Bihar Board - Class 12 Chemistry - Chapter 3: Electrochemistry Long Answer Question
Long Question Answer
Q. 1. How would you determine the standard electrode potential of the system Mg2+ /Mg ?
Ans⇒ The potential of the system Mg2+ /Mg is determined by constructing a cell in which standard hydrogen electrode is left and electrode and after half cell (Mg2+ /Mg) is the right hand electrode. The potential of the other half cell is equal to the potential of the cell. If the activity of all the participating species at the right hand electrode is unity, then the measured potential is called standard potential (E) of the electrode.
Q. 2. Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging.
Ans⇒ Recharging : During the charging process, hydrogen ions move to cathode and sulfate ions to anode and the following reactions take place.
At cathode :
PbSO4 + H2 Pb + H2SO4
At anode :
PbSO4 + SO4 + 2H2O PbO2 + 2H2SO4
Thus, during charging active materials namely Pb cut the cathode and PbO2 at the anode are formed. Sulphuric acid is formed and water is consumed. Due to this the specific gravity of sulphuric acid increases and emf of the cell goes up to 2.3 volt.
Q. 3. Explain how rusting of iron is envisaged as setting up of an electrochemical cell.
Ans⇒ At a particular spot an object made of iron oxidation takes place and that spot behaves as an anode.
At anode : 2Fe (s)2Fe+ 4e-
E(Fe2+ , Fe )=0.44 V
Electrons released at anode spot moves through the metal and go to another spot on the metal and reduce oxygen in presence of H+ (which is believed to be available from H2CO3 formed due to dissolution of carbon dioxide from air into water. Hydrogen ions in water may also be available due to dissolution of other acidic oxides from the atmosphere. This spot behaves as a cathode with the reaction.
At cathode :
O2 (g)+4H+ (aq) + 4e- 2H2O (l) (E=1.23 V] The over reaction being
2Fe(s) + O2 (g) + 4H+ (a) 2 Fe2+ (aq)+2 H2O (l) [E(cell)=1.67 V]
The ferrous ions are further oxidized by atmospheric oxygen to ferric ions which come out as rust in the form of hydrated ferric oxide (Fe2O3.x H2O)
Q. 4. Depict the galvanic cell in which the reaction
Zn(s)+2+ Ag (aq) Zn2+ (aq) + 2 Ag (s)
takes place. Further show
(i) Which of the electrodes is negatively charged ?
(ii) The carriers of the current in the cell.
(iii) Individual reaction at each electrode.
Ans⇒ The electrochemical cell can be depicted as
Zn(s) | Zn2+ (aq) }} Ag+ (aq) } Ag (s)
Anode Cathode
(- ve) (+ ve)
(i) Cathode reaction :
2Ag+ (aq) + 2e- 2Ag (s)
(positive electrode reduction)
Anode reaction :
Zn(s)Zn2+ (aq) +2e
(negative electrode oxidation)
(ii) Electrons move from anode (zinc electrode) to cathode (silver electrode) in the external circuit. zinc ions go into solution at anode and Ag+ ions get deposited cathode. Thus electrons in the external and metal ions in the internal circuit act as carrier of current in
Zn(s) +2Ag+ (aq)Zn2+ (aq) + 2 Ag (s)
(iii) Overall reaction is obtained by anode and cathode reactions.
Q. 5. What is the Nernst equation for the Potential of an electrode ? Can Nernst equation be applied to the cell relation ? Apply this equation to a general reaction.
aA+bBcC+dD
Ans⇒ Let us take a electrode reaction
Zn2+ + 2e - Zn
The Nernst equation of this electrode
E=E-2.303RTnFlogaproductareactant
Instead of activity, we can take molar concentration.
E=E-0.05916nlog[Zn][Zn2+]
For pure solid and liquid molar concentration is taken as unity.
E=E0.05916nlog1[Zn2+]
Yes, Nernst equation can be applied to the cell reaction.
aA+bBcC+dD
E=E-0.05916nlog[C]c [D]d[A]a [B]b
Q. 6. State and explain Faraday’s law of electrolysis.
Ans⇒ Faraday’s First law : The amount of substance produced at an electrode during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte.
W Q
or, W It
or, W ZIt
where W is the mass of substance produced at an electrode.
