is `200 ohm` . If cell constant of the conductivity cell is unity, then the equivalent conductance of the solution is: |
|
Answer» Correct Answer - `500 ohm^(-1)` Step I. Calculation of specific conductance Specific conductance (k)`=("cell constant")/("R")=((1" cm"^(-1)))/((200 ohm))=0.005" ohm"^(-1) cm^(-1)`. Step II. Calculation of equivalent conductance of solution `Lambda_(E)=(1000xxk)/(N)=((1000 cm^(3))xx(0.005" ohm"^(-1)cm^(-1)))/((0.01"g eq"))=500" ohm"^(-1)cm^(2)("g eq")^(-1)`. |
|
| 2614. |
The algebraic sum of potentials of two electrodes of a galvanic cell is calledA. potential differenceB. ionic differenceC. EMFD. electrode difference |
|
Answer» Correct Answer - C `E_("cell") = E_(OP) + E_(RP)` |
|
| 2615. |
In a cell reaction, `Cu_((s))+ 2Ag_((aq))^(+) to Cu_((aq))^(2+) + 2Ag_((s)) E_"cell"^@`=+0.46 V . If the concentration of `Cu^(2+)` ions is doubled then `E_"cell"^@` will beA. doubledB. halvedC. increased by four timesD. unchanged. |
|
Answer» Correct Answer - D `E_"cell"^@=E_"cathode"^@-E_"anode"^@` It will remain unchanged |
|
| 2616. |
NERNST EQUATIONA. `E_(RP) = E_(RP)^(@) - (0.059)/(n) "log" ("[oxidant]")/("[reductant]")`B. `E_(OP) = E_(OP)^(@) - (0.059)/(n) "log" ("[oxidant]")/("[reductant]")`C. `E_(OP) = E_(OP)^(@) - (0.059)/(n) "log" ("[reductant]")/("[oxidant]")`D. `E_(RP) = E_(RP)^(@) + (0.059)/(n) "log" ("[reductant]")/("[oxidant]")` |
|
Answer» Correct Answer - B A/c to Nernst equation . |
|
| 2617. |
In the cell reaction `Cu(S) + 2 Ag^(+) (aq) to Cu^(2+) (aq) + 2 Ag(s) , E_("cell")^(@) = 0.46` V By doubling the concentration of `Cu^(2+) , E_("cell")^(@)` isA. doubledB. halvedC. increases but less than doubleD. decreases by a small fraction . |
|
Answer» Correct Answer - C `E_("cell") = E_("cell")^(@) - (RT)/(nF) "ln" ([Cu^(2+)])/([Ag^(+)]^(2))` . Doubling `[Cu^(2+)]` decreases the EMF by a small fraction . |
|
| 2618. |
Assertion : `E_(Ag^(+)//Ag` increases with in concentration of `Ag^(+)` ions. Reason : `E_(Ag^(+)//Ag)` has a positive value.A. Both assertion and reaction are true and the reason is correct explanation for assertionB. Both assertion and reason are true and reason is not correct explanation for assertionC. ) Assertion is true but the reason is False.D. Both assertion and reason are false. |
|
Answer» Correct Answer - B Reason is not the correct explanation for assertion as concentation of `Ag^(+)` ions increases, more ions are reduced to the metallic from. Therefore, `E_(Ag^(+)//Ag)` increases. |
|
| 2619. |
Consider the Fig. 3.2 and answer the following questions. |
|
Answer» (i) Cell ‘B’ will act as electrolytic cell as it has lower emf ∴ The electrode reactions will be: Zn2+ + 2e– ⎯→ Zn at cathode Cu → Cu2+ + 2e– at anode (ii) Now cell ‘B’ acts as galvanic cell as it has higher emf and will push electrons into cell ‘A’. The electrode reaction will be: At anode : Zn → Zn2+ + 2e– At cathode : Cu2+ + 2e– ⎯→ Cu |
|
| 2620. |
In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (i) Both assertion and reason are true and the reason is the correct explanation of assertion. (ii) Both assertion and reason are true and the reason is not the correct explanation of assertion. (iii) Assertion is true but the reason is false. (iv) Both assertion and reason are false. (v) Assertion is false but reason is true.Assertion : EAg+ /Ag increases with increase in concentration of Ag+ ions.Reason : EAg+/Ag has a positive value. |
|
Answer» (ii) Both assertion and reason are true and the reason is not the correct explanation of assertion. |
