Important Questions for CBSE Class 12 Chemistry Chapter 2 - Electrochemistry 2024-25

Very Short Questions and Answers (1 Marks Questions)

1. What is the effect of temperature on molar conductivity?

Ans: Molar conductivity of an electrolyte increases with an increase in temperature.  

2. Why is it not possible to measure single electrode potential?

Ans: It is not possible to measure single electrode potential because the half cell containing a single electrode cannot exist independently, as the charge cannot flow on its own in a single electrode.

3. Name the factor on which emf of a cell depends:

Ans: Emf of a cell depends on the following factors

1. Nature of reactants.  

2. The concentration of the solution in two half cells. 

3. Temperature 

4. What are the units of molar conductivity?

Ans:  $c{{m}^{2}}oh{{m}^{-1}}mo{{l}^{-1}}$ or $Sc{{m}^{2}}mo{{l}^{-1}}$

5.Write Nernst equation or the general cell reaction $aA+bB\to cC+dD$ 

Ans: $Ecell\text{ }=\text{ }{{E}^{0}}cell-RT\ln \dfrac{{{\left[ C \right]}^{c}}{{\left[ D \right]}^{d}}}{nF{{\left[ A \right]}^{a}}{{\left[ B \right]}^{b}}}$ 

6. What is the EMF of the cell when the cell reaction attains equilibrium?

Ans: Zero

7. What is the electrolyte used in a dry cell?

Ans:  A paste of $N{{H}_{4}}Cl$ , $Mn{{O}_{2}}$ and Carbon.

8. How is cell constant calculated from conductance values?

Ans: $Cell\text{ }Constant=\dfrac{specific\text{ }conductance}{observed\text{ }conductance}$

9. What flows in the internal circuit of a galvanic cell? 

Ans: Ions 

10. Define electrochemical series. 

Ans: The arrangement of various electrodes in the decreasing or increasing order of their standard reduction potentials is called electrochemical series.

11. Why in a concentrated solution, a strong electrolyte shows deviations from Debye-Huckle- Onsagar equation? 

Ans: Because interionic forces of attractions are large. 

12. What is the use of Platinum foil in the hydrogen electrode?

Ans: It is used for the inflow and outflow of electrons. 

13. Corrosion of motor cars is of a greater problem in winter when salts are spread on roads to melt ice and snow. Why?

Ans: A short-circuited cell is formed when two metals are brought together under the surface of an electrolyte. Corrosion is more of an issue in the winter because the car contains metal like lead and chromium, and the salt sprinkled to melt ice is NaCl, which acts as an electrolyte forming a short circuit cell.  

14. Is it safe to stir AgNO3 solution with a copper spoon?

\[~\left( {{\mathbf{E}}^{\mathbf{0}}}_{\mathbf{Ag}+/\text{ }\mathbf{Ag}}=\text{ }\mathbf{0}.\mathbf{80}\text{ }\mathbf{Volt};\text{ }{{\mathbf{E}}^{\mathbf{0}}}_{\mathbf{Cu}+2/\text{ }\mathbf{Cu}}=\text{ }\mathbf{0}.\mathbf{34}\text{ }\mathbf{Volt} \right)\] 

Ans: No it is not safe because it reacts with $AgN{{O}_{3}}$ Solution ( Emf will be positive.)

15. Why is it necessary to use salt bridges in a galvanic cell?

Ans: To complete the inner circuit and to maintain the electrical neutrality of the solution.

Short Questions and Answers (2 Marks Questions)

1. How can you increase the reduction potential of an electrode? For the reaction?\[{{M}^{n+}}\text{(aq)+}n{{e}^{-1}}\to M(s)\] 

Ans: Nernst equation is:

\[{{E}^{0}}_{\dfrac{{{M}^{n+}}}{M}}={{E}_{\dfrac{{{M}^{n+}}}{M}}}\text{ }-\text{ }\dfrac{2.303RT}{nF}\log \dfrac{1}{[{{M}^{n+}}]}\] 

${{E}_{\dfrac{Mn+}{M}}}$  can be increased by

1. increase in concentration of ${{M}^{n+}}$  ions in solution

2. By increasing the temperature. 

2. Calculate emf of the following cell at $298K$.

\[Zn/Z{{n}^{2+}}\text{(}{{10}^{-4}}M)||\text{ }C{{u}^{2+}}\left( {{10}^{-2}}M \right)/Cu\] 

Given $E_{Z{{n}^{2+}}/Zn}^{0}\text{ }=-0.76V$ 

\[E_{C{{u}^{2+}}/Cu}^{0}=+0.34V~\] 

Ans: Cell reaction is as follows.