I is current in amperes
t is time in seconds for which current is passed.
Z is the electro-chemical equivalent of substance.
Faraday’s second law : It states that the masses of different substances liberated or dissolved by the same amount of electricity passed is directly proportional to their chemical equivalents.
Or in other words “the same quantity of electricity will produce or dissolve chemically equivalent quantities of all substances.”
Mathematically,
Mass of A depositedMass of B deposited = Equivalent mass of AEquivalent mass of B
Thus, we can say that the same quantity of Electricity is required to produce one equivalent of any hence. It is called Faraday, F. It is equal to 96500 coulombs, and is equal to the charge on one mole of electrons.
Q. 7. What are fuel cells ? Write the reaction of an oxygen hydrogen fuel cell.
Ans⇒ Fuel Cell : A galvanic cell in which the reactants are continuously fed into the cell and the products are continuously removed is called a fuel cell
The most important fuel cell is hydrogen oxygen fuel cell.
The reactions taking place in this fuel cell are
At cathode :
O2 + 2 H2O (I)+4e- 4OH- Ecathode=1.299 V
At anode :
2H2(g)+4OH- 4H2O + 4e- Ecathode =0.00 V
Net reaction :
O2+2H22H2O Ecell =1.299 V
Q. 8. What is mercury cell ? Give the electrode reactions.
Ans⇒ Mercury cell is a new type of dry cell. It consists of
Anode : Zinc-mercury
Cathode : Paste of HgO and carbon
Electrolyte : Paste of ZnO in KOH. The reactions are :
Anode : Zn (amalgam)+2OH- ZnO(s)+H2O+2e-
At cathode : HgO+H2O +2e- Hg (I) + 2OH-
Overall reaction : Zn (amalgam) +HgO (s)ZnO(s)+Hg (I)
Q. 9. Define the terms equivalent and molar conductivity. What are their physical significance ?
Ans⇒ Equivalent conductivity : The conductivity of a volume (V) of a solution containing one equivalent of electrolyte placed between two electrodes separated by unit distance apart and of large enough area of cross section to hold the entire volume (V) is called equivalent conductance. It is denoted by eq.
eq =K V
where K= Specific conductance
V=Volume of solution containing one equivalent of electrolyte.
Molar conductivity : It is the conductivity of volume (V) of a solution containing 1 mole of a dissolved electrolyte place between two electrodes separated by unit distance apart and of
enough area of cross-section to hold the entire volume V. It is denoted by Am .
m= KV=KV
where V=Volume of solution
Containing 1 mole of electrolyte
C=molarity of solution
K=Specific conductivity.
Q. 10. Define entropy. Why is it a state function ? Explain the effect of increased temperatures on the entropy of a substance.
Ans⇒ Entropy is a measure of the degree of randomness or disorder of a system. It is denoted by S. It is state function and a change in state of a system is accompanied by a change in entropy when disorder increases entropy also increases.
Entropy change in reversible process.
According to Carnot’s cycle
Q2 - Q1Q2=T2 -T1T2. Q1T1=Q2T2
or, Q2T2-Q1T1=0 on quantity Q2T2 and T2 are change in entropy and it is denoted by S2.
Similarly on Q2T2 and T1 are less in entropy and it is denoted by S1 .
∴ S2-S1 =0.
Hence in reversible reaction or process entropy of the system is constant .
Q. 11. (i) What is corrosion ? What are the factors which affect corrosion ?
(ii) CO2 is always present in natural water. Explain its effect (increases, stops or no effect) on rusting of iron.
(iii) Rusting of iron is quicker in saline water than in ordinary water. Explain.
(iv) We can use aluminum in place of zinc for cathodic protection of rusting, Comment.
(v) How is cathodic protection of iron different from its galvanisation?
Ans⇒ (i) The gradual eating away of metals exposed to the atmosphere is known as corrosion. The phenomenon of corrosion involves the destruction of metal in which metal is generally converted into oxide. Common examples are rusting of iron, tarnishing of silver and deposition of green coating on copper and bronze.
Factors affecting corrosion :
(a) Presence of oxygen, sulfur etc. (elements which gain electrons).
(b) presence of moisture.