|
| 2621. |
The reduction potential of hydrogen half cell will be negative if :A. ` P(H_(2))=1"atm and"[H^(+)]=1.0M`B. `p(H_(2))=2 "atm and"[H^(+)]=1.0M`C. `p(H_(2))=2"atm and"[H^(+)]=2.0M`D. `p(H_(2))=1"atm and"[H^(+)]=2.0M`. |
| Answer» Correct Answer - C | |
| 2622. |
In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (i) Both assertion and reason are true and the reason is the correct explanation of assertion. (ii) Both assertion and reason are true and the reason is not the correct explanation of assertion. (iii) Assertion is true but the reason is false. (iv) Both assertion and reason are false. (v) Assertion is false but reason is true.Assertion : Copper sulphate can be stored in zinc vessel.Reason : Zinc is less reactive than copper. |
|
Answer» (iv) Both assertion and reason are false. |
|
| 2623. |
Unit of ionic mobility is :A. `m^(2)"sec"^(-1)` voltB. `m s^(-1)`C. `m "sec"^(-1)` voltD. `m "sec"^(-1) "volt"^(-1)` |
| Answer» Correct Answer - A | |
| 2624. |
Assertion: Copper sulphate can be stored in zinc vessel. Reason: Zinc is less reactive than copper.A. Both assertion and reason are true and the reason is the correct explanation of assertion.B. Both assertion and reason are true and the reason is not the correct explanation of assertion.C. Assertion is true but the reason is false.D. Both assertion and reason are false. |
|
Answer» Correct Answer - D Correct Assertion. Copper sulphate solution cannot be stored in zinc vessel. Correct Reason. Zinc is more reactive than copper. |
|
| 2625. |
In the representation of galvanic cell, the ions in the same phase are separated by a :(a) single vertical line (b) comma (c) double vertical lines (d) semicolon |
|
Answer» Option : (b) comma |
|
| 2626. |
In an electrolytic cellA. oxidation takes place at the negative electrodeB. reduction is spontaneousC. reduction takes place at the positive electrodeD. reduction takes place at the negative electrode |
|
Answer» Correct Answer - D Electrodes are surfaces on which oxidation or reduction half-reactions occurs. They may or may not par-ticipate in the reactions. Those that do not react are called inert electrodes. Regardless of the kind of cell, electrolytic or voltaic, the electrodes are identified as follows: The cathode is defined as the electrode at which reduction occurs as electrons are gained by some species. The anode is the electrode at which oxidation occurs as electrons are lost by same species. Each of these can be either the positive or the negative elec-trode. The signs of the electrodes are opposite for the two kinds of cells. In a galvanic cell, the cathode is considered negative because it receives electrons from the external circuit, but in an electrolytic cell, the cathode is considered negative because electrons are pumped into it by the battery i.e. it is connected to the negative terminal of the battery. An electrolytic cell is an electrochemical cell in which an electric current drives an otherwise nonspontaneous reaction. The process of producing a chemical change in an electrolytic cell is called electrolysis. |
|
| 2627. |
How many faradays of electric charge is required to liberate `5600cm^(3)` of oxygen at STP? |
|
Answer» `O^(2-)to(1)/(2)O_(2)+2e^(-)` or `2O^(2-)toO_(2)+4e^(-)` Thus, 1 mole of `O_(2)`, i.e., 22400 `cm^(3)` requrie 4 faradays of charge. `therefore5600cm^(3)` of `O_(2)` will require charge `=(4)/(22400)xx5600=1` faraday. |
|
| 2628. |