\[~Zn(s)+\text{ }C{{u}^{2+}}(aq)\to \text{ }Z{{n}^{2+}}\text{(aq) }+\text{ }Cu(s)\]

n = 2 and T = 298 K

\[{{E}_{cell}}=\left( E_{C{{u}^{2+}}/Cu}^{0}-E_{Z{{n}^{2+}}/Zn}^{0} \right)-\dfrac{0.0591}{n}\log \text{ }\dfrac{\left[ Z{{n}^{2+}}(aq) \right]}{\left[ C{{u}^{+2}}(aq) \right]}\] 

  \[{{E}_{cell}}=0.34V-(-0.76)-\dfrac{0.0591}{2}\log \dfrac{{{10}^{-4}}}{{{10}^{-2}}}\] 

\[{{E}_{cell}}=1.10-0.02955\log {{10}^{-2}}\] 

\[=1.10+2\times 0.02955\] 

   \[\therefore {{E}_{cell}}=+1.1591V\] 

3. Electrolysis of $\mathbf{KBr}(\mathbf{aq})$ gives $\mathbf{B}{{\mathbf{r}}_{\mathbf{2}}}$ at anode but $\mathbf{KF}(\mathbf{aq})$ does not give  ${{\mathbf{F}}_{\mathbf{2}}}$ . Give a reason. 

Ans: Oxidation takes place at anode. Now higher the oxidation Potential, easier to oxidize. Oxidation potential of $Br-$, ${{H}_{2}}O$, $F-$  are in the following order. ${{Br}^{-}} > {{H}_{2}}O > {{F}^{-}}$

Therefore in the Aqueous Solution of KBr. ${{Br}^{-}}$ ions are oxidized to $B{{r}_{2}}$  in preference to ${{H}_{2}}O$ . On the other hand, in the aqueous solution of $KF$,  ${{H}_{2}}O$ is oxidized in preference to ${{F}^{-}}$ . Thus in this case oxidation of ${{H}_{2}}O$ at anode gives ${{O}_{2}}$  and no ${{F}_{2}}$ is produced.  

4. What happens when a piece of copper is added to

a. An aqueous solution of $\mathbf{FeS}{{\mathbf{O}}_{\mathbf{4}}}$ ?

Ans: Nothing will happen when the piece of copper is added to $FeS{{o}_{4}}$ because reduction potential $E_{C{{u}^{+2}}/Cu}^{0}\text{ }\left( -0.34V \right)$ is more than the reduction potential $E_{\left( F{{e}^{2+}}/Fe \right)}^{0}\left( -0.44V \right)$ therefore copper is less reactive than iron.

b. An aqueous solution of $\mathbf{FeC}{{\mathbf{l}}_{\mathbf{3}~}}~$?

Ans: copper will dissolve in an $aq$ solution of $FeC{{l}_{3}}$  because reduction potential $E_{(F{{e}^{3+}}/F{{e}^{2+}})}^{0}\left( +0.77V \right)$ is more than the reduction potential $E_{C{{u}^{+2}}/Cu}^{0}\text{ }\left( -0.34V \right)$ .

\[Cu\left( s \right)+\text{ }2FeC{{l}_{3}}\text{ }\left( aq \right)\text{ }\to \text{ }C{{u}^{+2}}\left( aq \right)\text{ }+\text{ }2\text{ }FeC{{l}_{2}}\left( aq \right)\] 

5. Define corrosion. Write the chemical formula of rust.

Ans: Corrosion is a process of deterioration of a metal as a result of its reaction with air and water surrounding it. It is due to the formulation of sulphides, oxides, carbonates, hydroxides, etc.

Formula of rust- $F{{e}_{2}}{{O}_{3}}.x{{H}_{2}}O$

6. Write short notes on reduction and oxidation potentials.

Ans: Oxidation potential: This is a measure of the tendency of a chemical species to lose electrons or to get oxidized at an electrode.

Reduction potential: This is a measure of the tendency of a chemical species to gain electrons or to get reduced at an electrode.

7. How are standard electrode potentials measured?

Ans: The standard electrode potential is measured by a galvanic cell which is constructed with a standard hydrogen electrode on one side and an unknown half cell on the other side. 

8. What is a cell constant? How is it determined?

Ans: Cell constant is defined as the ratio of the distance between the electrodes which is divided by the area of the cross-sectional of the electrode. Cell constant can be determined by calculating the resistance of a cell of known conductivity..

9. What is conductivity water?

Ans: Conductivity water is a measure of the ability of water to pass an electric current.

10. Why is it necessary to platinize the electrodes of a conductivity cell before it is used for conductance measurement?

Ans: Because platinization will increase the surface area of the electrode on which the hydrogen gas will contact with hydrogen ions. Due to this the reactions happen faster and lead to better conduction measurement.

11. Why does mercury cells give constant voltage? 

Ans: Mercury cells give constant voltage because overall cell reaction does not include any ion in the solution whose concentration changes during its lifetime.

12. What is a fuel cell? Write reaction involved in ${{\mathbf{H}}_{\mathbf{2}}}-{{\mathbf{O}}_{\mathbf{2}}}$ fuel cell.

Ans: An electrochemical cell in which the chemical energy of the fuel is directly converted into electrical energy is known as a Fuel cell.

