/ Class 12 Chemistry, NEB
Class 12 Chemistry Unit 5: Electrochemistry Notes (NEB New Syllabus) | Electrode Potential, Galvanic Cell, Batteries, Numericals | Electrochemistry Class 12 PDF
Importantedunotes.com
Back to Chemistry Notes

1. Syllabus

Unit 5: Electrochemistry

5.1 Electrode potential and standard electrode potential
5.2 Types of electrodes
5.3 Electrochemical series and its applications
5.4 Voltaic cell
5.5 Cell potential and its relationship with free energy
5.6 Commercial batteries and fuel cells

These notes follow the NEB Class 12 Chemistry syllabus (+2 Chemistry, Nepal). Practise the multiple choice questions first, then work through the solved numerical and subjective questions, and finish with the PDF notes for quick revision.

2. Multiple Choice Questions

Question 1. Electrode potential is

  • a) The potential difference across a cell
  • b) The potential of an electrode on connecting with another electrode
  • c) The ability of an electrode to attract ions
  • d) The measure of the ability of the electrode to gain or lose electrons

Correct Answer: d) The measure of the ability of the electrode to gain or lose electrons

Reason: When a metal is dipped in a solution of its own ions, an electrical double layer forms. The potential difference between the metal and its ions measures how strongly the electrode tends to gain electrons (reduction) or lose electrons (oxidation).

Question 2. The conditions required for electrode potential to be standard are:

  • a) \(25^\circ\mathrm{C}\), 1 atm pressure and 0.1 M concentration
  • b) \(0^\circ\mathrm{C}\), 1 atm pressure and 1 M concentration
  • c) \(25^\circ\mathrm{C}\), 1 atm pressure and 1 M concentration
  • d) \(0\ \mathrm{K}\), 1 atm pressure and 1 M concentration

Correct Answer: c) \(25^\circ\mathrm{C}\), 1 atm pressure and 1 M concentration

Reason: The standard conditions are 298 K (25 degrees Celsius), 1 atm pressure for gases and 1 M concentration of the ions in solution.

Question 3. The standard electrode potential of hydrogen is taken as:

  • a) +1 V
  • b) 0 V
  • c) -1 V
  • d) 1.5 V

Correct Answer: b) 0 V

Reason: By convention, the standard hydrogen electrode (SHE) is given a potential of exactly 0 V. The potentials of all other electrodes are measured against it.

Question 4. The standard hydrogen electrode is commonly used as:

  • a) An anode only
  • b) A cathode only
  • c) A reference electrode
  • d) A universal electrode

Correct Answer: c) A reference electrode

Reason: The SHE is the primary reference electrode. It can act as an anode or a cathode depending on the electrode connected to it, so it is not limited to one role.

Question 5. The calomel electrode is primarily used for:

  • a) Determining cell potential
  • b) As a standard reference electrode
  • c) To generate hydrogen gas
  • d) Measuring pH levels

Correct Answer: b) As a standard reference electrode

Reason: The saturated calomel electrode (+0.241 V at 298 K) is a secondary reference electrode. It is easier to handle than the SHE because it needs no supply of hydrogen gas.

Question 6. Metals in the electrochemical series are arranged on the basis of:

  • a) Atomic number
  • b) Atomic mass
  • c) Reduction potential
  • d) Ionization potential

Correct Answer: c) Reduction potential

Reason: The electrochemical series lists the elements in increasing order of their standard reduction potentials.

Question 7. Which of the following has the highest reduction potential?

  • a) Zinc
  • b) Copper
  • c) Lithium
  • d) Fluorine

Correct Answer: d) Fluorine

Reason: The standard reduction potentials are \(E^\circ(\mathrm{F_2/F^-})=+2.87\ \mathrm{V}\), \(E^\circ(\mathrm{Cu^{2+}/Cu})=+0.34\ \mathrm{V}\), \(E^\circ(\mathrm{Zn^{2+}/Zn})=-0.76\ \mathrm{V}\) and \(E^\circ(\mathrm{Li^+/Li})=-3.04\ \mathrm{V}\). Fluorine is the strongest oxidising agent.

Question 8. Which of the following would displace copper from a copper sulphate solution?