Daniell cell is :(a) Secondary cell (b) Irreversible cell (c) primary irreversible cell (d) primary reversible cell |
|
Answer» Option : (d) primary reversible cell |
|
| 2629. |
Calculalting the amount the product from the amount of charge in an electrolysis: A constant current of `0.452 A` is passed through an electrolytic cell containing molten `CaCl_(2)` for a time of `1.50` hours. Write the electrode reactions and calculate the quantity of products (in grams) formed at the electrodes. Also find the volume (at STP) of any gaseous product formed. Strategy: to convert the current and time to grams or litres of product, carry out the sequence of conversions in Figure 3.9. |
|
Answer» Since the only ions present in molten `CaCl_(2)` are `Ca^(2+)` and `Cl^(-)`, the reactions are `{:("Anode": 2Cl^(-)(1)rarr Cl_(2)(g)+2e^(-)),("Cathode": Ca^(2+)(1)+2e^(-)rarrCa(1)),(bar("Overall": Ca^(2+)(1)+2Cl^(-)(1)rarr Ca(1)+Cl_(2)(g))):}` The quantities of `Ca` metal and `Cl_(2)` gas formed depend on the number of electrons that pass through the electrolytic cell, which in turn depends on the current and time or charge. Step 1: Because electrons can be thought of as a reactant in the electrolysis process, the first step is to calculate the charge and the number of moles of electrons passed throgh the cell: `Q = It` Charge `= (0.452 A)(1.50h)((3600s)/(1h))((1 C)/(1 A.s))` `(0.452(C)/(s)) (1.50h) ((60 min)/(h)) ((60 s)/(min))` `= 2440.8 C = 2.44 xx 10^(3)C` Moles of `e^(-) = 2.44 xx 10^(3)C (("1 mol" e^(-))/(96,500C))` `= 0.025 mol e^(-)` Step 2 : The cathode reaction yield `1` mol of `Ca` per `2` mol of electrons, so `0.025//2` or `0.0125` mol of `Ca` will be obtained: `Ca^(2+) + 2e^(-) rarr Ca` Moles of `Ca = (0.025 mol e^(-)) ((1 mol Ca)/(2 mol e^(-)))` `= 0.0125 mol Ca` Step 3: Converting the number ofmoles of `Ca` to grams of `Ca` gives Grams of `Ca = (0.0125 mol Ca) ((40g Ca)/(mol Ca))` `= 0.5 g Ca` As a shortcut, the entire porcess of conversion of coulombs to grams can be carried out in one step: `? g Ca = (2.44 xx 10^(3)C)(1 "mole"^(-))/(96,500 C)(1 mol Ca)/(2 "mole"^(-))((40 g Ca)/(1 mol Ca))` `= 0.5 g Ca` Step 4: The anode reaction gives `1 mol` of `Cl_(2)` per 2 mol of electrons, so `0.0125 mol` of `Cl_(2)` will be obtained: `2Ci^(-)(l) rarr Cl_(2)(g) + 2e^(-)` Moles of `Cl_(2) = (0.025 "mole"^(-)) ((1 mol Cl_(2))/(2 mol E^(-)))` `= 0.0125 mol Cl_(2)` Step 5: Converting the number of moles of `Cl_(2)` to grams of `Cl_(2)` gives Grams of `Cl_(2) = (0.125 mol Cl_(2)) ((71 g Cl_(2))/(1 mol Cl_(2)))` `= 8.88 g Cl_(2)` Step 6: Since `1` mole of an ideal gas occupies `22.4 L` at STP, the volume of `Cl_(2)` obtained is Litres of `Cl_(2) = (0.125 mol Cl_(2))((22.4 L Cl_(2))/(1 mol Cl_(2)))` `= 0.28 L` As a shortcut, the entire sequence of conversions can be carried out in just one step. For example the volume of `Cl_(2)` produced at the anode is `(0.452(C)/(s))(1.50 h)((3600s)/(h))((1mol e^(-))/(96,500C))((1molCl_(2))/(2 "mole"^(-)))` `((22.4LCl_(2))/(1molCl_(2)) = 0.28 L` |
|
| 2630. |
Downs cell is used commercially to perform electrolysis of moltenA. `NaCl`B. `LiCl`C. `KCl`D. `RbCl` |
|