The Half cell reactions in Hydrogen-oxygen fuel cell as follows:

Oxidation at anode:

$2{{H}_{2}}(g)+4O{{H}^{-}}(aq)\to 4{{H}_{2}}O(l)+4{{e}^{-}}$ 

Reduction at cathode:

${{O}_{2}}(g)+2{{H}_{2}}O(l)+4{{e}^{-}}\to 4O{{H}^{-}}(aq)$

13. Why is Li the best reducing agent whereas Fluorine is the best oxidizing agent? but not to ${{E}_{cell}}$ . Explain.

Ans: according to the electrochemical series, Li ion has the lowest reduction potential which means it acts as the best reducing agent. Similarly, F has the highest reduction potential, thus fluorine oxidizes another substance readily. So, F is the best oxidizing agent. 

14. Why sodium metal is not obtained at cathode when $\mathbf{aq}\text{ }\mathbf{NaCl}$ is electrolysed with $\mathbf{Pt}$ electrodes but obtained when molten $\mathbf{NaCl}$ is electrolysed?

Ans: Because the reduction potential of $N{{a}^{+}}$ , $E_{N{{a}^{+}}/Na}^{0}=-2.7V$ is energetically more difficult than the reduction of water. 

15. Zn rod weighing $\mathbf{25}\text{ }\mathbf{g}$ was kept in $\mathbf{100}\text{ }\mathbf{mL}$ of $\mathbf{1M}$  copper sulphate solution. After certain time interval, the molarity of $\mathbf{C}{{\mathbf{u}}^{\mathbf{2}+}}$ was found to be $\mathbf{0}.\mathbf{8}\text{ }\mathbf{M}$ . What is the molarity of $\mathbf{S}{{\mathbf{O}}_{\mathbf{4}}}^{-\mathbf{2}}$ in the resulting solution and what should be the mass of $\mathbf{Zn}$ rod after cleaning and drying? 

Ans: weight of Zn rod = 25g

Number of moles that are required by $C{{u}^{2+}}$ to Cu = $Molarity\times volume$ 

\[=(1-0.8)\times \dfrac{100L}{1000}=0.2\times 0.1=0.02mole\] 

Molar mass of Zinc = 64.5g/mol

The zinc gets oxidized = $0.02\times 65.4=1.308g$ 

Hence, the weight of Zn rod left will be = $25-1.308=23.69gm$ 

16. Which will have greater molar conductivity and why? Sol A. 1 mol KCl dissolved in 200cc of the solution or Sol B. 1 mol KCl dissolved in 500cc of the solution. 

Ans: Sol B: 1 mol KCl has dissolved in 500c of the solution has greater molar conductivity. Because the number of ions in Sol B is greater than Sol A. 

Molar conductivity is the conduction of all moles produced from 1mole of KCl and which increases with dilution. 

Long Questions and Answers (3 Marks Questions)

1. Write any three differences between potential difference and e.m.f.

Ans:

2. Why does an electrochemical cell stop working after some time? 

Ans: The reduction potential of an electrode depends upon the concentration of a solution with which it is in contact. 

As the cell works, the concentration of reactants decreases. Then according to Le Chatelier’s principle, it will shift the equilibrium in the backward direction. On the other hand, if the concentration is more on the reactant side then it will shift the equilibrium in the forward direction. When a cell works, concentration in an anodic compartment in the cathodic compartment decreases, and hence $E_{cathode}^{0}$ cathode will decrease. Now EMF of the cell is $E_{cell}^{0}=\text{ }E_{cathode}^{0}\text{ - }E_{anode}^{0}$ 

A decrease in${{E}^{0}}_{cathode}$  and a corresponding increase in ${{E}^{0}}_{anode}$ will mean that EMF of the cell will decrease and will ultimately become zero i.e., the cell stops working after some time. 

3. For the standard cell, 

$Cu(s)/C{{u}^{+2}}(aq)||\text{ }A{{g}^{+}}(aq)/Ag(s)~~$

\[E_{C{{u}^{2+}}/cu}^{0}=+0.34V\] 

\[E_{A{{g}^{+}}/Ag}^{0}\text{ }=+0.80\text{ }V~~\] 

(i). Identify the cathode and the anode as the current is drawn from the cell.

Ans:

From the cell representation, $A{{g}^{+}}/Ag$ electrode is cathode and $Cu/C{{u}^{2+}}$ electrode is anode.  

(ii). Write the reaction taking place at the electrodes.

Ans:

At anode: 

$Cu(s)\to C{{u}^{2+}}(aq)+2{{e}^{-}}$ 

At cathode:

$2A{{g}^{+}}(aq)+2{{e}^{-}}\to Ag(s)$ 

(iii). Calculate the standard cell potential.