(c) Presence of carbon dioxide.
(ii) CO2 increases rusting of iron : CO2 present in water is in the form of H2CO3 , H2CO3 dissociates into H+and HCO3 .
H2CO3=HCO3-=H+
H2O=H+ + OH-
2H+ + 2e H2
Hydroxyl ion or bicarbonate ions attack the iron surface to form an ionic region in which iron loses electrons and passes on to ferrous sulfate which is further oxidized to ferric state by oxygen of the air. The released electrons moved towards the cathode region when H+ ions are converted into hydrogen gas.
Fe Fe2+ +2e-
O2 =2H2O+4e-4OH-
4 Fe +3O2 + 2 H2O 2Fe2O3 .XH2O
Thus CO2 increases rusting because H2CO3 gives H+ which gain electrons to form H2 gas. The electrons are released by iron.
(iii) Saline water contains greater number of ions Dissolved salts, NaCl than ordinary water and therefore electro-chemical reaction is quicker with saline water man when ordinary water is present. This increases the rate of corrosion of metal.
(iv) We can use aluminum in place of zinc for cathodic protection because aluminum is more reactive than Fe. When a plate of Al metal is buried inside the iron pipe or tank and connected to it by wires. The Al metal loses electrons easily and is easily sacrificed and protects from rusting
At anode : Al Al3+ + 3e-
At cathode : O2 + 2 H2O +4e- 4OH-
(v) In cathode protection, the iron object is made cathode by connecting it with a more reactive metal like Mg , Zn etc. Whereas in galvanisation, the iron object is protected from corrosion by coating it with zinc.
Q. 12. Explain Kohrausch’s law of independent migration of ions. Mention one application of this law
Ans⇒ At infinite dilution when the dissociation is complete each ion of the electrode makes a definite contribution of its own towards the molar conductivity of the electrolyte and it is quite independent of the presence of the other ion of the electrolyte. This the molar conductivity of an electron at infinite dilution is sum of the ionic conductive of the cations and anions i.e.
m=+
Q. 13. Why does conductivity of a solution decrease with dilution ?
Ans⇒ The conductivity of a solution is linked with the number of ions per unit volume; with dilution these decrease and the corresponding conductivity or specific conductance of the solution decreases.
Q. 14. Define Kohlrausch’s law. How does it help in
(i) calculating of for a weak electrolyte.
(ii) degree of dissociation of a weak electrolyte.
Ans⇒ “The molar conductivity of an electrolyte at infinite dilution is the sum of the ionic conductivities of the cations and anions each multiplied with the number of ions present in one formula unit of the electrolyte.”
Mathematically m for AxBy =xAy+ + yB x-
where m is the molar conductivity of the electrolyte at infinite dilution A y and B x- are the molar conductivities of the cation Ay+ and the anion Bx+ respectively at infinite dilution.
For example
m for NaCl = ∝ Na+ +∝ Cl-
m for BaCl2 =∝ Ba2+ +∝ Cl-
m for Al2(SO4)3 =2∝ Al3+ +3∝ SO42-
In terms of equivalent conductivities, Kohlrausch’s law is defined as follows. “The equivalent conductivity of an electrolyte at infinite dilution is the sum of two values-one depending upon the cation and the other upon anions.”
i.e., eq∝ =c∝ + a∝
where c∝ and a∝ are called the ionic conductivities at infinite dilution for the cation and anion respectively.
Applications of Kohlrausch’s Law : Some of the important applications of Kohlrausch’s law are :
1. Calculation of molar conductivities of weak electrolytes at infinite dilution : Kohlrausch’s law is helpful in determining the limiting molar conductivities of weak electrolytes. It has been discussed earlier the value of m for weak electrolytes cannot be obtained directly by the extrapolation of plot of m versus c . But this can be easily calculated with the help of Kohlrausch’s law.
For example, the value of m for acetic acid (CH3COOH) can be calculated from the knowledge of the molar conductivities at infinite dilution of strong electrolytes like CH3COONa , HCl and NaCl as follows :
m (CH3COOH) = CH3COO-∝ + H+∝
Now, add and subtract Na+ and Cl- to the expression on the right hand side and rearrange.