Answer» Correct Answer - A A Downs cell is a commerical electrochemical cell used to obtain `Na` metal by the electrolysis of molten sodium chloride. The cell is constructed to keep the products of the electrolysis separate, because they would otherwise react. Calcium chloride is added to the sodium chloride to lower the melting point from `801^(@)C` for `NaCl` to about `580^(@)C` for the mixture. (Remember that the melting point, or freezing point, of a substance is lowered by the addition of a solute). We obtain the cell reaction by adding the half-reaction: `{:(Na^(+)(l)+e^(-) rarr Na(l)),(Cl^(-)(l) rarr (1)/(2)Cl_(2)(g)+e^(-)))/(Na^(+)(l)+Cl^(-)(l) rarr Na(l)+(1)/(2)Cl_(2)(g))` A number of other reactive metals are obtained by the electrolysis of a molten salt or ionic compound. Lithium, magnesium, and clacium metals are all obtained by the electrolysis of the chlorides. The first commercial preparation of sodium metal adapted the method used by Humphry Davy when he discovered the ele-ment in `1807`. Davy electrolyzed molten sodium hydroxide, `NaOH` whose melting point `(318^(@)C)` is relatively low for an ionic compound. The half-reaction are Cathode: `Na^(+)(l)+e^(-) rarr Na(l)` Anode: `4OH^(-)(l) rarr O_(2)(g)+2H_(2)O(g)+4e^(-)` Many of the commercial uses of electrolysis involve aqueous solutions |
|
| 2631. |
Number of faradays of electricity required to liberate 12 g of hydrogen is :(a) 1 (b) 8 (c) 12 (d) 16 |
|
Answer» Option : (c) 12 |
|
| 2632. |
Calculating the amount of charge from the amount of product in an electrolysis: How many amperes must be passed through a Downs cell to produce sodium metal a rate of `30.0 Kg//h.` Strategy : Producted through a sequence of conversion similar to that in worked Example 3.26 but in reverse order `Na^(+1) + e^(-) rarr Na` The electrode equation for sodium says that `1` mol `Na` is equiva-lent to `1` mole `e^(-)`. We can use this in the conversion of grams of `Na`. The Faraday constant (which says that one mole of electrons is equivalent to `9.65 xx 10^(4)C`) converts mole of electrons to cou-lombs. The conversions are `g Na rarr mol Na rarr mol e^(-) rarr "coulombs" (C)` The current in ampres `(A)` equals the charge in coulombs divided by the time in seconds. |
|
Answer» Because the moalr mass of sodium is `23.0 g//mol`, the number of moles of sodium produced per hour is Moles of `Na = (30.0 Kg Na) ((1000 g)/(1 Kg)) (1 mol Na)/(23.0 gNa))` `= 1304.3 mol Na` ` 1.30 xx 10^(3) mol Na` To produce each mole of sodium, `1` mol of electrons must be passed through the cell : `Na^(+)(l) + e^(-) rarr Na(l)` Therefore, the charge passed per hour is Charge `=(1.30xx10^(3) "mol Na")((1mol e^(-))/(1 mol Na))((96,500)/(1 mol e^(-)))` `= 125450 xx 10^(3)C` `= 1.25 xx 10^(8)C` Since there are `3600` s in `1` h, the current required is Current `= (1.25 xx 10^(8)C)/(3600S)` `= 0.000347 xx 10^(8)C/s` `= 3.47 xx 10^(4)C/s` `= 34, 700 A` As a shortcut, the entire conversion of grams of `NA` to coulombs required to deposit this amount to sodium is `(30.0 Kg)((1000 g)/(1 Kg))((1 mol Na)/(23.0 g Na))((1 mol e^(-))/(23.0 g Na)) ((96,500 C)/(1 mol e^(-)))` `= 1.25 xx 10^(8)C` |
|
| 2633. |
On passing 1.5 F charge, the number of moles of aluminium deposited at cathode are [Molar mass of Al = 27 gram mol-3](a) 1.0 (b) 13.5 (c) 0.50 (d) 0.75 |
|
Answer» Option : (c) 0.50 |
|
| 2634. |
On calculating the strength of current in amperes if a charge of 840 C (coulomb) passes through an electrolyte in 7 minutes, it will be :(a) 1 (b) 2 (c) 3 (d) 4 |
|
Answer» Option : (b) 2 |
|
| 2635. |
Statement-I: Blocks of magnesium are ofter stapped to steel hulld of ocean going ships. Because Statement-II: Magnesium causes cathode protection of iron. |
| Answer» Correct Answer - A | |
| 2636. |
