3195 + Interview Questions in ELECTROCHEMISTRY IN GENERAL AWARENESS Page 36 InterviewSolution

1751.	In the electrolysis of an aqueous solution of NaF , the product obtained at the cathode and anode respectively areA. `Na , F_(2)`B. `Na, O_(2)`C. `H_(2) , O_(2)`D. `H_(2) , F_(2)`
Answer» Correct Answer - C At cathode , `H_(2)O` is more easily reduced than `Na^(+)` to give `H_(2)`

Discussion

1752.	z On the electrolysis of an aqueous solution of NaF using the gram equivalence of an electrolyteA. Na is obtained at cathodeB. `F_(2)` is obtained at anodeC. `H_(2)` is obtained at cathodeD. `O_(2)` is obtained at anode.
Answer» Correct Answer - C,D See comprehensive review for details. Answers to similar questions have already been discussed earlier.

Discussion

1753.	Electrolysis is the process in which electrical energy is converted to chemical energy. In electrolytic cell, oxidation takes place at anode and reduction at cathode. Electrode process depends on the electrode taken for electrolysis. Amount of substance liberated at an electrode is directly propertional to the amount of charge passed through it. The mass of substance liberated at electrode is calculated using the following relation : `m= ("ItE")/(96500)` Here, E represents the equivalent mass and 96500 C is called the Faraday constant. Faraday (96500 C) is the charge of 1 mole electron, i.e., `6.023 xx 10^(23)` electrons, it is used to liberate one gram equivalent of the substance. The quantity of electricity required to liberate 112 cc hydrogen at S.T.P. from acidified water is :A. 965 CB. 9650 CC. 96500 CD. 4825 C
Answer» Correct Answer - A

Discussion

1754.	Electrolysis is the process in which electrical energy is converted to chemical energy. In electrolytic cell, oxidation takes place at anode and reduction at cathode. Electrode process depends on the electrode taken for electrolysis. Amount of substance liberated at an electrode is directly propertional to the amount of charge passed through it. The mass of substance liberated at electrode is calculated using the following relation : `m= ("ItE")/(96500)` Here, E represents the equivalent mass and 96500 C is called the Faraday constant. Faraday (96500 C) is the charge of 1 mole electron, i.e., `6.023 xx 10^(23)` electrons, it is used to liberate one gram equivalent of the substance. The passage of current liberates `H_(2)` at cathode and `Cl_(2)` at anode. The solution is :A. copper chloride in waterB. NaCl in waterC. mercuric chloride in waterD. `AuCl_(3)` in water
Answer» Correct Answer - B

Discussion

1755.	Electrolysis is the process in which electrical energy is converted to chemical energy. In electrolytic cell, oxidation takes place at anode and reduction at cathode. Electrode process depends on the electrode taken for electrolysis. Amount of substance liberated at an electrode is directly propertional to the amount of charge passed through it. The mass of substance liberated at electrode is calculated using the following relation : `m= ("ItE")/(96500)` Here, E represents the equivalent mass and 96500 C is called the Faraday constant. Faraday (96500 C) is the charge of 1 mole electron, i.e., `6.023 xx 10^(23)` electrons, it is used to liberate one gram equivalent of the substance. The platinum electrodes were immersed in a solution of cupric sulphate `(CuSO_(4))` and electric current is passed through the solution. after sometime, it was observed that the colour of copper sulphate disappeared with evolution of a gas at the electrode. The colourless solution contains :A. platinum sulphateB. copper nitrateC. copper sulphateD. sulphuric acid
Answer» Correct Answer - D

Discussion

1756.	Electrolysis is the process in which electrical energy is converted to chemical energy. In electrolytic cell, oxidation takes place at anode and reduction at cathode. Electrode process depends on the electrode taken for electrolysis. Amount of substance liberated at an electrode is directly propertional to the amount of charge passed through it. The mass of substance liberated at electrode is calculated using the following relation : `m= ("ItE")/(96500)` Here, E represents the equivalent mass and 96500 C is called the Faraday constant. Faraday (96500 C) is the charge of 1 mole electron, i.e., `6.023 xx 10^(23)` electrons, it is used to liberate one gram equivalent of the substance. How many faradays are required to reduce 1 mol `BrO_(3)^(-)` to `Br^(-)` ?A. 3B. 5C. 6D. 4
Answer» Correct Answer - C

Discussion

1757.	Electrolysis is the process in which electrical energy is converted to chemical energy. In electrolytic cell, oxidation takes place at anode and reduction at cathode. Electrode process depends on the electrode taken for electrolysis. Amount of substance liberated at an electrode is directly propertional to the amount of charge passed through it. The mass of substance liberated at electrode is calculated using the following relation : `m= ("ItE")/(96500)` Here, E represents the equivalent mass and 96500 C is called the Faraday constant. Faraday (96500 C) is the charge of 1 mole electron, i.e., `6.023 xx 10^(23)` electrons, it is used to liberate one gram equivalent of the substance. On electrolysis of dilute sulphuric acid using platinum electrodes, the product obtained at the anode will be :A. hydrogenB. oxygenC. hydrogen sulphideD. sulphur oxide
Answer» Correct Answer - B