Thus, (CH3COOH)=CH3COO- +H++Na + -Na++ +Cl- -Cl-
=[CH3COO- +Na+] + [H+ Cl-] - [Na++ Cl-]
m (CH3COOH)+ m(HCl)- m(NaCl)
In the similar way ,
m(HN4OH)=m(NH4Cl)++m(NaOH)- - m(NaCl)
2. Calculation of degree of dissociation of weak electrolyte : Molar conductivity of an electrolyte depends upon its degree of dissociation. Higher the degree of dissociation, higher is the molar conductivity. As dilution increases, the degree of dissociation of weak electrolyte also increases and consequently, molar conductivity of electrolyte increases. At infinite dilution molar conductivity becomes maximum because degree of dissociation approaches unity.
Thus, if mc= molar conductivity of solution at any concentration.
m = molar conductivity at infinite dilution
= degree of dissociation
= mcm
Q. 15. Distinguish between Electrolytic cell and Electrochemical cell.
Ans⇒ distinction between Electrolytic and Electrochemical cell or (Galvanic cell) :
Galvanic cell | Electrolyte cell |
1. In galvanic cells,electrical energy is produced. | 1. In an electrolytic cell, electrical energy is Consumed. |
2. In a galvanic cell,reaction taking place is spontaneous. | 2. In an electrolytic cell, reaction taking place is non-spontaneous. |
3. The two half cells are set up in different containers and are connected through a salt bridge or porous partition. | 3. Both the electrodes are placed in the solution or| molten electrolyte in the same container. |
4. In a galvanic cell, anode 1s negative and cathode is positive. | 4. In an electrolytic cell, the anode is positive and the cathode is negative. |
5. The electrons move from anode to cathode in an external circuit. | 5. The electrons are supplied by the external Source. They enter through cathode and come out through anode. |
Q.16. Distinguish between E.M.F. and Potential difference.
Ans⇒ Distinction between Potential difference and E.M.F. of the cell:
Potential difference | E.M.F. |
1. It is the difference of potential between the two electrodes when the cell is producing current iieThe circuit is closed. | 1. It is the difference of potential between the two electrodes when no net current flows through the cell i.e. when the circuit is open. |
2. It is always less than the e.mn.f. of the cell. | 2. It is the maximum potential difference between the two electrodes of the cell. |
3 It does not have a definite value and varies as the cell produces electricity. The work calculated | 3. It has a definite value of the cell |
4. from the potential difference is always less than the maximum work obtainable from the cell. | 4. The maximum that can be obtained from a cell is calculated from its e.m.f. |
Q.17. Distinguish between Metallic conduction and Electrolytic conduce conduction.
Ans⇒ Distinction between Metallic conduction and Electrolytic conduction
Metallic Conduction | Electrolytic Conduction |
1. Conduction through metals is carried by the movement of electrons. | 1 Conduction through electrolytes is carried by movement of ions. |
2. No change is observed 2. in the chemical properties of the conductors. | 2. It involves chemical change resulting in the decomposition of electrolytes. |
3. It does not involve any transfer of matter. | 3. It involves the transfer of matter as ions. |
4. The resistance of metals is due to obstacles of vibrating kernels of metal atoms. | 4. The resistance of solution of electrolyte is due to factors like interionic attraction, viscosity of solvent etc. |
5. It shows an increase in resistance with the increase in temperature. | 5. It shows a decrease in resistance with the increase in temperature. |
6. The conductivity of metals is generally high. | 6. The conductivity of electrolytic solutions is generally low. |
Q.18. The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is 1500 . What is the cell constant if the conductivity of 0.001 M KCl solution at 298 K is 0.146 10-3 S cm-1.
Ans⇒Given,
Conductivity, k=0.146 10-3 S cm-1
Resistance, R=1500
Cell constant = k R
=0.146 10-3 1500
=0.219 cm-1
Q.19: The conductivity of the 0.20 M solution of KCl at 298 K is 0.0248 S cm-1 . Calculate its molar conductivity.
Ans⇒ Given, k=0.0248 S cm-1
c=0.20 M
Molar conductivity,
m=k1000c
= 0.0248 10000.2
=124 S cm2 mol-1
Q.20. Given the standard electrode potentials.
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
Arrange these metals in their increasing order of reducing power.