  • a) Silver
  • b) Gold
  • c) Nickel
  • d) Platinum

Correct Answer: c) Nickel

Reason: A metal displaces another metal from its salt solution only if its reduction potential is lower. Nickel (\(-0.25\ \mathrm{V}\)) is below copper (\(+0.34\ \mathrm{V}\)), but silver, gold and platinum are above copper.

Question 9. The reaction that occurs at the cathode in a Zn-Cu voltaic cell is:

  • a) \(Zn\rightarrow Zn^{2+}+2e^-\)
  • b) \(Cu\rightarrow Cu^{2+}+2e^-\)
  • c) \(Zn^{2+}+2e^-\rightarrow Zn\)
  • d) \(Cu^{2+}+2e^-\rightarrow Cu\)

Correct Answer: d) \(Cu^{2+}+2e^-\rightarrow Cu\)

Reason: Reduction takes place at the cathode. Copper has the higher reduction potential, so \(\mathrm{Cu^{2+}}\) ions gain electrons and deposit as copper metal.

Question 10. The standard reduction potential of a Zn-Cu cell is:

  • a) 0.34 V
  • b) 1.10 V
  • c) 0.76 V
  • d) 0.42 V

Correct Answer: b) 1.10 V

Reason: \(E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}=0.34-(-0.76)=1.10\ \mathrm{V}\).

Question 11. If the cell potential of a galvanic cell is positive, the reaction is:

  • a) Reversible
  • b) At equilibrium
  • c) Spontaneous
  • d) Non-spontaneous

Correct Answer: c) Spontaneous

Reason: A positive cell potential gives a negative free energy change (\(\Delta G=-nFE<0\)), which is the condition for a spontaneous reaction.

Question 12. The relationship between cell potential and free energy is:

  • a) \(\Delta G=nFE\)
  • b) \(\Delta G=-nFE\)
  • c) \(\Delta G=\dfrac{E}{nF}\)
  • d) \(\Delta G=\dfrac{nF}{E}\)

Correct Answer: b) \(\Delta G=-nFE\)

Reason: The maximum work obtained from a cell is \(w_{max}=nFE\), and this equals the decrease in free energy, so \(-\Delta G=nFE\).

Question 13. The electrolyte in a lead storage battery is:

  • a) Potassium hydroxide
  • b) Sulphuric acid
  • c) Lead hydroxide
  • d) Hydrochloric acid

Correct Answer: b) Sulphuric acid

Reason: The lead storage battery has a lead anode, a lead dioxide cathode and dilute \(\mathrm{H_2SO_4}\) as the electrolyte. The overall discharge reaction is \(\mathrm{Pb+PbO_2+2HSO_4^-+2H^+\rightarrow2PbSO_4+2H_2O}\).

Question 14. The battery which is commonly used in smartphones is:

  • a) Alkaline battery
  • b) Ni-Cd battery
  • c) Lead storage battery
  • d) Li-ion battery

Correct Answer: d) Li-ion battery

Reason: The Li-ion battery is rechargeable, light and has a high energy density, which suits portable devices.

Question 15. The byproduct of a hydrogen/oxygen fuel cell is:

  • a) Carbon dioxide
  • b) Oxygen
  • c) Water
  • d) Sulphur dioxide

Correct Answer: c) Water

Reason: The overall reaction is \(\mathrm{2H_2+O_2\rightarrow2H_2O}\). Water is the only product, so the cell is pollution free.

Question 16. Standard conditions are necessary during measurement of standard electrode potential because:

  • a) They maximize the reaction speed
  • b) They ensure consistent comparison across electrodes
  • c) They reduce the potential difference
  • d) They prevent corrosion

Correct Answer: b) They ensure consistent comparison across electrodes

Reason: The electrode potential changes with concentration, temperature and pressure. Fixing them at standard values lets us compare different electrodes fairly.

Question 17. If the standard reduction potential of a metal is -0.76 V, on comparing it with the standard hydrogen electrode, it indicates:

  • a) Greater tendency to gain electrons than hydrogen
  • b) Greater tendency to lose electrons than hydrogen
  • c) It is more stable than hydrogen
  • d) It has a lower density than hydrogen

Correct Answer: b) Greater tendency to lose electrons than hydrogen

Reason: A negative standard electrode potential means the metal is oxidised more easily than hydrogen, so it is a stronger reducing agent than hydrogen.