Discussion

1758.	Conductivity of an aqueous solution of `0.1M HX` (a weak mono-protic acid) is `5 xx 10^(-4) Sm^(-1)` Given: `^^_(m)^(oo) [H^(+)] = 0.04 Sm^(2) mol^(-1): ^^_(m)^(oo) [X^(-)] = 0.01 Sm^(2)mol^(-1)` Find `pK_(a)[HX]`
Answer» Correct Answer - 9 `HX hArr H^(+) +X^(-)` `0.1(1-alpha) 0.1 alpha 0.1 alpha` `K_(a) = (0.1alpha^(2))/(1-alpha)` `alpha =(^^_(m))/(^^_(m)^(oo)) = 0.05` `^^_(m) = (K)/(M) = (5 xx 10^(-4)Sm^(-1))/((0.1mol)/(10^(-3)m^(3)))`

Discussion

1759.	A current of 0.4 ampere is passed for 30 minutes through a voltameter containing `CuSO_(4)` solution. The weight of Cu dpeosited will beA. 3.18gB. 0.318gC. 0.296gD. 0.150g
Answer» Correct Answer - C `Q=1xxt=0.5xx30xx60=900C` `Cu^(2+)+2e^(-)rarrCu` `2xx96500C` deposit `Cu=63.5g` `therefore 900C` will deposit Cu `=(63.5)/(2xx96500)xx900=0.296g`

Discussion

1760.	Electrolysis is the process in which electrical energy is converted to chemical energy. In electrolytic cell, oxidation takes place at anode and reduction at cathode. Electrode process depends on the electrode taken for electrolysis. Amount of substance liberated at an electrode is directly propertional to the amount of charge passed through it. The mass of substance liberated at electrode is calculated using the following relation : `m= ("ItE")/(96500)` Here, E represents the equivalent mass and 96500 C is called the Faraday constant. Faraday (96500 C) is the charge of 1 mole electron, i.e., `6.023 xx 10^(23)` electrons, it is used to liberate one gram equivalent of the substance. Calculate the volume of gas liberated at the anode at STP during the electrolysis of a `CuSO_(4)` solution by a current of 1 A passed for 16 minutes and 5 seconds :A. 224 mLB. 56 mLC. 112 mLD. 448 mL
Answer» Correct Answer - B At anode: `2OH^(-) rarr H_(2)O+1/2 O_(2) +2e^(-)` (Oxygen gas is evolved) Equivalent volume `V_(e)` of oxygen =5.6 litre `:. V=(ItV_(e))/96500=(1xx965xx5.6xx10^(3))/96500=56 mL`

Discussion

1761.	The conducativity of water is `7.6 xx 10^(-2)Sm^(-1)` and the con-ductivity of `0.1 M` aqueous solution of `KCl` is `1.639 Am^(-1)`. A cell has a resistance of `33.20 Omega` when filled with `0.1 M Kcl` solution and `300 Omega` when filled with `0.1` M `CH_(3)CO_(2)CO_(2)H` solution. The molar conductivity of `CH_(3)CO_(2)CO_(2)H` isA. `5.3 xx 10^(-4)Sm^(2)mol^(-1)`B. `4.7 xx 10^(-4)Sm^(2)mol^(-1)`C. `6.7 xx 10^(-4)Sm^(2)mol^(-1)`D. `7.5 xx 10^(-4)Sm^(2)mol^(-1)`
Answer» Correct Answer - A First calculate conductivity due to `CH_(3)COOH`. `kappa = K_("cell")//R` i.e. `K prop 1//R` Thus, we have `(kappa(CH_(3)COOH))/(kappa(KCl)) = (R(KCl))/(R(CH_(3)COOH))` or `kappa(CH_(3)COOH) = (R(KCl))/(R(CH_(3)COOH)) xx kappa(KCl)` `= ((33.20 Omega)(1.1639 Sm^(-1)))/((300 Omega))` `= 0.129 Sm^(-1)` Note that conductivity of sq. soln of `CH_(3)COOH` alos contains a contribution due to water. Thus, true value of conductivity of acetic acid will be `kappa(CH_(3)COOH) = 0.129 S m^(-1) - 7.6 xx 10^(-2) S m^(-1)` `= 0.053 s m^(-1)` Finally `Lambda_(m)(CH_(3)COOH) = (kappa)/(C) = (5.3 xx 10^(-2) S m^(-1))/(0.1 xx 10^(3) mol m^(-3))` `= 5.3 xx 10^(-4)S m^(2) mol^(-1)`

Discussion

1762.	ELECTROCHEMICAL CELLA. chemical energy to electrical energyB. electrical energy to chemical energyC. chemical energy to mechanical energyD. electrical energy to mechanical energy .
Answer» Correct Answer - B Electrolytic cell converts electrical energy to chemical energy .