Ans⇒ The reducing power increases with the lowering of the reduction potential. In order of given standard electrode potential (increasing order):
K+/K <Mg2+/Mg <Cr3+/Cr <Hg2+/Hg <Ag+/Ag
Thus, in the order of reducing power, we can arrange the given metals as Ag<Hg<Cr<Mg<K
Q.21 . If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire ?
Solution:
Q (coulomb)=1 (ampere) t (sec)
Q =0.5 ampere 2 60 60
=3600 C
A flow of IF, i.e. 96500 C is equivalent to flow of 1 mole of electrons
i. e., =6.023 1023 electrons
3600 C is equivalent to flow of electrons
=6.023 102396500 3600
=2.246 1022 electrons
Q.22. Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.
Ans⇒ Conductivity: The conductivity of a solution is defined as the conductance of a solution of 1 cm length and having 1 sq. cm as the area of cross-section.
Molar conductivity: Molar conductivity of a solution at a dilution (V) is the conductance of all the ions produced from one mole of the electrolyte dissolved in V cm3 of the solution when the electrodes are one cm apart and the area of cross-section of the electrodes is so large that the whole of the solution is contained between them. It is usually represented by m .
Variation with concentration: The conductivity of a solution (Both for strong and weak electrolytes) decreases with a decrease in the concentration of the electrolyte i.e., on dilution. This is due to the decrease in the number of ions per unit volume of the solution on dilution. The molar conductivity of a solution increases in the decrease in the concentration of the electrolyte i.e., on dilution. This is due to the decrease in the number of ions per unit volume of the solution on dilution. The molar conductivity of a solution increases with decrease in the concentration of the electrolyte. This is because both the number of ions, as well as mobility of ions, increases with dilution. When concentration approaches zero, the molar conductivity is known as limiting molar conductivity.
Q.23: What do you understand about oxidation-reduction reactions ?
Ans⇒ Oxidation-Reduction reactions: Chemical reactions in which valency of elements changes are known as oxidation-reduction reactions. In this process both oxidation and reduction reactions occur simultaneously, in which one of the substances is oxidized and the other substance is reduced. Like
In this reaction FeCl3 is reduced to FeCl2 and SnCl2 is oxidized to SnCl4. In the reaction valency of Fe decreases and valency of Sn increases.
Q.24 : What are the factors which influence the electrical conductance of electrolytes ?
Ans⇒Factor which influence electrical conductivity of electrolytes :
The main factor which influence the electrical conductivity are following:
1. Temperature: It influences the following interactions.
(a) Interionic attractions: It depends upon the solute-solute interactions. Which is found between the ions of solute.
(b) Solvation of ions: It depends upon solute-solvent interactions. It is the relation between ions of solute and solvent molecules.
(c) Viscosity of solvent: It depends upon solvent-solvent interactions. Solvent molecules are related to each other.
With increase in temperature all these three effects decrease and average kinetic energy of ions increases. Thus, with increase of temperature, resistance of solution decreases and hence conductance increases.
2. Nature of electrolyte : The conductance of solution depends upon the nature of electrolyte. On the basis of conductance measurement electrolytes are classified as strong electrolyte and weak electrolyte. Strong electrolytes have high value of conductance even at higher concentration also.
3. Dilution or concentration : It is the main factor which influences electrical conductance. Effect of dilution or concentration can be studied individually in equivalent conductance, specific conductance and molar conductance. But for a general concept of electrical conductance of solution, as the concentration is lowered or dilution increases, electrical conductance of the whole solution increases.
Q .25 :On what factors does the various conductivities of an electrolytic solution depend ?
Ans⇒ Conductivities of electrolytic solution depend on the following factors :
(1) Dilution : On increasing dilution, value of specific conductance of a solution decreases, value of equivalent conductance and molar conductance increases.
(2) Nature of solvent: A solvent with high dielectric constant has high conductivity and with low dielectric constant has low conductivity.
(3) Number of ions present in solution: Conductivity of strong electrolytes is higher than the conductivity of weak electrolytes.
(4) Size of ions : In aqueous solution, small ions are heavily hydrated due to which their conductivity decreases.
(5) Effect of Temperature: With the increase in temperature conductivity increases.
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