Question 18. In a voltaic cell the function of salt bridge is:

  • a) To provide electrons for the reaction
  • b) To keep the solution neutral by allowing flow of ions
  • c) To increase the voltage of the cell
  • d) To prevent the anode from corroding

Correct Answer: b) To keep the solution neutral by allowing flow of ions

Reason: Positive ions build up near the anode and negative ions near the cathode. The salt bridge supplies ions to neutralise this charge and completes the internal circuit.

Question 19. For the construction of a voltaic cell using two different metals, what should be the first step in predicting the cell potential?

  • a) Comparison of atomic masses of metals
  • b) Identification of ionization potential of metals
  • c) Identification of reduction potentials of metals
  • d) Identification of electrolyte in salt bridge

Correct Answer: c) Identification of reduction potentials of metals

Reason: The cathode and the anode are decided from the reduction potentials, and the cell potential is \(E^\circ_{cathode}-E^\circ_{anode}\).

Question 20. In real-world contexts the use of electrochemical series is best explained by:

  • a) Determining the quality of metals for construction
  • b) Determining the color of metal compounds
  • c) Comparing the market price of metals
  • d) Selecting metals for rechargeable batteries

Correct Answer: d) Selecting metals for rechargeable batteries

Reason: The series gives the voltage that two electrode materials can produce, which guides the choice of electrodes in batteries.

Question 21. Engineers choose materials with high reduction potential for designing commercial battery to:

  • a) Ensure a lower energy density
  • b) Achieve a higher cell potential for greater power output
  • c) Make the battery heavier and more stable
  • d) Decrease the overall cost

Correct Answer: b) Achieve a higher cell potential for greater power output

Reason: A cathode with a high reduction potential combined with an anode of low reduction potential gives a larger cell potential, and so more electrical energy per unit charge.

Question 22. When metal X and Y with +0.80 V and -0.76 V respectively are used in a galvanic cell then for spontaneity of cell:

  • a) Electrons flow from X to Y
  • b) Electrons flow from Y to X
  • c) Current flow from Y to X
  • d) X should be the anode and Y should be the cathode

Correct Answer: b) Electrons flow from Y to X

Reason: Y has the lower reduction potential, so it is the anode and loses electrons. X is the cathode. Electrons move from the anode (Y) to the cathode (X), and the cell potential is \(0.80-(-0.76)=1.56\ \mathrm{V}\).

Question 23. If a voltaic cell has a cell potential +1.0 V, then

  • a) \(\Delta G\) is positive indicating non-spontaneous reaction
  • b) \(\Delta G\) is negative indicating spontaneous reaction
  • c) \(\Delta G\) is zero indicating the reaction at equilibrium
  • d) The value of \(\Delta G\) depends upon the concentration of reactant

Correct Answer: b) \(\Delta G\) is negative indicating spontaneous reaction

Reason: \(\Delta G=-nFE\), so a positive E gives a negative \(\Delta G\).

Question 24. To prevent marine corrosion in an iron structure the metal suitable for sacrificial metal is

  • a) Copper
  • b) Zinc
  • c) Gold
  • d) Aluminium

Correct Answer: b) Zinc

Reason: A sacrificial metal must have a lower reduction potential than iron (\(-0.44\ \mathrm{V}\)), so that it is oxidised in place of iron. Zinc (\(-0.76\ \mathrm{V}\)) does this, while copper and gold do not. Aluminium also qualifies by potential, and zinc is the answer in the textbook.

3. Numerical and Subjective Questions

Exercise 5.2 Questions

Question 1. Given: \(E^\circ_{\mathrm{Na^+/Na}}=-2.71\ \mathrm{V}\) and \(E^\circ_{\mathrm{Au^{3+}/Au}}=+1.5\ \mathrm{V}\)

a. Which ion is a better reducing agent: Na or Au?

A reducing agent is the species that loses electrons. The more negative the reduction potential, the greater the tendency to lose electrons. Since \(-2.71<+1.5\), Na (the metal of the \(\mathrm{Na^+/Na}\) couple) is the much better reducing agent. The ions \(\mathrm{Na^+}\) and \(\mathrm{Au^{3+}}\) cannot be oxidised further, and \(\mathrm{Au^{3+}}\) is the stronger oxidising agent.

b. Can \(\mathrm{Au^{3+}}\) oxidize Na to \(\mathrm{Na^+}\)? Write the reaction if feasible.