Ans: 

\[E_{cell}^{0}=\text{ }E_{cathode}^{0}\text{ - }E_{anode}^{0}\]  \[E_{cell}^{0}=E_{A{{g}^{+}}/Ag}^{0}-E_{C{{u}^{2+}}/Cu}^{0}\]

The standard cell potential,$E_{cell}^{0}=+0.80V-(+0.34V)=+0.46V$

4. Can we store copper sulphate in Zinc vessels and silver vessels? Give reasons. 

Given $E_{C{{u}^{2+}}/Cu}^{0}\text{ }=\text{ }+0.34V;E_{Z{{n}^{2+}}/Zn}^{0}=\text{ }-0.76V;\text{ }E_{A{{g}^{+}}/Ag}^{0}\text{ }=\text{ }+0.80V~$ 

Ans: A metal having lower reduction potential can displace a metal having higher reduction potential from solution of its salt of $C{{u}^{2+}}(E_{C{{u}^{2+}}/Cu}^{0}=+0.34V)$ . Since standard reduction potential of $Z{{n}^{2+}}\left( E_{Z{{n}^{2+}}/Zn}^{0}=-0.76V \right)$ is less than the standard reduction potential of $C{{u}^{2+}}(E_{C{{u}^{2+}}/Cu}^{0}=+0.34V)$  ,$Zn$ can displace copper from copper sulphate solution. Thus, $CuS{{o}_{4}}$ solution cannot be stored in a zinc vessel.   

Since the reduction potential of silver is more than copper so silver is less reactive than copper and will not be able to displace copper. Hence CuSO4 solution can be stored in a silver vessel.

5. How many grams of chlorine can be produced by the electrolysis of matters $\mathbf{NaCl}$ with a current of $\mathbf{1}.\mathbf{02}\text{ }\mathbf{A}$ for $\mathbf{15}$ min? 

Ans: 

\[2NaCl\left( l \right)\text{ }\to 2N{{a}^{+}}\left( l \right)+2C{{l}^{-}}\left( l \right)~\] 

\[{{\underset{2mole}{\mathop{2\text{ }Cl}}\,}^{-}}\to \underset{1mole}{\mathop{C{{l}_{2}}\left( g \right)}}\,\text{ }+\text{ }2{{e}^{~}}\] 

\[Q=\text{ }nf\] 

\[Q=\text{ }2\text{ }\times \text{ }96500\text{ }C/mol=\text{ }1.93\text{ }\times \text{ }{{10}^{5}}C~\]  

Quantity of electricity used = $I\times t=1.02A\times (15\times 60)s=900C$ 

 Molar mass of $C{{l}_{2}}$ = $2\times 35.5=71g$ 

$\therefore 1.93\times {{10}^{5}}C$ of charge produce chlorine = 71gm

     900C of charge produce chlorine = $\dfrac{71\times 900}{1.93\times {{10}^{5}}}$ =0.331gm

Hence, 0.331 grams of chlorine can be produced by the electrolysis of matter $\mathbf{NaCl}$ with a current of $\mathbf{1}.\mathbf{02}\text{ }\mathbf{A}$ for $\mathbf{15}$ min.

6. What is understood by a normal hydrogen electrode? Give its significance.

Ans: Normal hydrogen electrode is also known as Standard Hydrogen Electrode which is used for reference on half cell potential reactions.

By using normal hydrogen electrodes, we can calculate cell potentials using different electrodes and this is a standard measurement of electrode potential for the thermodynamic scale of redox potential.

7. Define electrode potential. Why the absolute value of the reduction potential of the electrode cannot be determined?

Ans: The voltage or potential difference of a cell assembled from a standard hydrogen electrode is known as electrode potential for a given electrode.

The potential is determined by using a standard hydrogen electrode as a reference and so absolute potential cannot.

8. Write the equation showing the effect of concentration on the electrode potential.

Ans: The equation showing the effect of concentration on the electrode potential is known as Nernst Equation.

\[{{E}_{cell}}=E_{cell}^{0}-\dfrac{RT}{nF}\ln Q\] 

Where, ${{E}_{cell}}$ = the reduction potential at current conditions

$E_{cell}^{0}$ = the standard reduction potential relative to hydrogen’s reduction potential at ${{25}^{0}}C$ 

R = universal gas constant 

T =temperature in K

n = the moles of electrons transferred between the positive and negative terminals of an electrochemical system.

F =faraday’s constant

Q = reaction quotient

9. Derive the relationship between Gibbs’s free energy change and the cell potential. 

Ans:  The electrical work of electrical energy produced is equal to the product of e.m.f of a cell. (${{E}_{cell}}$)

\[{{W}_{electrical}}=-{{E}_{cell}}\times q\] 

The negative sign indicates the work done by the system on the surroundings. 

Therefore, ${{W}_{electrical}}=-nF{{E}_{cell}}$ 

Since, $\Delta G=-W(\text{non-expansion})$ 

In this case non-expansion work done is electrical work.

Therefore, $\Delta G=-nF{{E}_{cell}}$ 

Considering standard electrode potential, the standard gibbs free energy as follows,

\[\Delta {{G}^{o}}=-nF{{E}^{o}}_{cell}\]

10. How can the Nernst equation be applied in the calculation of the equilibrium constant of any cell reaction? 

Ans:  Nernst equation, ${{E}_{cell}}=E_{cell}^{0}-\dfrac{RT}{nF}\ln Q$

For any cell reaction in equilibrium, then $Q={{K}_{c}}$ 

Where ${{K}_{c}}$ = equilibrium constant

\[{{E}_{cell}}=E_{cell}^{0}-\dfrac{RT}{nF}\ln {{K}_{c}}\] 

11. The cell reaction as written is spontaneous if the overall EMF of the cell is positive. Comment on this statement.

Ans:   If the e.m.f of a cell is positive, then the Gibbs free energy of the overall reaction is less than zero.

 \[\Delta G=-nF{{E}_{cell}}\] 

Therefore, if ${{E}_{cell}}$ is positive, then the cell reaction is spontaneous.