Discussion

1763.	The conductance of a 0.0015M aqueous solution of a weak monobasic acid was determined by using a conductivity acid was determined by using a cunductivity cell consisting of platinised pt electrodes. The distance between the electrodes is 120cm with an area of croos section of 1 `cm^(2)`. The conductance of this solution was found to be `5xx10^(-7)` S. the pH of the solution is 4. The value of limiting moalr conductivity `(A_(m)^(@))` of this weak monobasic acid in aqueous solution is `Zxx10^(2)S cm^(-1) "mol"^(-1)`. the value of Z is
Answer» Correct Answer - `6xx10^(2)S cm^(-1) "mol"^(-1)`

Discussion

1764.	Which of the following statements is correct ?A. Oxidation number of oxygen in `K_(2)O` is + 1B. The specific conductance of an electrolyte solution decreases with increase in dilutionC. `Sn^(2+)` oxidises `Fe^(3+)`D. `Zn\|ZnSO_(4)` is reference electrode .
Answer» Correct Answer - B (a) O.N. of O = `-2` (c) `Fe^(3+)` oxidises `Sn^(2+)` (d) `Zn \|ZnSO_(4)` is not reference electrode .

Discussion

1765.	Measurement of the coonductivity of electrolytic solutions is done with the helpof Wheatstone bridge using a/anA. batteryB. induction coilC. conductivity cellD. both (2) and (3)
Answer» Correct Answer - D Induction coil, an alternating current `(A)` source of power is used to nullify the electrolysis of the electrolyte while conductivity cell is used to connet the electrolytic solution to the bridge like a matallic wire or other solid conductor. Galvanometer is replaced by an earphone bacause when `AC` source is used, ordinary galvanometer fails to detect the null point.

Discussion

1766.	Which one of the following relationship is not correct ?A. Specific conductance (K) = Observed conductance (C) `xx` cell constantB. Eq. conductance `(Lambda) = (k xx 1000)/("Normality")`C. Degree of dissociation `\|(alpha)\| = (Lambda_(0))/(Lambda_(c))`D. `E_("cell")^(@) = (0.0591)/(n) "log" K_(c)` at 298 K
Answer» Correct Answer - C `alpha = (Lambda_(c))/(Lambda_(o))`

Discussion

1767.	The conductance of 0.0015 M aqueous solution of a weak monobasic acid was determined by using a conductivity cell consisting of platinized Pt electrodes. The distance between the electrodes is 120 cm with an area of cross-section of `1 cm^(2)`. The conductance of this solution was found to be `5xx10^(-7) S`. The pH of the solution is 4. Calculate the value of limiting molar conductivity.
Answer» Correct Answer - 6 `[H^(+)]=10^(-pH)=10^(-4)M` `Lambda_(m)=kappaxx1000/M` `=[C xx l/A]xx1000/M` `=5xx10^(-7)xx120/1xx1000/0.0015=40" S cm"^(2)"mol"^(-1)` `[H^(+)]=c alpha=0.0015 xx Lambda_(m)/Lambda_(m)^(@)` `10^(-4)=0.0015xx40/(Zxx 10^(2))` `Z=6`

Discussion

1768.	A conductivity cell is platinized toA. avoid temperature effectB. avoid capacitance of the cellC. avoid polarization effectD. prolonf its life
Answer» Correct Answer - C Electrodes of the cell are made of platinum and are coatedx freshly electrolytically with finely divided platinum black before use to minimize the polarization effect (i.e. change of resistance of solution near the electrodes due to change of concentration of electrolyte).

Discussion

1769.	If x is the conduction of 1 `cm^(3)` of the solution and C is the concentration of the solution . The molar conductivity of the solution is given byA. `(1000 C)/(x)`B. `(1000 x)/(N)`C. `(C)/(x)`D. `(x)/(C)`
Answer» Correct Answer - B Molar conductivity ` = (1000 xx "conductance")/("concentration")`

Discussion

1770.	A conductivity cell is platinized toA. to prolong its lifeB. to avoid polarisation effectC. to avoid capacitance of the cellD. to avoid temperature effect .
Answer» Correct Answer - B Conductivity cell is platinized to avoid polarisation effects .

Discussion

1771.	Which one of the following statements is wrong ?A. `K^(+)` has different molar ionic conductivity at infinite dilution in KCl and KBrB. `Lambda^(@)` for `BaCl_(2) = Lambda^(@) (Ba^(2+)) + 2 lambda^(@) m (Cl^(-))`C. Degree of dissociation , `alpha = (Lambda_(c))/(Lambda_(0))`D. Kohlrausch law can be used for finding `Lambda_(0)` for weak electrolytes .
Answer» Correct Answer - A A particular ion has a definite value of `Lambda^(@)` irrespective of the electrolyte .