Yes. \(\mathrm{Au^{3+}}\) has the high positive potential, so it gains electrons easily, and Na loses electrons easily. The cell potential is positive:

\[ E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}=1.5-(-2.71)=+4.21\ \mathrm{V} \]

Oxidation: \(\mathrm{Na\rightarrow Na^++e^-}\) (multiply by 3). Reduction: \(\mathrm{Au^{3+}+3e^-\rightarrow Au}\).

\[ \mathrm{Au^{3+}(aq)+3Na(s)\rightarrow Au(s)+3Na^+(aq)} \]

c. Why is gold found in nature in its elemental form?

Gold has a high positive reduction potential (\(+1.5\ \mathrm{V}\)), so it has very little tendency to lose electrons and be oxidised. It does not react with oxygen, water or other substances in the earth, so it stays as the free metal (native state).

Question 2. The electrochemical series has many applications. It can also help in the protection of metal from corrosion.

a. Why does iron corrode faster in saline water? Use the electrochemical series to explain.

Iron has a negative reduction potential (\(E^\circ_{\mathrm{Fe^{2+}/Fe}}=-0.44\ \mathrm{V}\)), while oxygen has a high positive one, so iron is oxidised by oxygen in the presence of water. Corrosion is an electrochemical process, in which small anodic and cathodic areas on the iron surface form tiny galvanic cells:

\[ \mathrm{Fe\rightarrow Fe^{2+}+2e^-}\ (\text{anode}),\qquad \mathrm{O_2+4H^++4e^-\rightarrow2H_2O}\ (\text{cathode}) \]

Saline water contains dissolved salts such as NaCl, which are strong electrolytes. The ions raise the conductivity of the water and complete the circuit between the anodic and cathodic areas, so the current flows more easily and iron corrodes faster.

b. Suggest why aluminum is used as sacrificial anode for iron ships. Even though \(E^\circ_{\mathrm{Al^{3+}/Al}}

The lower reduction potential of aluminium (\(-1.66\ \mathrm{V}\) against \(-0.44\ \mathrm{V}\) for iron) is exactly the reason. A metal with a lower reduction potential is oxidised more easily. When aluminium is connected to the iron hull, aluminium acts as the anode and is oxidised, and iron becomes the cathode and is protected. The aluminium is used up (sacrificed) and is replaced from time to time, which is cheaper than replacing the ship.

c. Write the oxidation half-reaction for aluminum acts as a sacrificial anode.

\[ \mathrm{Al(s)\rightarrow Al^{3+}(aq)+3e^-} \]
Question 3. A galvanic cell consists of the following half-cells:
\[ \mathrm{Ni^{2+}+2e^-\longrightarrow Ni}\qquad E^\circ=-0.25\ \mathrm{V} \]
\[ \mathrm{Cr^{3+}+3e^-\longrightarrow Cr}\qquad E^\circ=-0.74\ \mathrm{V} \]
Answer the following questions:

a. Identify the anode and cathode. Write cell notation.

The electrode with the lower reduction potential is the anode. Anode: Cr (\(-0.74\ \mathrm{V}\)). Cathode: Ni (\(-0.25\ \mathrm{V}\)).

\[ \mathrm{(-)\ Cr(s)\,|\,Cr^{3+}(aq)\,||\,Ni^{2+}(aq)\,|\,Ni(s)\ (+)} \]

b. Write cell reaction showing the electrode reactions.

\[ \text{Anode (oxidation): }\mathrm{Cr(s)\rightarrow Cr^{3+}(aq)+3e^-}\quad(\times2) \]
\[ \text{Cathode (reduction): }\mathrm{Ni^{2+}(aq)+2e^-\rightarrow Ni(s)}\quad(\times3) \]
\[ \mathrm{2Cr(s)+3Ni^{2+}(aq)\rightarrow2Cr^{3+}(aq)+3Ni(s)} \]

Six electrons are transferred in the balanced equation (\(n=6\)).

c. Calculate standard emf of the cell.

\[ E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}=-0.25-(-0.74)=+0.49\ \mathrm{V} \]

d. What will happen to the cell potential if the concentration of \(\mathrm{Ni^{2+}}\) is increased to 2 M while \(\mathrm{Cr^{3+}}\) remains at 1 M?

