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2351.

Give an example of secondary cell.

Answer»

Lead storage battery.

2352.

Calculate the emf of the cell in which the following reactions takes placeNi(s) + 2 Ag+ (0.002M) → Ni2+ (0.16 M) + Ag(s) Given that Eo cell = 1.05 V.

Answer»

E cell =  E0 -0.0591/n x log 0.16/(2 x 10-3 ) 2

= 1.05 -0.0591/2 x log 16 x 104 /4

=0.91 volt

2353.

Define electrode.

Answer»

Electrode : The arrangement consisting of a metal rod dipping in an aqueous solution or molten electrolyte containing ions and conduct electric current due to oxidation or reduction half reactions occurring on its surface is called an electrode.

The electrodes which take part in the reactions are called active electrodes while those which do not take part in the reactions are called inert electrodes.

2354.

What would happen if the protective tin coating over an iron bucket is broken in some places? 

Answer»

Iron will corrode faster as the oxidation potential of Fe is higher than that of tin.

2355.

Define : (a) Anode (b) Cathode.

Answer»

(a) Anode : An electrode of an electrochemical cell, at which oxidation half reaction occurs due to the loss of electrons from some species is called an anode.

(b) Cathode : An electrode of an electrochemical cell at which reduction half reaction occurs due to gain of electrons by some species is called a cathode.

2356.

Ostwald dilution law of weak electrolyte `HA` can be given as :A. `K_(a) = (C.alpha^(2))/((1-alpha))`B. `K_(a) = C.alpha^(2) if alpha rarr 0`C. `K_(a) = (CLambda^(2))/(Lambda_(0)(Lambda_(0)-Lambda))`D. All of these

Answer» Correct Answer - D
`K_(a) = (Calpha)^(2)/(1-alpha)`
If `alpha = (Lambda)/(Lambda_(0))`
`K_(a) = (CLambda^(2)//Lambda_(0)^(2))/((1 - Lambda//Lambda_(0)) ) = (CLambda)/((Lambda_(0) - Lambda).Lambda_(0))`
Also if `alpha rarr 0, (1 - alpha) = 1`
`:. K_(a) = C.alpha^(2)`
2357.

What are the types of electrochemical cells?

Answer»

There are two types of electrochemical cells as follows : 

1. Electrolytic cells 

2. Voltaic or galvanic cells.

2358.

`Zn` acts as sacrifical or cathodic protect iont to prevent rusting of iron becauseA. `E_(OP)^(@)` of `ZnltE_(OP)^(@)`of `Fe`B. `E_(OP)^(@)` of `ZngtE_(OP)^(@)`of `Fe`C. `E_(OP)^(@)` of `Zn=E_(OP)^(@)`of `Fe`D. `Zn` is cheaper than iron

Answer» Correct Answer - B
`Zn` is oxidised in place of iron and iron acts as cathode if `Zn` plating or galvanisation of iron is made and thus called cathodic protection. Since `Zn` is oxidised and thus also called sacrificial protection.
2359.

Define : (1) Electrolytic cell (2) Voltaic or galvanic cell.

Answer»

(1) Electrolytic cell : An electrochemical cell in which a non-spontaneous chemical reaction is forced to occur by passing direct electric current into the solution from the external source and where electrical energy is converted into chemical energy is called an electrolytic cell. 

E.g. voltameter, electrolytic cell for deposition of a metal.

(2) Voltaic or galvanic cell : An electrochemical cell in which a spontaneous chemical reaction occurs producing electricity and where a chemical energy is converted into an electrical energy is called voltaic cell or galvanic cell. 

E.g. Daniell cell, dry cell, lead storage battery, fuel cells, etc.

2360.

Which metals can be used in the cathodic protection of `Fe` against rusting.

Answer» A metal which is placed above iron in the reactivity series.
2361.

Define : Galvanic cell or voltaic cell.

Answer»

Galvanic or voltaic cell : An electrochemical cell which is used to produce electrical energy by a spontaneous chemical reaction inside it is called an electrochemical cell. In this chemical energy is converted into electrical energy. 

Example : Daniell cell.

2362.

What is Cathodic protection?

Answer»

Instead of coating more reactive metal or iron, the use of such metal is made as sacrificial anode. Connecting Iron with more reactive metal with the help of a wire.

2363.

Define : Half cell or Electrode.

Answer»

Half cell or Electrode : It is a metal electrode dipped in the electrolytic solution and capable of establishing oxidation reduction equilibrium with one of the ions of electrolyte solution and develop electrode potential. 