Very Long Questions and Answers (5 Marks Questions)

1. Explain the term electrolysis. Discuss briefly the electrolysis of 

(i). molten $\mathbf{NaCl}$ 

 Ans:   Electrolysis: This is a process of decomposition of a chemical compound in aqueous solutions or in a molten state by a chemical change during electric current

The electrolysis of molten NaCl when sodium chloride is melted above ${{801}^{0}}C$ then two electrodes are inserted into the melt and electric current is passed through the molten salt. The following chemical reactions take place at the electrodes.

At cathode: $N{{a}^{+}}+{{e}^{-}}\to Na$

At anode: $C{{l}^{-}}\to \dfrac{1}{2}C{{l}_{2}}+{{e}^{-}}$ 

the overall reaction is, $2NaCl\to 2Na(s)+C{{l}_{2}}(g)$ 

(ii). aqueous sodium chloride solution 

Ans:  Sodium chloride dissolves as $N{{a}^{+}}$ and $C{{l}^{-}}$ into the water and produces sodium and hydrogen at respective electrodes.

At cathode: ${{H}_{2}}O(l)+2{{e}^{-}}\to {{H}_{2}}(g)+2O{{H}^{-}}$ 

At anode: $C{{l}^{-}}\to \dfrac{1}{2}C{{l}_{2}}+{{e}^{-}}$ 

the overall reaction, 

\[NaCl(aq)+{{H}_{2}}O(l)\to N{{a}^{+}}(aq)+O{{H}^{-}}(aq)+{{H}_{2}}(g)+\dfrac{1}{2}C{{l}_{2}}(g)\] 

(iii). molten lead bromide 

Ans:  Molten lead bromide $PbB{{r}_{2}}$  is an electrolyte.

At cathode: $P{{b}^{+2}}+2{{e}^{-}}\to Pb(s)$ 

At anode: $2B{{r}^{-}}\to B{{r}_{2}}+2{{e}^{-}}$ 

(iv). water

Ans: The process of water is decomposed into oxygen and hydrogen gas when an electric current is passed through is known as the electrolysis of water.

At cathode: $2{{H}_{2}}O(l)+2{{e}^{-}}\to {{H}_{2}}(g)+2O{{H}^{-}};{{E}^{0}}=-0.42V$ 

At anode: $2{{H}_{2}}O\to {{O}_{2}}(g)+4{{H}^{+}}+4{{e}^{-}};{{E}^{o}}=+0.82V$ 

The overall reaction of electrolysis of water as follows,

\[2{{H}_{2}}O(l)\to {{O}_{2}}+2{{H}_{2}};{{E}^{o}}=-1.24V\] 

2. State and explain Faraday’s laws of electrolysis. What is the electrochemical equivalent? 

Ans: First Law of Electrolysis: The mass deposited on any electrode is directly proportional to the quantity of electricity passing through it during electrolysis.

W = zct

Where, W = deposited mass on the electrode

c = current and t = time in sec

If 1 amp current is passed through a solution in one second then the deposited mass of substance on the electrode is equal to its electrochemical equivalent. Here, z is known as the electrochemical equivalent. 

Second Law of Electrolysis: The deposited masses on the electrodes is directly proportional to their chemical equivalents when the same quantity of electricity is passed through the different electrolytic cells. 

Consider, ${{W}_{1}},{{W}_{2}}\And {{E}_{1}},{{E}_{2}}$be the amount of mass deposited and their chemical equivalents respectively.

According to faraday’s second law of electrolysis,

${{W}_{1}}\alpha {{E}_{1}}$ , and ${{W}_{2}}\alpha {{E}_{2}}$ 

Then,  $\dfrac{{{W}_{1}}}{{{W}_{2}}}=\dfrac{{{E}_{1}}}{{{E}_{2}}}$ 

3. What do you understand by ‘electrolytic conduction’? What are the factors on which electrolyte conduction depends? What is the effect of temperature on electrolytic conduction? 

Ans:  When electric current passes through the electrolytic solutions then the ability of the solutions to allow the current is known as electrolytic conductance.

Factors affecting electrolytic conductance,

1. Concentration of ions

2. Nature of electrolyte

3. Temperature

Effect of temperature: when the temperature changes the electrolyte gets dissolved in solution. Hence, the temperature increases the solubility of electrolytes and increases the electrolytic conduction.

4. How is electrolytic conductance measured experimentally? 

Ans: The resistance between two nodes will give information about the conductance of electrolytes. When electricity passes through electrolyte solution dissociates into positive ions and negative ions. The conductance of electrolytes can be measured by using galvanic cells or the method of electrolysis.