Use the Nernst equation at 298 K with \(n=6\):

\[ E=E^\circ-\dfrac{0.0592}{n}\log\dfrac{[\mathrm{Cr^{3+}}]^2}{[\mathrm{Ni^{2+}}]^3} \]
\[ E=0.49-\dfrac{0.0592}{6}\log\dfrac{(1)^2}{(2)^3}=0.49+0.00987\times0.903=0.499\ \mathrm{V} \]

The cell potential increases slightly, from 0.49 V to about 0.499 V. A higher concentration of the reactant \(\mathrm{Ni^{2+}}\) (at the cathode) makes the forward reaction more favourable (Le Chatelier's principle), so the driving force of the cell grows.

Question 4. To construct a galvanic cell of zinc and iron, following reaction is given:
\[ \mathrm{Zn(s)+Fe^{2+}(aq)\longrightarrow Zn^{2+}(aq)+Fe(s)} \]

a. Convert the above reaction in to cell notation.

Zn is oxidised (anode, left) and \(\mathrm{Fe^{2+}}\) is reduced (cathode, right).

\[ \mathrm{(-)\ Zn(s)\,|\,Zn^{2+}(aq)\,||\,Fe^{2+}(aq)\,|\,Fe(s)\ (+)} \]

b. Calculate the \(\Delta G^\circ\) for the using the standard cell potential \(E^\circ_{cell}=0.32\ \mathrm{V}\).

Given: \(n=2\), \(F=96485\ \mathrm{C\,mol^{-1}}\), \(E^\circ_{cell}=0.32\ \mathrm{V}\)
\[ \Delta G^\circ=-nFE^\circ_{cell}=-2\times96485\times0.32=-61750\ \mathrm{J\,mol^{-1}} \]
\[ \Delta G^\circ\approx-61.75\ \mathrm{kJ\,mol^{-1}} \]

c. Is this reaction spontaneous under standard conditions? Justify.

Yes. The standard cell potential is positive (\(+0.32\ \mathrm{V}\)) and \(\Delta G^\circ\) is negative (\(-61.75\ \mathrm{kJ/mol}\)). Both conditions of spontaneity are met. The potentials also agree: \(-0.44-(-0.76)=0.32\ \mathrm{V}\).

d. What happens to the mass of zinc electrode as the reaction proceeds?

The mass of the zinc electrode decreases. Zinc is the anode, so zinc atoms are oxidised and pass into the solution as \(\mathrm{Zn^{2+}}\) ions, and the electrode dissolves.

Question 5. The following standard reduction potentials are given: \(E^\circ_{\mathrm{Mg^{2+}/Mg}}=-2.37\ \mathrm{V}\) and \(E^\circ_{\mathrm{Ni^{2+}/Ni}}=-0.25\ \mathrm{V}\)

a. Design a galvanic cell using Mg and Ni. Label the anode, cathode and electron flow.

Mg has the lower reduction potential (\(-2.37\ \mathrm{V}\)), so Mg is the anode. Ni has the higher value (\(-0.25\ \mathrm{V}\)), so Ni is the cathode. Put a Mg rod in 1 M \(\mathrm{MgSO_4}\) and a Ni rod in 1 M \(\mathrm{NiSO_4}\), join the two solutions with a salt bridge (KCl or \(\mathrm{KNO_3}\)) and connect the rods with a wire and a voltmeter. Electrons flow through the wire from the Mg electrode (anode, negative) to the Ni electrode (cathode, positive); the current flows in the opposite direction.

b. Write cell notation and cell reaction

\[ \mathrm{(-)\ Mg(s)\,|\,Mg^{2+}(aq)\,||\,Ni^{2+}(aq)\,|\,Ni(s)\ (+)} \]
\[ \text{Anode: }\mathrm{Mg\rightarrow Mg^{2+}+2e^-}\qquad\text{Cathode: }\mathrm{Ni^{2+}+2e^-\rightarrow Ni} \]
\[ \mathrm{Mg(s)+Ni^{2+}(aq)\rightarrow Mg^{2+}(aq)+Ni(s)} \]

c. Calculate cell potential.

\[ E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}=-0.25-(-2.37)=+2.12\ \mathrm{V} \]

d. Which metal acts as oxidizing agent? Explain

The oxidising agent is the species that is reduced, so it is the \(\mathrm{Ni^{2+}}\) ion (the species of the nickel couple). It takes the electrons from Mg and is reduced to Ni. The metals themselves are not oxidising agents here: Mg is the reducing agent because it loses electrons.