Discussion

1772.	The equivalent conductance is the conductance of a solution provided the following conditions are statisfied . Which one of the following is wrong ?A. The solution should contain 1 g equivalent of the electrolyteB. The distance between the electrodes should be 1 cmC. The area of the electrodes should be so large that the solution should touch them completelyD. The volume of the solution should be very small.
Answer» Correct Answer - D By definition of equivalent conductance .

Discussion

1773.	The variation of `^^_(m)` of acetic acid with concentration is correctly represented byA. B. C. D.
Answer» Correct Answer - C For weak electrolyte , `Lambda` first decreases slowly and then rapidly with increase in conc.but does not become constant .

Discussion

1774.	The degree of ionization increasesA. with increase in concentration of the solutionB. on addition of excess of water to the solutionC. on decreasing the temperature of the solutionD. on stirring the solution vigorousoly.
Answer» Correct Answer - B Degree of ionization increases with dilution .

Discussion

1775.	In terms of SI base units , ohm stands forA. `(kg m^(2)) //(As)`B. `(kg m^(2))//(A^(2) s^(2))`C. `(kg m^(2))//(A^(2) s^(2))`D. `(kg m^(2)) // (A^(2) s^(3))`
Answer» Correct Answer - D In SI units , `Omega = (kg m^(2))// (A^(2) s^(3))` . `Omega = (V)/(A) = ("Work")/("charge") xx (1)/(A) = (F xx l )/(A xx s) xx (1)/(A)` `= (m xx a xx l)/(A^(2) s) = (kg ms^(-2) xx m)/(A^(2) s) = (kg. m^(2))/(A^(2) .s^(3))`

Discussion

1776.	What is Ohm’s law?
Answer» Ohm’s law : According to Ohm’s law, The electrical resistance R of a conductor is equal to the electric potential difference, V divided by the electric current, I. R = \(\frac{V}{I}\)ohm

1776.

What is Ohm’s law?

Answer»

Ohm’s law :

According to Ohm’s law,

The electrical resistance R of a conductor is equal to the electric potential difference, V divided by the electric current, I.

R = \(\frac{V}{I}\)ohm

Discussion

1777.	In terms of SI base a units, ohm `(Omega)`=_____.
Answer» `kg" "m^(2)s^(-2)A^(-2)`

Discussion

1778.	Unit of ionic mobility is :A. `m^(2)sec^(-1)"volt"^(-1)`B. `msec^(-1)`C. `msec^(-1)`voltD. m `sec^(-1)"volt"^(-1)`
Answer» Correct Answer - A `u=(v)/((dv)/(dx))=(msec^(-1))/(Vm^(-1))=m^(2)V^(-1)sec^(-1)` v=Velocity (m `sec^(-1)`) `(dv)/(dx)=`potential gradient (V `m^(-1)`)

Discussion

1779.	What are SI units of :(a) electrical resistance (b) potential and (c) electric current?
Answer» (a) The SI unit of electrical resistance is Ohm denoted by Ω (omega). (b) The SI unit of potential is volt denoted by V. (c) The SI unit of electric current is ampere denoted by A.

1779.

What are SI units of :(a) electrical resistance (b) potential and (c) electric current?

Answer»

(a) The SI unit of electrical resistance is Ohm denoted by Ω (omega).

(b) The SI unit of potential is volt denoted by V.

(c) The SI unit of electric current is ampere denoted by A.

Discussion

1780.	What is specific conductance or conductivity?
Answer» The reciprocal of specific resistance or resistivity is called specific conductance or conductivity. If ρ is the resistivity then, conductivity = \(\frac{1}{resistivity}\) = \(\frac{1}{ρ}\) Conductivity is denoted by κ (kappa), where κ = \(\frac{1}{ρ}\) It is the conductance of a conductor that is 1m in length and 1m² in cross section area in SI units. (In C.G.S. units, it is the resistance of a conductor that is 1 cm in length and 1 cm² in cross section area.) It is the conductance of a conductor of volume 1m³ (or in C.G.S. units, the volume of 1 cm³).

1780.

What is specific conductance or conductivity?

Answer»

The reciprocal of specific resistance or resistivity is called specific conductance or conductivity.

If ρ is the resistivity then,

conductivity = \(\frac{1}{resistivity}\) = \(\frac{1}{ρ}\)

Conductivity is denoted by κ (kappa), where κ = \(\frac{1}{ρ}\)

It is the conductance of a conductor that is 1m in length and 1m² in cross section area in SI units.

(In C.G.S. units, it is the resistance of a conductor that is 1 cm in length and 1 cm² in cross section area.)

It is the conductance of a conductor of volume 1m³

(or in C.G.S. units, the volume of 1 cm³).