5. Describe normal hydrogen electrodes and their applications.

Ans: An electrode that is used for reference on all half cell potential reactions is known as a normal hydrogen electrode or standard hydrogen electrode.

The normal hydrogen electrode potential is zero and this is used to calculate the potentials of different half cells and different concentrations.

Key Questions for Class 12 Exams: Focus on 3 or 5 Marks Weightage for Success!

1. What do you mean by

(i). Negative standard electrode potential 

Ans: Negative standard electrode potential means the tendency to get reduce is less than hydrogen or greater ease of oxidation compared to that of hydrogen  

(ii). Positive standard electrode potential?

Ans: Positive standard electrode potential means the tendency to get reduce more than hydrogen.

2. Which cell is generally used in hearing aids? Name the material of the anode, cathode, and electrolyte. Write the reactions involved.

Ans: Mercury cells are used in hearing aids.

At cathode: $HgO+{{H}_{2}}O+2{{e}^{-}}\to Hg+2O{{H}^{-}}$ 

At anode: $Zn+O{{H}^{-}}\to ZnO+{{H}_{2}}O+2{{e}^{-}}$ 

3. Iron does not rust even if Zinc coating is broken in a galvanized iron pipe but rusting occurs much faster if tin coating over iron is broken. Explain.

Ans: Fe is less electropositive than Zn. So, as long as Zn is on the surface of the Fe pipe then Zn acts as an anode and the Fe pipe acts as a cathode. As a result of this, rusting of Fe is prevented by Zn. 

On the other hand, Zn is more electropositive than Sn, which protects iron until the Sn coating is unbroken. If any pores or breaks are observed, even tin is there the exposed iron gets rusted.

4. Corrosion is an electrochemical phenomenon, explain. 

Ans: When a metal surface is exposed to a wet environment, due to the presence of moisture or air oxidation takes place.

5. Calculate the pH of following cell: $Pt,{{H}_{2}}(1atm)|{{H}^{+}}({{H}_{2}}S{{O}_{4}})$, if its electrode potential is $\mathbf{0}.\mathbf{03V}$ .

Ans:

Nernst equation for the given cell,

\[{{E}_{{{H}_{2}}/{{H}^{+}}}}=E_{{{H}_{2}}/{{H}^{+}}}^{0}-\dfrac{0.0591}{2}\log \dfrac{{{[{{H}^{+}}]}^{2}}}{{{P}_{{{H}_{2}}}}}\] 

\[0.03=0.0-0.0591\log \dfrac{[{{H}^{+}}]}{1}\] 

\[0.03=0.0591pH\] 

\[pH=\dfrac{0.03}{0.0591}=0.5076\] 

6. A cell contains two hydrogen electrodes. The negative electrode is in contact with a solution of ${{10}^{-5}}{{H}^{+}}$ ions. The emf of the cell is $\mathbf{0}.\mathbf{118}\text{ }\mathbf{V}$ at $\mathbf{298}\text{ }\mathbf{K}$ . Calculate the concentration of the ${{H}^{+}}$  ions at the positive electrode. 

Ans:

The cell reaction can be represented as,

$Pt|{{H}^{+}}({{10}^{-5}}M)1atm\parallel {{H}^{+}}(aM)|1atm|Pt$ 

At anode: ${{H}_{2}}\to 2{{H}^{+}}+2{{e}^{-}}$ 

Negative polarity, $[{{H}^{+}}]={{10}^{-5}}M$ 

At cathode: $2{{H}^{+}}+2{{e}^{-}}\to {{H}_{2}}$ 

\[{{E}_{cell}}=\dfrac{0.0591}{2}\log \dfrac{[{{H}^{+}}]}{[{{10}^{-5}}]}\] 

\[0.118=0.0591\log \dfrac{[{{H}^{+}}]}{[{{10}^{-5}}]}\] 

\[\therefore [{{H}^{+}}]={{10}^{-3}}M\]

The concentration of the ${{H}^{+}}$  ions at the positive electrode = ${{10}^{-3}}M$ 

7. Crude copper-containing $\mathbf{Fe}$ and $\mathbf{Ag}$ as contaminations were subjected to electrorefining by using a current of 175A for 6.434 min. The mass of anode was found to decrease by 22.260g, while that of the cathode was increased by 22.011g. Estimate the % of copper, iron, and silver in crude copper. 

Ans: 

$22.260-22.011=0.249g$ of impurity present.

A current of 175A for 6.434 min(386.04sec) corresponds to $175\times 386.04=67551.75C$ of electricity.

The equivalent of Cu is 31.77g

The mass of Cu that should be deposited = $\dfrac{67551.75}{96500}\times 31.77=22.239g$ 

Increasing in a mass of anode = 22.011g

$22.239-22.011=0.228g$ corresponds to the mass of Cu that is not deposited. This can be equated to the mass of iron that passed into the solution.