E.g. Zn in ZnSO4 solution.

2364.

Molar conductance of 2 M H2A acid is 10 S cm2 mol–1. Molar conductance of H2A at infinite dilution is 400 S cm2 mol–1. Which statement is/are correct? (1) Degree of dissociation is 2.5% and pH of solution is 1.3 (2) Degree of dissociation is 4 and pH of solution is 1.4 (3) Dissociation constant of H2A is 6.24 × 10–5 (4) Dissociation constant of H2A as per H2H A ⇋ is 2.56 x 10–4

Answer»

(1) Degree of dissociation is 2.5% and pH of solution is 1.3 

2365.

Differences between electrochemical reaction and electrolysis. 

Answer»
Electrochemical reactionElectrolysis
Chemical reaction produce electricity.Electricity causes chemical reaction to take place.
It is spontaneous, ∆G is -veIt is non-spontaneous, ∆G is +ve

2366.

Which of following is/are correct? (1) The metallic conduction is due to the movement of electrons in the metal (2) The electrolytic conduction is due to the movement of ions in the solution (3) The metallic conduction increases with increase in temperature whereas electrolytic conduction decreases with increase in temperature (4) None of these

Answer»

Correct option  (1, 2)

Explanation:

The metallic conduction is due to the presence of electrons in the metal and electrolytic conduction is due to the movement of ions in the solution.

2367.

What are the functions of a salt bridge?

Answer»

The functions of a salt bridge are :

1. It maintains the electrical contact between the two electrode solutions of the half cells.

2. It prevents the mixing of electrode solutions.

3. It maintains the electrical neutrality in both the solutions of two half cells by a flow of ions.

4. It eliminates the liquid junction potential.

2368.

Write any two functions of salt bridge.

Answer»

The functions of a salt bridge are :

  • It maintains the electrical contact between the two electrode solutions of the half cells. 
  • It prevents the mixing of electrode solutions. 
  • It maintains the electrical neutrality in both the solutions of two half cells by a flow of ions.
  • It eliminates the liquid junction potential.
2369.

1. Write the relation between molar conductance and molarity of an electrolyte solution.2. λm∞ of NaCl, HCl and CH3COONa are 126.4, 425.9 and 91 S cm2mol-1 respectively. Calculate the λm∞ of acetic acid. 

Answer»

1. Λm\(\frac{1000k}{M}\)

2.  Λ° CH3COOH = Λ° CH3COONa +  Λ° HCl  -  Λ° Nacl

= 91 + 425.9 – 126.4

= 390.5 Ω-1 cm2 mol-1

2370.

For electrolyte Ax By which is/are not correct relation between molar conductivity (ΛM) and equivalent conductivity (Λeq) (1) ΛM = xy Λeq (2) Λeq = xyΛM (3) xΛM = yΛeq (4) yΛM = xΛeq

Answer»

Correct option (2, 3, 4) 

Explanation:

For the electrolyte Ax By

n-factor = xy 

ΛM = (xy) Λeq

Only 1st option is correct & others are incorrect option.

2371.

State the Functions of Salt Bridge.

Answer»

Functions of Salt Bridge :

1. Salt Bridge maintains electrical neutrality in solutions. 

2. It completes circuit internally. 

2372.

What are the types of Commercial cells?

Answer»

These are of 4 types :

(1) Dry cell (primary cell)

(2) Mercury cell/Button Cell

(3) Lead storage battery

(4) Fuel cell

2373.

Every conductivity cell has a definite value for its cell constant.1. What is cell constant? 2. How is it related with specific conductance? 3. Calculate the cell constant of a conductivity cell containing 0.05M KCl solution, whose resistance is 100 ohm and conductivity is 0.0075 ohm-1 cm-1

Answer»

1. The quantity ℓ/A is called cell constant.

A = Area of cross section of the electrodes.

ℓ = distance of separation of two electrodes.

2. Specific conductance = Cell constant × Conductance

3. Cell constant = Specific conductance × Resistance

= 0.0075 ohm 1 cm-1 × 100 ohm = 0.75 cm-1

2374.

Consider the equation, λm∞HCl = λm∞H+ + λm∞Cl-1. Is the above equation correct?2. Explain the law.3. Write the application of this law.

Answer»

1. The above equation is correct.

2. Kohlrausch’s law. Molar conductance at infinite dilution of an electrolyte is the sum of the molar ionic conductivities at infinite dilution of the cations and anions present in the solution.