Discussion

1781.	Conductivity of 0.01M NaCl solution is 0.00147 `ohm^(-1) cm ^(-1)`. What happen to this conductivity if extra 100m L of `H_(2)O` will be added to the above solution?A. Increases because of the electrolyte conducts betterB. Decreases because of the increased heatC. Remains unchangedD. First increases and then decreases
Answer» Correct Answer - B The conductivity given is specific conductivity. It decreases with dilution.

Discussion

1782.	How is electrical conductance of a solution denoted ? What are its units?
Answer» The electrical conductance of a solution is denoted by G and it is the reciprocal of resistance, R. G = \(\frac{1}{R}\) The unit of G is siemens denoted by S or Ω^-1. Hence we can write, S = Ω^-1 = AV^-1 = CV^-1S^-1 Where A is ampere and C is coulomb.

1782.

How is electrical conductance of a solution denoted ? What are its units?

Answer»

The electrical conductance of a solution is denoted by G and it is the reciprocal of resistance, R.

G = \(\frac{1}{R}\)

The unit of G is siemens denoted by S or Ω^-1.

Hence we can write,

S = Ω^-1 = AV^-1 = CV^-1S^-1

Where A is ampere and C is coulomb.

Discussion

1783.	What is electrical conductance? What are its units?
Answer» The reciprocal of the electrical resistance of a solution is called the conductance. It is represented by G. ∴ Conductance (G) = \(\frac{1}{Resistance}\) = \(\frac{1}{R}\) The conductance has units of reciprocal of ohm (Ω^-1,ohm^-1 or mho). In SI units, conductance has units as Siemens, (S). (1S = 1Ω^-1 = 1ohm^-1 = 1 mho = AV^-1 = CV^-1S, where C represents electric charge in coulomb, and A represents current strength)

1783.

What is electrical conductance? What are its units?

Answer»

The reciprocal of the electrical resistance of a solution is called the conductance. It is represented by G.

∴ Conductance (G) = \(\frac{1}{Resistance}\) = \(\frac{1}{R}\)

The conductance has units of reciprocal of ohm (Ω^-1,ohm^-1 or mho).

In SI units,

conductance has units as Siemens, (S).

(1S = 1Ω^-1 = 1ohm^-1 = 1 mho = AV^-1 = CV^-1S, where C represents electric charge in coulomb, and A represents current strength)

Discussion

1784.	Conductivity of 0.01M NaCl solution is 0.00147 `ohm^(-1) cm ^(-1)`. What happen to this conductivity if extra 100m L of `H_(2)O` will be added to the above solution?A. IncreaseB. DecreasesC. Remains unchangedD. First increases and then decreases.
Answer» Correct Answer - B On dilution, number of ions/`cm^(3)` decreases. Hence, conductivity decreases.

Discussion

1785.	The potential of hydrogen electrode having a pH=10 isA. 0.59VB. 0.00VC. `-0.59V`D. `-0.059V`
Answer» Correct Answer - C `H^(+)+erarr(1)/(2)H_(2)` `E=E^(@)-(0.0591)/(1)"log"(1)/([H^(+)])=-0.0591xxpH` `=0-0.0591xx10=-0.591V`.

Discussion

1786.	When same quantity of electricity is passed for half an hour, the amount of Cu and Cr deposited are respectivley 0.375g and 0.30g. Radio of electrochemical equivalents of Cu and Cr isA. 0.8B. 1.25C. 2.5D. 1.62.
Answer» Correct Answer - B Weights deposited are in the ratio of their eq. wts. `((W_(2))/(W_(2))=(E_(1))/(E_(2)))` Now Z= E/96500 i.e. Z`prop` E `therefore (W_(1))/(W_(2))=(Z_(1))/(Z_(2))=(0.375)/(0.300)=1.25`

Discussion

1787.	Units of conductance areA. ohmB. mhoC. siemenD. `ohm^(-1)`
Answer» Correct Answer - B,C,D Units of conductance are mho, siemen and `ohm^(-1)`.

Discussion

1788.	A salt bridgeA. allows the flow of current by completing the electrical circuit.B. allows easy intermixing of the ions in the two half cellsC. elimination liquid junction potentialD. maintains the electrical neutrility of the two h alf cells.
Answer» Correct Answer - A,C,D See comprehensive review for details. Answers to similar questions have already been discussed earlier.

Discussion

1789.	When electricity is passed through an electrolyteA. only cations migrateB. only anions migrateC. both cations and anions migrateD. only the solvent molecules migrate.
Answer» Correct Answer - A,C When elecricity is passed through an electrolyte either only cations migrate or both cations and anions migrate . As anions are much bigger than cations, only anions cannot migrate.

Discussion

1790.	An electric current of 100 mA is passed through an electrolyte for 2 hours, 20 minutes and 20 seconds. Find the quantity of electricity passed.
Answer» Given : Electric current = I = 100 mA = 100 × 10^-3A = 0.1 A Time = t = 2 hrs + 20 min + 20 s = 2 × 60 × 60 + 20 × 60 + 20 = 8420 s The quantity of electricity = Q = ? Q = I × t = 0.1 × 8420 = 842 C ∴ Quantity of electricity passed, Q = 842 C

1790.