The equivalent mass of Fe =27.75g

The mass of Fe = $0.228\times 27.75=0.199g$ 

The percentage of Fe = $\dfrac{0.199\times 100}{22.26}=0.90%$ 

8. Zinc electrode is constituted at 298 K by placing $\mathbf{Zn}$ rod in $\mathbf{0}.\mathbf{1}\text{ }\mathbf{M}$ aqueous solution of zinc sulphate which is $\mathbf{95}\text{ }%$ dissociated at this concentration. What will be the  electrode potential of the electrode given that $\text{E}_{Z{{n}^{2+}}/Zn}^{0}=-0.76V$ 

Ans:  $Z{{n}^{+2}}+2{{e}^{-}}\to Zn(s)$ 

\[[Z{{n}^{+2}}]=0.1\times \dfrac{95}{100}=0.095M\] 

According to Nernst equation,

\[{{E}_{Z{{n}^{+2}}/Zn}}=E_{Z{{n}^{+2}}/Zn}^{0}-\dfrac{0.0591}{2}\log \dfrac{1}{[Z{{n}^{+2}}]}\] 

 \[=-0.76V-\left( \dfrac{0.0591}{2}\log \dfrac{1}{0.095} \right)\] 

\[=-0.76-\dfrac{0.0591}{2}\times 1.0223=-0.7902V\] 

9. At what pH will hydrogen electrode at $\mathbf{298}\text{ }\mathbf{K}$ show an electrode potential of -0.118V, when Hydrogen gas is bubbled at 1atm pressure?

Ans:

From Nernst equation, ${{E}_{cell}}=E_{cell}^{o}+0.0591pH$ 

for hydrogen electrode, $E_{cell}^{0}=\pm 0.0V$ 

Then, ${{E}_{cell}}=0.0591pH$ 

Given, ${{E}_{cell}}=-0.118V$ 

pH = 0.5008

10. Electrolysis of the solution of $\mathbf{MnS}{{\mathbf{O}}_{4}}$ in aqueous sulphuric acid is a method for the  preparation of $\mathbf{Mn}{{\mathbf{O}}_{2}}$ as per the chemical reaction  

$\mathbf{M}{{\mathbf{n}}^{+2}}(aq)+\mathbf{2}{{\mathbf{H}}_{2}}\mathbf{O}\to \mathbf{Mn}{{\mathbf{O}}_{2}}\text{+ }\mathbf{2}{{\mathbf{H}}^{+}}(aq)+{{\mathbf{H}}_{2}}~~$ 

Passing a current of $\mathbf{27}\text{ }\mathbf{A}$ for $\mathbf{24}$  Hrs gives $\mathbf{1}\text{ }\mathbf{kg}$ of $\mathbf{Mn}{{\mathbf{O}}_{2}}$ . What is the current efficiency? What are the reactions occurring at anode and cathode?  

Ans: According to faraday’s second law of electrolysis,

\[weight\text{ }in\text{ }grams=\dfrac{equivalent\text{ }weight\times current\text{ }in\text{ }amperes\times time\text{ }in\text{ }\sec onds}{96500}\] 

Molecular weight of $Mn{{O}_{2}}$ = 87g

The oxidation number of Mn in $M{{n}^{+2}}$ =+2

The oxidation number of Mn in $Mn{{O}_{2}}$ =+4

Therefore, the change in the oxidation number = +4-(+2)=+2

Hence, the equivalent weight of $Mn{{O}_{2}}$ = $\dfrac{87g}{2}=43g$

Given, weight of $Mn{{O}_{2}}$ = 1Kg =1000g

Current = 27A

Time = 24 hrs = $(24\times 60\times 60)\sec $ 

Substitute the values in the faraday’s second law equation,

\[1000g=\dfrac{43.5\times i\times (24\times 60\times 60)}{96500}\] 

i =25.67A

The value of current efficiency during the electrolysis of $MnS{{O}_{4}}$ can be calculated as,

Current efficiency = $\dfrac{25.67}{27}\times 100=95.074%$ 

The reaction takes place at cathode,

$2{{H}^{+}}+2{{e}^{-}}\to {{H}_{2}}$ 

the reaction takes place at anode,

$M{{n}^{+2}}\to M{{n}^{+4}}+2{{e}^{-}}$ 

Electrochemistry

1. What do you mean by Kohlrauch’s law: from the following molar conductivities at infinite dilution  $\lambda _{m}^{\infty }\text{ }Ba{{\left( OH \right)}_{2}}=457.6\text{ }{{\Omega }^{-1}}c{{m}^{2}}mo{{l}^{-1}}~$  

\[\lambda _{m}^{\infty }\text{ }Ba\text{ }C{{l}_{2}}=\text{ }240.6\text{ }{{\Omega }^{-1}}c{{m}^{2}}mo{{l}^{-1}}~\] 

\[\lambda _{m}^{\infty }\text{ }N{{H}_{4}}Cl=\text{ }129.8\text{ }{{\Omega }^{-1}}c{{m}^{2}}mo{{l}^{-1}}~\] 

Calculate $\lambda _{m}^{\infty }~$ for $N{{H}_{4}}OH$  

Ans: The Kohlrauch’s law states that,

“An infinite dilution each ion migrates independently of it co-ion and makes its own contribution to the total molar conductivity of an electrolyte irrespective of nature.”