3. The application of this law :

  • To find the molar conductivity at infinite dilution (Λ°m) of weak electrolytes.
  • To find the degree of dissociation (α) of weak electrolytes. α = \(\frac{Λ_m}{Λ^0_m} \)
  • To determine the dissociation constant of weak electrolytes. Ka = \(\frac{c^2\alpha}{1-\alpha}\)  
2375.

Give the Applications of electrochemical series. 

Answer»

(i) To compare the relative oxidizing and reducing powers- A chemical species with higher value of E° will be stronger oxidizing agent and a chemical species with more – ve E° will be stronger reducing agent.

(ii) To compare the relative activities of metals- A metal with more –ve E° value will be more reactive 

(iii) To calculate the E°cell of given cell- 

E0cell = E°cathode - E°anode or E°cell = E°RHS - E°LHS 

N.B. - E° for SHE is taken 0 (Zero) 

(iv) To predict whether a metal will displace hydrogen from its compound or not- All the chemical species which have –ve E° value will displace hydrogen from its compounds.

M → Mn+ + ne- (electron lost is gained by H+ ion to undergo reduction and H2 is formed) 

Usefulness of Nernst equation to calculate Ecell at different conc. And temp. 

Definition of strong electrolyte, weak electrolyte, limiting molar conductivity.

2376.

1. Explain the working of a H2 – O2 fuel cell. 2. What are the advantages of fuel cells over other cells?

Answer»

1. In H2 – O2 fuel cell, hydrogen and oxygen are bubbled through porous carbon electrodes into concentrated aqueous NaOH solution, which acts as the electrolyte. The electrode reactions are,

Anode : 2H2(g) + 4 OH-(aq) → 4 H2O(l) + 4\(\bar e\)

Cathode : O2(g) + 2 H2O(l) + 4\(\bar e\) → 4 OH-(aq)

Overall reaction : 2H2(g) + O2(g) → 2 H2O(l)

2. The advantages of fuel cells over other cells 

  • They are pollution free. 
  • More efficient than conventional methods. 
  • Runs continuously as long as the reactants are supplied. 
  • Electrodes are not affected. 
2377.

Give the Advantages of Fuel Cells.

Answer»

Advantages of Fuel Cells : 

1. As works as long as we supply Hydrogen & oxygen gas.

2. It causes no pollution 

3. Its efficiency is 70% 

4. Its By product (Water) is also useful.

2378.

Fuel cells are special types of galvanic cells. (a) (i) Whataregalvaniccells?(ii) Write any two advantages of fuel cells.(b) Write the electrode reactions is H2 – O2 fuel cells.

Answer»

(a) (i) These are devices for converting chemical energy into electrical energy. The decrease in free energy in a spontaneous chemical process appears as electrical energy. e.g., Daniell oeil.
(ii) (1) Fuel cells are pollution free.

(2) Fuel cells are highly efficient (about 70%) compared to thermal plants (about 40%)

(3) Fuel cells run continuously as long as the reactants are supplied. 

(b) At cathode:

O2(g) + 2H2O(l) + 4e → 4OH(aq)

At anode:

2H2(g) + 4OH(aq) → 4H2O(l)

2379.

List the  Advantages of fuel cell.

Answer»

Advantages of fuel cell:

  • Long lasting, 
  • More efficient, 
  • Eco friendly, 
  • Give useful by products
2380.

Iron does not rust even if zinc coating on its surface is broken but the same is not true when coating is of tin.

Answer» The `E^(@)` values of these metals are listed as follows :
`Zn^(2+)(aq)+2e^(-) to Zn(s) , " " E^(@)=-0.76" V"`
`Fe^(2+)(aq)+2e^(-) to 2Fe(s) , " "E^(@)=-0.44" V"`
`Sn^(2+)(aq)+2e^(-) to Sn(s) , " "E^(@)=-0.14" V"`
The data shows that zinc is more reactive than iron. This means that if a crack appears on the surface of iron coated with zinc even then zinc will take part in the redox reaction and not iron.In other words, zinc will be corroded in preference to iron. But same is not the case with tin. It is less reactive than iron. If a crack appears on the surface of iron coated with tin, then iron will take part in the redox reaction and not tin. Therefore, iron will be corroded under these circumstances.
2381.

What are the Corrosion?