Consider the following cell reaction: `Fe_((s))+O_(2(g))+4H_((aq))^(+)to2Fe_((aq))^(2+)+2H_(2)O_((l))` `E^(@)=1.67V` `At[Fe^(2+)]=10^(-3)M,P_((O_(2)))^(0)=0.1atm and pH=3,` the cell potential at `25^(@)C` isA. 1.47VB. 1.77VC. 1.87VD. 1.57V |
|
Answer» Correct Answer - D `E=E^(o)-(0.059)/(4)"log"([Fe^(2+)]^(2))/([H^(+)]^(4)P_(O_(2)))` `=1.67-(0.06)/(4)"log"((10^(-3))^(2))/((10^(-3))^(4)xx0.1)=1.67-(0.03)/(2)"log"10^(7)` `=1.67-(0.03)/(2)xx7=1.67-0.105=1.565=1.57V` |
|
| 2637. |
In a galvanic cell,t he salt bridgeA. Does not participiate chemically in the cell reationB. stops the diffusion of ions from one electrode to anotherC. Is necessary for the occurrence of the cell reactionD. Ensures mixing of the two electrolytic solution |
|
Answer» Correct Answer - A Salt bridge is introduced to keep the solutions of two electrodes separate, such that the ions in electrode do not mix freely with each other. But it cannot stop the process of diffusion. It oes not participiate in the chemical reaction. However, it is not necesary for occurrence of cell reaction, as we know that designs like lead accumulator, there was no salt bridge, but still reaction takes place. |
|
| 2638. |
Neglecting the liquid-liquid junction potential, calculate the emf of the following cell at `25^(@)C` `H_(2) (1 atm)|0.5 M HCOOH|| 1M CH_(3)COOH | (1 atm) H_(2)` `K_(a) " for " HCOOH and CH_(3) COOH` are `1.77 xx 10^(-4) and 1.8 xx 10^(-5)` respectively. |
|
Answer» `[H^(+)] " in " HCOOH = sqrt(C xx K_(a)) = sqrt(0.5 xx 1.77 xx 10^(-4))` `[H^(+)]` in `CH_(3)COOH = sqrt(C xx K_(a)) = sqrt(1 xx 1.8 xx 10^(-5))` `= 4.2426 xx 10^(-3)M` `E_("cell") = 0.059 log. ([H^(+)]_(RHS))/([H^(+)]_(LHS)) = 0.059 log. (4.2426 xx 10^(-3))/(0.9407 xx 10^(-2))` `= -0.0204` volt |
|
| 2639. |
A solution of `M(NO_(3))_(2)`was electrolysed by passing a current of 2.5 A and 3.06 g of the metal was deposited in 35 minutes. Determine the molar mass of the metal. |
|
Answer» Quantity of charge passed (Q) `=Ixxt=(2.5 A)xx(35xx60 s)=5250 As =5250 C` No.of faradays of charge passed `=((5250C))/((96500C))=0.0544.` 0.0544 F of charge deposited metal (M)=3.06 g 1 F of charge deposited metal (M) `=((3.06g))/(0.0544 F)xx(1F)=56.25 g` Equivalent mass of metal =56.25 g=56.25 amu. Molar mass of the metal `="Equivalent mass" xx"valency"` `=(56.25 amu)xx(2)=112.5" amu"` |
|
| 2640. |
A current of 3.7 ampere is passed for 6 hrs. between Ni electrodes in 0.5 litre of 2 M solution of Ni(NO3)2. What will be the molarity of solution at the end of electrolysis? |
|
Answer» The electrolysis of Ni(NO3)2 in presence of Ni electrode will bring in following changes: At anode : Ni -----> Ni2+ + 2e At cathode : Ni2+ + 2e -----> Ni Eq. of Ni2+ formed = Eq. of Ni2+ lost Thus, there will be no change in conc. of Ni(NO3)2 solution during electrolysis i.e., It will remain 2 M. |
|
| 2641. |
A 100 watt, 110 volt incandescent lamp is connected in series with an electrolytic cell containing cadmium sulphate solution. What mass of cadmium will be deposited vy the current flowing for 10 hours ? |
|
Answer» We kanoe that, Watt = ampere `xx` volt 100= ampere `xx 110` Ampere `=100/10` Quantity of charge = ampere `xx` second `=100/110xx10xx60xx60` coulomb The cathodic reaction is : `underset(112.4 g)(Cd^(2+))+underset(2xx 96500 C)(2e^(-)) rarr Cd` Mass of cadmium deposited by passing `100/110xx10xx60xx60` coulomb charge `=112.4/(2xx96500)xx100/110xx10xx60xx60=19.0598 g` |
|
| 2642. |
For the cell reaction `Zn_((s)) + Cu_("0.1 M")^(2+) to Zn_(0.1 M)^(2+) + Cu_((s))` if the standard EMF of the cell is `E^(@)` , thenA. `E gt E^(@)`B. `E lt E^(@)`C. `E = E^(@)`D. `E le E^(@)` |