Question 6. During the construction of a galvanic cell no voltage was observed.

a. List three possible reasons for this observation.

The salt bridge is missing, dried up or not dipped in both solutions, so the circuit is incomplete.
The external circuit is open: a loose or broken wire, or a faulty voltmeter connection.
Both electrodes are of the same metal, or both half-cells are identical, so there is no potential difference between them.

b. How can a person overcome the problem to ensure proper functioning?

Use a fresh salt bridge (a U-tube of KCl or \(\mathrm{KNO_3}\) in agar, or soaked filter paper) with both ends dipped in the solutions.
Check and tighten all the connections and test the voltmeter.
Use two different metals, each dipped in a 1 M solution of its own ions, so that their reduction potentials differ.

c. Explain the role of salt bridge in maintaining cell operation.

The salt bridge completes the internal circuit by providing an ionic path. As the cell works, \(\mathrm{Zn^{2+}}\) type ions build up in the anode compartment and positive charge accumulates, while the cathode compartment loses positive ions. The salt bridge supplies anions to the anode half-cell and cations to the cathode half-cell, so both stay electrically neutral. It also prevents the solutions from mixing and reduces the liquid junction potential. Without it, the charge build-up stops the flow of electrons and the voltage drops to zero.

Question 7. Five different Metals P, Q, R, S and T show the following reactivity: R displaces P and Q from their solutions while S does not react with any of the metal solutions. T displaces hydrogen from 1 M acid but not R. P displaces Q from its solution.

a. Arrange the metals in an activity series with hydrogen.

Each statement gives one comparison:

R displaces P and Q, so R is more reactive than both P and Q.
P displaces Q, so P is more reactive than Q. Together: R > P > Q.
T displaces hydrogen from acid, so T is above hydrogen. T does not displace R, so R is more reactive than T. Together: R > T > H.
S does not react with any of the metal solutions, so S is the least reactive of the five metals and is placed at the bottom.

The statements place R at the top and S at the bottom, and keep the orders R > P > Q and R > T > H. The position of T relative to P and Q is not fixed by these statements alone. The series is therefore: R > (P > Q) and (T) > H > S.

Note: the potentials given in part b (\(E^\circ_{\mathrm{T^+/T}}=+0.34\ \mathrm{V}\)) place T above hydrogen in reduction potential, so they are used only for the cell calculation in part b.

If the standard reduction potentials of the following electrodes are: \(P^{2+}+2e^-\rightarrow P,\ E^\circ=-0.44\ \mathrm{V}\) and \(T^++e^-\rightarrow T,\ E^\circ=+0.34\ \mathrm{V}\)

b. i. Construct a Galvanic cell and indicate the anode and cathode.

P has the lower reduction potential, so P is the anode (oxidation). T has the higher value, so T is the cathode (reduction).

\[ \mathrm{(-)\ P(s)\,|\,P^{2+}(aq)\,||\,T^{+}(aq)\,|\,T(s)\ (+)} \]
\[ \mathrm{P\rightarrow P^{2+}+2e^-}\qquad\mathrm{2T^++2e^-\rightarrow2T} \]
\[ \mathrm{P(s)+2T^+(aq)\rightarrow P^{2+}(aq)+2T(s)} \]

c. ii. Calculate the EMF of the cell with 1M ion solutions.

\[ E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}=0.34-(-0.44)=+0.78\ \mathrm{V} \]

d. iii. Predict whether the reaction \(P^{2+}+T\rightarrow P+T^+\) will occur. Explain

In this reaction \(\mathrm{P^{2+}}\) is reduced and T is oxidised, so P is the cathode and T is the anode:

\[ E^\circ=E^\circ_{(P^{2+}/P)}-E^\circ_{(T^+/T)}=-0.44-(+0.34)=-0.78\ \mathrm{V} \]

The potential is negative, so the free energy change is positive and the reaction will not occur spontaneously. It is the reverse of the cell reaction in part b, which does occur: P displaces \(\mathrm{T^+}\) from solution, but T cannot displace P from its solution.