Answer»

It is a process of eating away of metals on their surfaces, it is an unwanted process as it results in loss of mass of metals. In this process metal surface reacts with atmospheric oxygen to form a layer of oxide. It is an electrochemical reaction.

2382.

Arrange the following metals in order of their increasing reducing power. K+ /K = - 2.93 V, Ag+ /Ag = 0.80 V, Hg2+/Hg = 0.79 V, Mg2+/Mg = - 2.37 V Cr3+/Cr = - 0.74 V. 

Answer»

Lower the reduction potential, easily if gets oxidized hence greater is the reducing power, therefore the increasing order of reducing power is Ag < Hg < Cr < mg < K. 

2383.

Given the `E_(Fe^(3+)//Fe)^(@)` and `E_(Fe^(2+)//Fe)^(@)` are `-0.36 V` and `0.-439 V`, respectively. The value of `E_(Fe^(3+)//Fe^(2+)) ^(@)` would be :A. `(-0.36 - 0.439)V`B. `[3(-0.36) + 2(-0.439)]V`C. `(0.36 + 0.439)V`D. `[3(-0.36) - 2(-0.439)]V`

Answer» Correct Answer - D
`{:(Fe^(3+)+3e rarr Fe l-Delta G_(1)^(@)=-0.36 xxF xx3),(underset(-)(Fe^(3+))+ underset(-)(2e) rarr underset(-)(Fe),-DeltaG_(2)^(@)=underset(+)(-)0.439 xxFxx2),(bar(Fe^(3)+e rarr Fe^(2+),-Delta G^(@)=[3 xx(0.36 F)+2xx(0.439 F)])),( :. +1xxF xxE^(2)=[3xx(-0.36F)+2xx (0.439F)]),(or E^(@)=[3xx(-0.36)+2xx0.439]V):}`
`:. +1 xx F xx E^(@) = [3 xx (-0.36F) + 2 xx (0.439F)]` ltbRgt or ` E^(@) = [3 xx (-0.36) + 2 xx 0.439]V`
2384.

Which one of the following statement is true for a electrochemical cellA. `H_(2)` is cathode and Cu is anodeB. `H_(2)` is anode and `Cu` is cathodeC. Reduction occurs at `H_(2)` electrodeD. Oxidation occurs at Cu electrode

Answer» Correct Answer - B
In electrochemical cell `H_(2)` releases at anode and Cu is deposited at the cathode.
2385.

The acid used in lead storage battery is:-A. `H_(2)SO_(4)`B. `H_(3)PO_(4)`C. `HCl`D. `HNO_(3)`

Answer» Correct Answer - A
Dil. `H_(2)SO_(4)` is used in lead in lead storage battery as electrolyte.
2386.

The process used in purifying metals by electrolysis is called _____A. electroplatingB. electrometallurgyC. electrorefiningD. electrodeposition

Answer» Correct Answer - C
2387.

When the total cell emf of a voltaic cell is greater than zero, which of the following is true about the reaction quotient Q and free energy change `DelatG` for the cell reaction ?A. Q is less than one and `DeltaG` is greater than zeroB. Q greater than one and `DeltaG` is greater than zero.C. Q is less than one and `DeltaG` is less than zero.D. Q is greater than one and `DeltaG` is less than zero.

Answer» Correct Answer - C
(c ) `DeltaG=DeltaG^(@)+RT" In "Q`
It `E_(cell)` is +ve,`DeltaG` is -ve
(Cell working is feasible)
Similarly `E_(cell)` is -ve, `DeltaG^(@)` is -ve.
`:. RT " In "`Q is also -ve
or `Q lt t`.
2388.

Emf of a voltaic cell can be given by ______A. SRP of anode - SRP of cathodeB. SRP of cathode + SRP of anodeC. SOP of catode - SOP of anodeD. SOP of cathode + SOP of anode

Answer» Correct Answer - A
2389.

In the electroplating of iron by nickel, nickel sulphate solution is taken as an electrolyte. What is the acid that is added to the electrolyte during the process ?A. HClB. HCNC. `HNO_(3)`D. `H_(2)SO_(4)`

Answer» Correct Answer - D
2390.