An electric current of 100 mA is passed through an electrolyte for 2 hours, 20 minutes and 20 seconds. Find the quantity of electricity passed.

Answer»

Given :

Electric current = I = 100 mA

= 100 × 10^-3A

= 0.1 A

Time = t = 2 hrs + 20 min + 20 s

= 2 × 60 × 60 + 20 × 60 + 20

= 8420 s

The quantity of electricity = Q = ?

Q = I × t

= 0.1 × 8420

= 842 C

∴ Quantity of electricity passed,

Q = 842 C

Discussion

1791.	Obtain a charge on one electron from Faraday’s value.
Answer» One Faraday is the electric charge on one mole of electrons (6.022 × 10²³ electrons). 1 Faraday = 96500 (per mol of electrons). Hence the charge on one electron is, change on one electron = \(\frac{96500}{6.022\times 10^{23}}\) = 1.602 × 10^-9C.

Discussion

1792.	Define Faraday.
Answer» Faraday : It is defined as the quantity of the electric charge carried by one mole of electrons. It has value, 1F = 96500 C/mol

1792.

Define Faraday.

Answer»

Faraday : It is defined as the quantity of the electric charge carried by one mole of electrons.

It has value,

1F = 96500 C/mol

Discussion

1793.	Define and explain the following electrical units : (1) Coulomb (2) Ampere (3) Volt (4) Joule (5) Ohm.
Answer» (1) Coulomb : It is a quantity of electricity obtained when one ampere current flows for one second. It is the unit of quantity of electricity. Q = I × t Coulomb (C) Where Q is the charge or quantity of electricity in coulombs. (2) Ampere : It is a strength of an electric current obtained when one coulomb of electricity is passed through a circuit for one second. ∴ I = Q/t (3) Volt : It is the potential difference between two points of an electric conductor required to send a current of one amphere through a resistance of one ohm. ∴ V = I × R Where V is the potential difference in volts and R is the resistance of a conductor in ohms. (4) Joule : It is the electrical work or energy produced when one coulomb of electricity is passed through a potential difference of one volt. ∴ Electrical work = Q × V J Where Q is electrical charge in coulombs and V is the potential difference. (5) Ohm : It is the resistance of an electrical conductor across which when potential difference of 1 volt is applied, a current of one ampere is obtained. It has units, Ω or per siemens.

1793.

Define and explain the following electrical units : (1) Coulomb (2) Ampere (3) Volt (4) Joule (5) Ohm.

Answer»

(1) Coulomb : It is a quantity of electricity obtained when one ampere current flows for one second.

It is the unit of quantity of electricity.

Q = I × t Coulomb (C)

Where Q is the charge or quantity of electricity in coulombs.

(2) Ampere : It is a strength of an electric current obtained when one coulomb of electricity is passed through a circuit for one second.

∴ I = Q/t

(3) Volt : It is the potential difference between two points of an electric conductor required to send a current of one amphere through a resistance of one ohm.

∴ V = I × R

Where V is the potential difference in volts and R is the resistance of a conductor in ohms.

(4) Joule : It is the electrical work or energy produced when one coulomb of electricity is passed through a potential difference of one volt.

∴ Electrical work = Q × V J

Where Q is electrical charge in coulombs and V is the potential difference.

(5) Ohm : It is the resistance of an electrical conductor across which when potential difference of 1 volt is applied, a current of one ampere is obtained. It has units, Ω or per siemens.

Discussion

1794.	For which of these oxidation/reduction pairs will the reduction potential vary with pH? I `AmO_(2)^(2+)//AmO_(2)^(+)` II. `AmO_(2)^(2+)//Am^(4+)` III `Am^(4+) // Am^(2+)`A. I onlyB. II onlyC. I and II onlyD. I, II only III
Answer» Correct Answer - B `4H^(+) + AmO_(2)^(2+) + 2e^(-) rarr Am^(4+) + 2H_(2)O` It includes `H^(+)` ions, hence the electrode potential depends on pH

Discussion

1795.	`Pt \|{:((H_(2))),(1atm):}:\| pH = 2:\|:\|:pH =3 \|:{:((H_(2))Pt),(1atm):}:\|`. The cell reaction for the given cell is:-A. spontaneousB. non-spontaneousC. equilibriumD. none is correct
Answer» Correct Answer - B `E_("cell")=0.0591` `(P_(H_(1))-P_(H_(2)))=0.0591(2-3)-0.0591` = non-spontaneous

Discussion

1796.	`M^(2+) +2e rarr M. 0.275 g` of metal `M` is deposited at the cathode due to passage of `1A` of current for `965s.` Hence atomic weight of the metal `M` is:-A. `27.5`B. `55.0`C. `110.0`D. `13.75`
Answer» Correct Answer - B W = Z I t , `0.275=(at xx14xx965)/(2xx96500)` `0.550xx100 = at` , at = 55