\[\lambda _{m(N{{H}_{4}}OH)}^{\infty }=\lambda _{NH_{4}^{+}}^{\infty }+\lambda _{O{{H}^{-}}}^{\infty }\]

 \[=(\lambda _{NH_{4}^{+}}^{\infty }+\lambda _{C{{l}^{-}}}^{\infty })+\dfrac{1}{2}(\lambda _{B{{a}^{2+}}}^{\infty }+2\lambda _{O{{H}^{-}}}^{\infty })-\dfrac{1}{2}(\lambda _{\mathbf{B}{{a}^{2+}}}^{\infty }+2\lambda _{C{{l}^{-}}}^{\infty })\]

\[=\lambda _{m(N{{H}_{4}}Cl)}^{\infty }+\dfrac{1}{2}\left( \lambda _{m(Ba{{(OH)}_{2}})}^{\infty } \right)-\dfrac{1}{2}\left( \lambda _{m(BaC{{l}_{2}})}^{\infty } \right)\]

\[\therefore \lambda _{m(N{{H}_{4}}OH)}^{\infty }=129.8+\dfrac{1}{2}\times 457.6-\dfrac{1}{2}\times 240.6\] = $238.3\text{ }{{\Omega }^{-1}}c{{m}^{2}}mo{{l}^{-1}}~$

2. Calculate the equilibrium constant for the reaction  

$Zn\text{ }+\text{ }C{{d}^{2+}}\leftrightarrow Z{{n}^{2+}}+Cd~~$ 

If $E_{C{{d}^{+2}}/Cd}^{0}=-0.403V$

\[E_{Z{{n}^{+2}}/Zn}^{0}\text{=-}0.763V\] 

\[Antilog\text{ }12.1827~\] 

Ans:\[E_{cell}^{0}=E_{C{{d}^{+2}}/Cd}^{0}-E_{Z{{n}^{+2}}/Zn}^{0}=-0.403V-(-0.763V)=0.360V\]

    n = 2, then the Nernst equation of equilibrium constant and emf of cell

   \[\log {{K}_{c}}=\left( \dfrac{nE_{cell}^{0}}{0.059} \right)=\left( \dfrac{2\times 0.360}{0.059} \right)=12.20\]

   \[{{K}_{c}}=\text{antilog}(12.20)=1.585\times {{10}^{12}}\]

3. Predict the products of electrolyzing of the following  

a. A dil. Solution of ${{H}_{2}}S{{O}_{4}}$ with Pt electrode

Ans: oxygen gas is liberated at anode and hydrogen gas at the cathode.

Reaction at cathode,$2{{H}^{+}}+2{{e}^{-}}\to {{H}_{2}}$ 

Reaction at anode, ${{H}_{2}}O\to 2{{H}^{+}}+\dfrac{1}{2}{{O}_{2}}+2{{e}^{-}}$ 

b. An aqueous solution of $\mathbf{AgN}{{\mathbf{O}}_{3}}$ with a silver electrode.

Ans: At the cathode: silver ions will be deposited in preference to hydrogen ions

At anode :$Ag\to A{{g}^{+}}+{{e}^{-}}$ 

Important Questions of Chapter 2 Chemistry Class 12

Class 12 Chemistry Chapter 2 Important Questions - Topics Covered

The Important Questions of Electrochemistry of Class 12 NCERT will cover the following topics:

• Electrochemical cells

• Galvanic cells

• Measurement of electrode potential

• Nernst equation

• The equilibrium constant from the Nernst equation

• Electrochemical cells and Gibbs free energy of the reaction

• The conductance of the electrolytic cells

• Measurement of the conductivity of the ionic solution

• Variation in conductivity and molar conductivity

• Kohlrausch's law and batteries.

Electrochemistry Class 12 Important Questions solution has been specially designed to provide a better understanding of the student for the preparation of their board exam. Along with these Extra Questions of Electrochemistry Class 12 are provided to the student for the preparation of competitive exams like NEET and JEE (Main). In Class 12 Chemistry Chapter 2 Important Questions the experts at Vedantu have provided the questions with a solution to build up a good conceptual understanding. As such with the main solution, an alternative method of solution (tricks) to solve a particular problem has also been provided.

Conclusion

Our aim is to provide the student with all type of important questions and answers, using a logical approach that can help them build a good base with the fundamentals of the subject. These Important Questions of Chapter 2 Chemistry Class 12 are specially designed to cover all the types of subjective and objective questions and answers which helps the student to write concise answers for the board examination and to solve objective questions for the preparation of competitive exams.

The important questions for CBSE Class 12 Chemistry Chapter 2 Electrochemistry are curated and solved by the subject matter experts at Vedantu as per the latest guidelines for CBSE. Students can solve these important questions on their own and verify their answers by comparing them with the solutions given here. In this way, they will be able to revise all topics and sub-topics covered in the Class 12 Chapter 2 Electrochemistry. This, in turn, will help them analyse their weak and strong areas from this chapter. Thus they will be able to streamline their revision process and prepare for their exam effectively. 

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