Ratio of number of Faradays of electricity required to deposit magnesium, aluminium and sodium in equimolar ratio on electrolysis of their respective molten salt is _____A. `2 : 3 : 2`B. `2 : 1 : 1`C. `2 : 3 : 1`D. `4 : 6 : 1`

Answer» Electrode reactions involved are
`Mg^(+2) + 2e^(-) rarr Mg`
`Al^(+3) + 3e^(-) rarr Al`
`Na^(+) + e^(-) rarr Na`
1 mole of Mg requires 2 Faradays
1 mole of Al requires 3 Faradays.
1 mole of sodium requires 1 Faraday
Ratio of number of Faradays `= 2 : 3: 1`
2391.

Equimolar solution of zinc sullphate and ferric sulphate are subjected to electrolysis. What is the ration of Faradays of electricity requried for depositoin of one mole each of zinc and iron ?A. `1 : 2`B. `2 : 3`C. `1 : 3`D. `3 : 2`

Answer» Correct Answer - B
2392.

100 mL of 0.05 M aqueous `CuSO_(4)` solution is subjected to electrolysis. Calculate the quantity of electricity required for the deposition of entire copper at the cathode.A. 0.005 FB. 0.01 FC. 0.1 FD. 0.05 F

Answer» `(W_(Cu))/(63.5) xx (1000)/(100) = 0.5 M`
`rArr W_(Cu) = (0.05 xx 63.5)/(10) = 0.3175 g`
NO. of equivalents `= (0.3175)/(31.75) = 0.01`
`:.` No of Faradays is 0.01 F
2393.

What is the current efficiency of an electro-deposition of copper is deposited of copper metal from `CuSO_(4)` solution in which `9.80 g` copper is deposited by the passage of `5` ampere current for `2` hour?

Answer» `(w)/(E) = (i.t)/(96500)`
`(9.8)/(63.5//2) = (i xx 2 xx 60 xx 60)/(96500)`
`i = 4.14` ampere
`:.` Current efficiency `= ("Theoretical valule of I")/("Practical value of I") xx 100`
`= (4.14)/(5) xx 100 = 82.8%`
2394.

Calculating The Degree of ionization: The molar conductance of an infinitely dilute solution `NH_(4)Cl` is `150 S cm^(2) mol^(-1)` and the limiting ionic conductacnes of `Cl^(-)` and `OH^(-)` ions are `76` and `198 S cm^(2) mol^(-1)` respectively. If the molar conductivity of a `0.01 M` solution of `NH_(4)OH` is ` 9.6 S cm^(2) mol^(-1)`, what will be its degree of ionization Strategy: At any concentration `C`, if `alpha` is the degree of ionization then it can be approximented to the ratio of molar conductivity `Lambda_(m)^(c)` at the concentration `C` to limiting molar conducting, `Lambda_(m)^(0)`. We are given `Lambda_(m)^(c)` but we need to find `Lambda_(m)^(0)`.

Answer» According to Kohlrausch law
`Lamda_(m)^(0)(NH_(4)Cl)=lamda_(NH_(4)^+)^(0)+lamda_(Cl^(-))^(0)`
Therefore
`lamda_(NH_(4)^(+))^(0)=Lamda_(m)^(0)(NH_(4)Cl)-lamda_(Cl^(-))^(0)`
`=(150-76)S cm^(2) mol^(-1)`
`=74 S cm^(2)mol^(-1)`
According to Kohlrausch law
`Lamda_(m)^(0)(NH_(4)OH)=lamda_(NH_(4)^(+))^(0)+lamda_(OH^(-))^(0)`
`=(74+198)S cm^(2) mol^(-1)`
`= 272 S cm^(2) mol^(-1)`
According to Equation (3.30)
`alpha=(Lamda_(m)^(c))/(Lamda_(0)^(m))`
`=(9.6 "S cm"^(2) "mol"^(-1))/(272 "S cm"^(2) "mol"^(-1))`
`=0.0353`
2395.

The equivalent conductances of an infinitely dilute solution `NH_(4)CI` is 150 and the ionic conductances of `OH^(-)` and `CI^(-)` ions are 198 and 76 respectively. What will be the equivalent conductance of the solution of `NH_(4)OH` at infinite dilution. if the equivalent conductance of a `0.01N` solution `NH_(4)OH` is `9.6, what will be its degree of dissociation.

Answer» Correct Answer - `272,0.0353`
`Lamda_(NH_(4)Cl)^(@) = lamda_(NH_(4)^(+))^(@)+lamda_(Cl^(-))^(@)`
`:. lamda_(NH_(4)^(+))^(@) =Lamda _(NH_(4)Cl)^(@) -lamda_(Cl^(-))^(@)=150 -76 =74`
`:. Lamda_(NH_(4)OH)^(@) =lamda_(NH_(4)^(+))^(@) + lamda_(OH^(-))^(@) =74 +198 =272`
Further,
`= (9.6)/(272)=0.0353`.
2396.