Discussion

1797.	For the concentration cell `Cu\|Cu^(2+)(C_(1))\|\|Cu^(2+)(C_(2))\|Cu,DeltaG` will be negative ifA. `C_(1)=C_(2)`B. `C_(1)gtC_(2)`C. `C_(2)gtC_(1)`D. None of these.
Answer» Correct Answer - C A concentration cell (reversible w.r.t. cation) is spontaneous `(DeltaGlt 0"or" E_("cell"gt0)` if `C_(2)gtC_(1)` `E_("cell")=(0.0591)/(2)"log"(C_(2))/(C_(1)))` If `C_(2)gtC_(1).E_("cell")gt0`

Discussion

1798.	For `M^(2+)+2e^(-)rarrM:0.275` g of metal M is depoited at cathode on passing 1 A of current for 965 s. the atomic mass of metal M isA. 55B. 27.5C. 13.75D. 110
Answer» Correct Answer - A Let at, mass of `M=m` `M^(2+)+ 2e^(-)rarrm` `2xx96500C mg` `q=1xxt=1xx965=965C` `2xx96500C` deposit M=mg 965C will deposit `M=(m)/(200)g` A.T.Q. `(m)/(200)=0.275` `m=27.5xx2=55.0g` `therefore` At. Mass of M=55

Discussion

1799.	All the energy released from the reaction `XrarrY,Delta_(r)G^(@)=-193kJ"mol"^(-1)` is used for oxidising `M^(@)` as `M^(+)rarrM^(3+)+2e^(-),E^(@)=-0.25V`. Under standard conditions, the number of moles of `M^(+)`oxidised when one mole of X is converted to Y is `[F=96500C "mol"^(-1)]`
Answer» Given data is : `X to Y , Delta_(r)G^(@)=-193" kJ "mol^(-1)` `M^(+) to M^(3+)+2e^(-) , E^(@)=-0.25" V"` `193" kJ"` of energy is used to oxidise x mol of `M^(+)` ions to `M^(3+)` ions. For 1 mole of `M^(+)` `DeltaG^(@)=-nFE^(@)=(- mol)xx96500" C")xx(0.25" V")` `=48250" J "mol ^(-1)=48.25" kJ "mol^(-1)` `48.25" kJ"` of energy is used oxidise =1 mol 193.0 kJ of energy is used to oxidise `=(1 mol)xx((193.0" kJ"))/((48.25" kJ"))=4 mol " ":. x=4`

Discussion

1800.	Consider the reaction : `H_(2)(g),Pt\|H^(+)(aq) , E=0.1" V"` The pH of the solution is
Answer» `H_(2)(g)+2e^(-) to2H^(+)(aq)` `E=E^(@)-(0.0591)/(2)"log"[H^(+)]^(2)` `0.1=0-(0.0591)/(2)2log[H^(+)]` `0.1=0-0.0591" log " [H^(+)]` `-log[H^(+)]=(0.1)/(0.0591)=1.69` `pH=-log[H^(+)]=1.69=2`

Discussion

Explore topic-wise InterviewSolutions in .

In the electrolysis of an aqueous solution of NaF , the product obtained at the cathode and anode respectively areA. `Na , F_(2)`B. `Na, O_(2)`C. `H_(2) , O_(2)`D. `H_(2) , F_(2)`

z On the electrolysis of an aqueous solution of NaF using the gram equivalence of an electrolyteA. Na is obtained at cathodeB. `F_(2)` is obtained at anodeC. `H_(2)` is obtained at cathodeD. `O_(2)` is obtained at anode.

Conductivity of an aqueous solution of `0.1M HX` (a weak mono-protic acid) is `5 xx 10^(-4) Sm^(-1)` Given: `^^_(m)^(oo) [H^(+)] = 0.04 Sm^(2) mol^(-1): ^^_(m)^(oo) [X^(-)] = 0.01 Sm^(2)mol^(-1)` Find `pK_(a)[HX]`

A current of 0.4 ampere is passed for 30 minutes through a voltameter containing `CuSO_(4)` solution. The weight of Cu dpeosited will beA. 3.18gB. 0.318gC. 0.296gD. 0.150g

ELECTROCHEMICAL CELLA. chemical energy to electrical energyB. electrical energy to chemical energyC. chemical energy to mechanical energyD. electrical energy to mechanical energy .

Which of the following statements is correct ?A. Oxidation number of oxygen in `K_(2)O` is + 1B. The specific conductance of an electrolyte solution decreases with increase in dilutionC. `Sn^(2+)` oxidises `Fe^(3+)`D. `Zn|ZnSO_(4)` is reference electrode .