4. View PDF Notes

i. Handwritten Notes

ii. Typed PDF Notes

5. Frequently Asked Questions

What is electrochemistry? (Class 12 Chemistry)
Electrochemistry is the branch of chemistry that studies the relationship between electrical energy and chemical reactions, mainly redox (oxidation-reduction) reactions. It explains how chemical energy is converted into electrical energy in batteries (galvanic cells) and how electrical energy drives non-spontaneous reactions in electrolysis. Its applications include batteries, fuel cells, electroplating and corrosion prevention. It is Unit 5 of NEB Class 12 Chemistry.
What topics are included in Unit 5 Electrochemistry of NEB Class 12 Chemistry?
5.1 Electrode potential and standard electrode potential; 5.2 types of electrodes (standard hydrogen electrode and calomel electrode); 5.3 electrochemical series and its applications; 5.4 voltaic (galvanic) cell; 5.5 cell potential and its relationship with free energy; 5.6 commercial batteries and fuel cells.
What is the difference between a galvanic cell and an electrolytic cell?
A galvanic (voltaic) cell converts chemical energy into electrical energy through a spontaneous redox reaction. Its two electrodes dip in different electrolyte solutions joined by a salt bridge, the anode is negative and the cathode is positive. An electrolytic cell uses electrical energy to drive a non-spontaneous reaction. Both electrodes dip in the same electrolyte, the anode is positive and the cathode is negative. In both cells oxidation occurs at the anode and reduction at the cathode.
What is electrode potential and standard electrode potential?
When a metal is dipped in a solution of its own ions, an electrical double layer forms and a potential difference is set up between the metal and its ions at the interface. This is the electrode potential. It depends on the concentration of the ions, the temperature and the nature of the metal. The standard electrode potential (E0) is measured under standard conditions: 1 M ion concentration, 298 K (25 degrees Celsius) and 1 atm pressure, against the standard hydrogen electrode, which is assigned 0 V. A positive E0 means a greater tendency to gain electrons than hydrogen, and a negative E0 means a greater tendency to lose electrons.
What is the standard hydrogen electrode (SHE)? What are its limitations?
The SHE is the primary reference electrode, assigned an electrode potential of exactly 0 V at all temperatures. It consists of a platinum foil coated with finely divided platinum (platinum black) dipped in 1 M H+ solution (for example HCl), with pure hydrogen gas at 1 atm bubbled around it: H2 gives 2H+ + 2e-. Its limitations are that it needs very pure hydrogen and costly platinum, and that it is difficult to set up and use for routine work.
What is a calomel electrode? Why is it preferred as a secondary reference electrode?
The calomel electrode is a secondary reference electrode made of mercury covered with a paste of mercurous chloride (Hg2Cl2, calomel) in contact with a saturated KCl solution, with a platinum wire for the electrical contact. The electrode reaction is Hg2Cl2(s) + 2e- gives 2Hg(l) + 2Cl-(aq), and the potential of the saturated calomel electrode is +0.241 V at 298 K. It is preferred because it is compact, reliable, easy to maintain and needs no continuous gas supply, though mercury is toxic and must be handled carefully.
What is the electrochemical series and what are its applications?
The electrochemical series (activity series) arranges elements in increasing order of their standard reduction potentials, measured against the SHE. Applications: to identify oxidising agents (high E0) and reducing agents (low E0); to calculate the cell potential; to predict the feasibility of a galvanic cell; to predict metal-metal displacement reactions; to predict corrosion and choose a sacrificial anode; to predict the products of electrolysis; and to decide whether a metal can liberate hydrogen from a non-oxidising acid (only the metals with negative E0, above hydrogen in the series, can).
What is the Daniel cell? Write its reactions and cell notation.