The ioisation constant of a weak electroluytes is `25 xx 10^(-6)` and the equivalent conductance of its ` 0.01M` solution is `19.6 S cm^2 eq^(-1)`. The equivalent conductance at infinited dilution of the electrolyte is ` S cm^2 sq^(-1)`. is .A. ` 50`B. ` 196`C. `392`D. ` 384`

Answer» Correct Answer - C
`{:(HA,hArr,H^(+),+,A^(-)),(C,,0,,0),(C-Calpha,,Calpha,,Calpha):}`
Or `K= C alpha^2`
Or `25 xx 10^(-6) = 10^(-2) . alpha^2`
Or `alpha = 5 xx 10^(-2)`
also `alpha= (lambda_v)/(lambda_("infty")) = (19.6)/(lambda_("infty")) = (19.6)/(5 xx10^(-2)) = 392`.
2397.

For `H^(+)` and `Na^(+)` the values of `lambda^(oo)` are `349.8` and `50.11`. Calculate the mobilities of these ions and their velecities if they are in a cell in which the electrodes are 5 cm apart and to which a potential of 2 volts is applied.

Answer» Correct Answer - `mu^(@)H^(+) = 3.62 xx 10^(-3) cm^(2) volt^(-1)sec^(-1), mu_(Na)^(@) = 5.20 xx 10^(-14)cm^(2) volt^(-1) sec^(-1)`
We have
`u_(H^(+))^(@)=(lamda_(H^(+))^(@))/(F)=(349.8)/(96500)=3.62 xx 10^(-3) 10^(-3) cm^(2) "volt"^(-1) s^(-1)`
`u_(Na^(+))^(@)=(lamda_(Na^(+))^(@))/(F)=(50.11)/(96500)=5.20 xx 10^(-4) "cm"^(2) "volt"^(-1) s^(-1)`
Further,we know that
`u^(@)=("ionic velocity(cm/s)")/("pot. diff.(volt)/distance between the electrodes (cm)")`
`:.` velocity of `H^(+)=3.62 xx10^(-3) xx(2)/(5) =1.45 xx 10^(-3) "cms"^(-1)`
velocity of `Na^(+)=5.20 xx 10^(-4) xx (2)/(5) =2.08 xx 10^(-4) "cms"^(-1)`.
2398.

Which of the following is correct ?A. Conductivity of a solution decrease with dilution, whereas molar conductivity inceases with dilutionB. Conductivity of a solution increases with dilution, whereas molar conductivity decreases with dilutionC. Both conductivity and molar conductivity decrase wihdilutionD. Both conductivity and molar conductivity decrase wihdilution

Answer» Correct Answer - A
Conductivity of a solution decreases with dilution due to decrease in concentration of the electtrolyte. On the other hand, molar conductivity increases because of more ionisation of weak electrolyte or decrease in interactions hetqeen ions of a strong electrolyte.
2399.

For `H^(+) ` the value of `lambda^(infty)=349.8 S cm^2 eq^(-1)`. Calculate the mobilities of `H^(+)` ion and its velocity is it is in a cell in which the electrodes are `5 cm` apart and to which a potential of `2` volts is applied .A. `3.575 xx 10^(-3) cm^2 "volt"^(-1) sec^(-1), . 43 xx 10^(-3) cm sec^(-1)`B. ` 5.2 xx 10^(-4) cm^2 "volt"^(-1) sec^(-1) , 2. 08 xx 10^(-4) cm sec^(-1)`C. `3.2 xx 10^(-4) cm^2 "voly"^(-1) sec^(-1), 1.05 xx 10^(-3) cm sec^(-1)`D. None of these

Answer» Correct Answer - A
Velocity `= 1/(2 xx 349.8) = 1.43 xx 10^(-3) cm//sec`
Mobility ` =1,42 xx 10^(-3) xx 5/2 = 3.575 xx 10^(-3)`.
2400.

Which of the following ion is expected to have highest value of molar conductivity at infinity at infinite dilution is the aqueous solution ?A. `Na^+`B. ` K^+`C. ` 1/2 Ca^(+)`D. `H^+`

Answer» Correct Answer - D
Among the cations ` H^+` has the maximu connductivity due to Grothus conduction..