Measurement of the coonductivity of electrolytic solutions is done with the helpof Wheatstone bridge using a/anA. batteryB. induction coilC. conductivity cellD. both (2) and (3)

A conductivity cell is platinized toA. avoid temperature effectB. avoid capacitance of the cellC. avoid polarization effectD. prolonf its life

If x is the conduction of 1 `cm^(3)` of the solution and C is the concentration of the solution . The molar conductivity of the solution is given byA. `(1000 C)/(x)`B. `(1000 x)/(N)`C. `(C)/(x)`D. `(x)/(C)`

A conductivity cell is platinized toA. to prolong its lifeB. to avoid polarisation effectC. to avoid capacitance of the cellD. to avoid temperature effect .

The variation of `^^_(m)` of acetic acid with concentration is correctly represented byA. B. C. D.

The degree of ionization increasesA. with increase in concentration of the solutionB. on addition of excess of water to the solutionC. on decreasing the temperature of the solutionD. on stirring the solution vigorousoly.

In terms of SI base units , ohm stands forA. `(kg m^(2)) //(As)`B. `(kg m^(2))//(A^(2) s^(2))`C. `(kg m^(2))//(A^(2) s^(2))`D. `(kg m^(2)) // (A^(2) s^(3))`

What is Ohm’s law?

In terms of SI base a units, ohm `(Omega)`=_____.

Unit of ionic mobility is :A. `m^(2)sec^(-1)"volt"^(-1)`B. `msec^(-1)`C. `msec^(-1)`voltD. m `sec^(-1)"volt"^(-1)`

What are SI units of :(a) electrical resistance (b) potential and (c) electric current?

What is specific conductance or conductivity?

How is electrical conductance of a solution denoted ? What are its units?

What is electrical conductance? What are its units?

Conductivity of 0.01M NaCl solution is 0.00147 `ohm^(-1) cm ^(-1)`. What happen to this conductivity if extra 100m L of `H_(2)O` will be added to the above solution?A. IncreaseB. DecreasesC. Remains unchangedD. First increases and then decreases.

The potential of hydrogen electrode having a pH=10 isA. 0.59VB. 0.00VC. `-0.59V`D. `-0.059V`

When same quantity of electricity is passed for half an hour, the amount of Cu and Cr deposited are respectivley 0.375g and 0.30g. Radio of electrochemical equivalents of Cu and Cr isA. 0.8B. 1.25C. 2.5D. 1.62.

Units of conductance areA. ohmB. mhoC. siemenD. `ohm^(-1)`

A salt bridgeA. allows the flow of current by completing the electrical circuit.B. allows easy intermixing of the ions in the two half cellsC. elimination liquid junction potentialD. maintains the electrical neutrility of the two h alf cells.

When electricity is passed through an electrolyteA. only cations migrateB. only anions migrateC. both cations and anions migrateD. only the solvent molecules migrate.

An electric current of 100 mA is passed through an electrolyte for 2 hours, 20 minutes and 20 seconds. Find the quantity of electricity passed.

Obtain a charge on one electron from Faraday’s value.

Define Faraday.

Define and explain the following electrical units : (1) Coulomb (2) Ampere (3) Volt (4) Joule (5) Ohm.

For which of these oxidation/reduction pairs will the reduction potential vary with pH? I `AmO_(2)^(2+)//AmO_(2)^(+)` II. `AmO_(2)^(2+)//Am^(4+)` III `Am^(4+) // Am^(2+)`A. I onlyB. II onlyC. I and II onlyD. I, II only III

`Pt |{:((H_(2))),(1atm):}:| pH = 2:|:|:pH =3 |:{:((H_(2))Pt),(1atm):}:|`. The cell reaction for the given cell is:-A. spontaneousB. non-spontaneousC. equilibriumD. none is correct

`M^(2+) +2e rarr M. 0.275 g` of metal `M` is deposited at the cathode due to passage of `1A` of current for `965s.` Hence atomic weight of the metal `M` is:-A. `27.5`B. `55.0`C. `110.0`D. `13.75`

For the concentration cell `Cu|Cu^(2+)(C_(1))||Cu^(2+)(C_(2))|Cu,DeltaG` will be negative ifA. `C_(1)=C_(2)`B. `C_(1)gtC_(2)`C. `C_(2)gtC_(1)`D. None of these.

For `M^(2+)+2e^(-)rarrM:0.275` g of metal M is depoited at cathode on passing 1 A of current for 965 s. the atomic mass of metal M isA. 55B. 27.5C. 13.75D. 110

All the energy released from the reaction `XrarrY,Delta_(r)G^(@)=-193kJ"mol"^(-1)` is used for oxidising `M^(@)` as `M^(+)rarrM^(3+)+2e^(-),E^(@)=-0.25V`. Under standard conditions, the number of moles of `M^(+)`oxidised when one mole of X is converted to Y is `[F=96500C "mol"^(-1)]`

Consider the reaction : `H_(2)(g),Pt|H^(+)(aq) , E=0.1" V"` The pH of the solution is