|
Answer» Correct Answer - C `E = E^(@) - 0.591 " log " (0.01)/(0.1)` `E = E^(@) + 0.059` V |
|
| 2643. |
How many Faradays are required to reduce `1 mol of BrO_(3)^(c-)` to `Br^(c-)` in basic medium ?A. 6B. 5C. 4D. 3 |
|
Answer» Correct Answer - a `BrO_(3)^(c-)+6H^(o+)+6e^(-) rarr Br^(c-)+3H_(2)O` `1 mol `of `BrO_(3) ^(c-)=6 `Faradays. |
|
| 2644. |
Which of the following is the reaction half- reaction for a storage battery ?A. `PbO_(2 (s)) + SO_(4)^(2-) + 4 H^(+) + 2e^(-) to PbSO_(4 (s)) + 2 H_(2)O`B. `Pb_((s)) + PbO_(2 (s)) + 2 H_(2)SO_(4) to 2 Pb SO_(4 (s)) + 2 H_(2)O`C. `Pb_((s)) + H_(2)SO_(4) to PbSO_(4 (s)) + 2 H^(+) + 2e^(-)`D. `2 PbSO_(4 (s)) + 2 H_(2)O to Pb_((s)) + PbO_(2(s)) + 2 H_(2)SO_(4)` |
|
Answer» Correct Answer - A Reduction half- cell reaction . `PbO_(2 (S)) + 4H^(+) + SO_(4)^(2-) +2 e^(-) to PbSO_(4 (S)) + 2 H_(2)O` Oxidation half -cell reaction . `Pb_((S)) + SO_(4)^(2-) + 2 e^(-)` |
|
| 2645. |
How many Faradays are required to reduce `1 mol of BrO_(3)^(c-)` to `Br^(c-)` in basic medium ? |
|
Answer» Correct Answer - 6 `6e^(-)+BrO_(3)^(c-)(aq)+3H_(2)Orarr Br^(c-)(aq)+6overset(c-)(O)H(aq)` `1 mol of BrO_(3)^(c-)=6F=6 mol` of electrons |
|
| 2646. |
During charging of lead storage cellA. external emf is greater than the emf of cellB. external emf is equal to the emf of cellC. external emf is less than emf of the cellD. none of these |
|
Answer» Correct Answer - A During charging of lead accumulator , electrical energy is supplied to it from external field , i.e, it acts as an electrical cell . |
|
| 2647. |
In a secondary cell ,A. the electrical energy obtained is first handB. magnitude of emf depends on chargingC. the electrical energy is developed within the cellD. we can draw a current of high magnitude |
|
Answer» Correct Answer - B During charging concentration of `H_(2)SO_(4)` increases and the emf depends on it . In a secondary cell electrical energy is first stored as chemical energy . Hence high magnitude current is not available . |
|
| 2648. |
The total number of Faradays required to oxidize the following separately`:` `a. 1 mol of S_(2)O_(3)^(2-)` in acid medium `b. 1 ` Equivalent of `S_(2)O_(3)^(2-)` in acid medium `c. 1 mol of S_(2)O_(3)^(2-)` in basic medium. |
|
Answer» Correct Answer - 10 `a. S_(2)O_(3)^(2-)rarr (1)/(2)S_(4)O_(6)^(2-)+e^(-)[1F]` `b. S_(2)O_(3)^(2-)+5H_(2)Orarr2HSO_(4)^(c-)+8H^(o+)+8e^(-)` `[1 mol =8F` but `1Eq=1F]` `c. S_(2)O_(3)^(2-) +10overset(c-)(O)Hrarr 2SO_(4)^(2-)+5H_(2)O+8e^(-)` `[1mol=8F]` Total Faradays in `(i),(ii), `and `(iii)=1+1+8=10F` |
|
| 2649. |
During the charging of storage cellA. concentration of lead is decreasedB. lead sulphate concentration is increasedC. concentration of `H_(2)SO_(4)` is increasedD. concentration of `PbO_(2)` is decreased |
|
Answer» Correct Answer - C During the charging of storage cell following reaction takes place . `2 PbSO_(4 (S)) + 2H_(2)O (l) to Pb_((S)) + PbO_(2 (S)) + 2 H_(2)SO_(4)` Concentration of lead sulphate decreases while the concentration of `H_(2)SO_(4) , PbO_(2)` and `Pb ` increases . |
|
| 2650. |
Which of the following statement is false for fuel cell?A. they are light weightB. they are efficientC. they cause no pollutionD. they cannot work continuously |
|
Answer» Correct Answer - D During charging of a lead storage cell , electrical energy is supplied to it from external source thereby the electrode reactions are reserved . |
|