The Daniel cell is the Zn-Cu galvanic cell. Anode (oxidation): Zn(s) gives Zn2+(aq) + 2e-. Cathode (reduction): Cu2+(aq) + 2e- gives Cu(s). Cell reaction: Zn(s) + Cu2+(aq) gives Zn2+(aq) + Cu(s). Cell notation: (-) Zn(s) | Zn2+(aq, 1 M) || Cu2+(aq, 1 M) | Cu(s) (+). The standard cell potential is E0cell = E0cathode - E0anode = 0.34 - (-0.76) = +1.10 V. Electrons flow from Zn to Cu through the wire, so the current flows from Cu to Zn.
What is the function of the salt bridge?
The salt bridge (a U-tube of an inert electrolyte such as KCl or KNO3 in agar gel, or filter paper soaked in it) completes the electrical circuit by providing an ionic path between the two half-cells. It keeps both solutions electrically neutral by letting anions move towards the anode and cations towards the cathode, it prevents direct mixing of the solutions, and it minimises the liquid junction potential.
How is cell potential related to free energy?
The maximum electrical work a cell can do is w(max) = nFE(cell), and the maximum useful work equals the decrease in free energy, so the change in Gibbs free energy is delta G = -nFE(cell), and under standard conditions delta G0 = -nFE0(cell). Here n is the number of electrons transferred and F = 96485 C/mol (Faraday constant). A positive cell potential gives a negative delta G, which means the reaction is spontaneous.
What are the types of commercial batteries? Give examples.
Primary batteries cannot be recharged, for example the dry (Leclanche) cell with a zinc anode, a graphite cathode and a paste of MnO2, NH4Cl and ZnCl2. Secondary batteries are rechargeable, for example the lead storage battery (Pb anode, PbO2 cathode, dilute H2SO4 electrolyte; overall reaction Pb + PbO2 + 2HSO4- + 2H+ gives 2PbSO4 + 2H2O), the nickel-cadmium battery (Cd anode, NiO(OH) cathode, KOH electrolyte) and the Li-ion battery used in smartphones. A fuel cell, for example the hydrogen-oxygen fuel cell, makes electricity from a fuel supplied continuously, and its only product is water.
Where can I find the Electrochemistry formula sheet for Class 12?
E(cell) = E(cathode) - E(anode); E0(cell) = E0(cathode) - E0(anode); delta G = -nFE(cell); delta G0 = -nFE0(cell); w(max) = nFE(cell); F = 96485 C/mol; Nernst equation at 298 K: E = E0 - (0.0592/n) log Q. SHE = 0 V; saturated calomel electrode = +0.241 V. Spontaneous cell: E0(cell) > 0 and delta G < 0. The notes, solved exercise and PDF on this page cover every formula of the unit.
Where can I find Electrochemistry Class 12 numericals with solutions?
This page gives step-by-step solutions to all questions of Exercise 5.2: reducing power of Na and Au, corrosion and the sacrificial anode, the Ni-Cr cell with a concentration change, delta G of the Zn-Fe reaction, the Mg-Ni cell, troubleshooting a galvanic cell and the activity series of five metals.
Where can I get Class 12 Chemistry Electrochemistry notes in PDF? Is there a book or PPT?
This page includes handwritten notes and typed PDF notes of Unit 5 Electrochemistry that you can view online free of cost, along with solved multiple choice questions and the exercise as per the NEB Class 12 Chemistry syllabus 2083. The typed PDF works as a short chapter book, and its short headings can be used like a PPT for quick revision before the exam.
Where can I get Class 12 Chemistry notes of NEB (handwritten, PDF, Nepal)?
The Class 12 Chemistry notes of the NEB syllabus 2083 are available on this website unit by unit, in handwritten and typed PDF form. Students who look for notes in the style of Note Mandir will find this unit in the same way: syllabus, solved exercise, handwritten notes and typed notes. These are NEB Class 12 Chemistry notes for Chapter 5, Electrochemistry.

Students also search for:

Electrochemistry class 12 Electrochemistry pdf Electrochemistry notes Electrochemistry notes pdf Electrochemistry Class 12 Notes Electrochemistry PPT Electrochemistry book Class 12 chemistry notes neb 2083 Class 12 Chemistry notes PDF Nepal Class 12 Chemistry notes NEB Class 12 chemistry notes chapter 5 Class 12 chemistry notes handwritten Class 12 chemistry notes note mandir Class 12 Chemistry syllabus NEB PDF

Also Read: NEB Class 12 Notes

Related Posts

Scroll to Top