Chapter 5: Electrochemistry
Exhaustive Theory, Nernst Equation, Cell Reactions, Numericals & Solved PYQs
1. Introduction and Types of Conductors
Electrochemistry is the branch of physical chemistry which deals with the relationship between electrical energy and chemical energy, and the interconversion of one form into the other.
Metallic vs. Electrolytic Conductors
| Property | Metallic (Electronic) Conductors | Electrolytic (Ionic) Conductors |
|---|---|---|
| Charge Carriers | Flow of free electrons. | Movement of ions in molten/aqueous state. |
| Chemical Change | No chemical change occurs during conduction. | Chemical decomposition occurs at electrodes. |
| Transfer of Matter | No transfer of matter takes place. | Ions move towards opposite electrodes. |
| Effect of Temp. | Conductivity decreases with increase in temperature. | Conductivity increases with increase in temperature. |
2. Electrical Conductance and Molar Conductivity
According to Ohm's law, electrical resistance ($R$) is directly proportional to length ($l$) and inversely proportional to area of cross-section ($a$).
$$R = \rho \frac{l}{a}$$Where $\rho$ (rho) is resistivity.
Conductivity ($\kappa$ or kappa): It is the reciprocal of resistivity ($\kappa = 1/\rho$). The unit of conductivity is $\Omega^{-1} \text{m}^{-1}$ or $\text{S m}^{-1}$ (Siemens per meter). Frequently used unit is $\text{S cm}^{-1}$.
Therefore, $\kappa = \frac{1}{R} \frac{l}{a}$. The quantity $(l/a)$ is called the Cell Constant ($b$).
Molar Conductivity ($\Lambda_m$)
It is defined as the conducting power of all the ions produced by dissolving 1 mole of an electrolyte in a given volume of solution.
$$\Lambda_m = \frac{\kappa}{C}$$If $\kappa$ is in $\text{S cm}^{-1}$ and concentration ($C$) is in $\text{mol L}^{-1}$, then:
$$\Lambda_m = \frac{1000 \times \kappa}{C}$$
Unit of $\Lambda_m$: $\text{S cm}^2 \text{mol}^{-1}$ or $\Omega^{-1} \text{cm}^2 \text{mol}^{-1}$
3. Kohlrausch's Law of Independent Migration of Ions
At infinite dilution, dissociation is complete, and all inter-ionic attractions disappear. The molar conductivity reaches its maximum limiting value, denoted by $\Lambda_0$ (Molar conductivity at zero concentration).
Statement: At infinite dilution, each ion migrates independently of its co-ion and makes its own definite contribution to the total molar conductivity of the electrolyte, irrespective of the nature of the other ion with which it is associated.
$$\Lambda_0 = \nu_+ \lambda_0^+ + \nu_- \lambda_0^-$$Where $\lambda_0^+$ and $\lambda_0^-$ are the molar ionic conductivities of cation and anion at infinite dilution, and $\nu_+$ and $\nu_-$ are the number of cations and anions produced by 1 formula unit of the electrolyte.
Example: For $BaCl_2$, $\Lambda_0(BaCl_2) = \lambda_0(Ba^{2+}) + 2\lambda_0(Cl^-)$
Applications of Kohlrausch's Law
- To calculate $\Lambda_0$ for weak electrolytes, which cannot be determined experimentally by extrapolation. (e.g., $\Lambda_0(CH_3COOH) = \Lambda_0(CH_3COONa) + \Lambda_0(HCl) - \Lambda_0(NaCl)$).
- To calculate degree of dissociation ($\alpha$) of weak electrolytes: $\alpha = \frac{\Lambda_c}{\Lambda_0}$.
4. Electrochemical Cells
Devices used to convert chemical energy into electrical energy or vice-versa.
| Electrolytic Cell | Galvanic (Voltaic) Cell |
|---|---|
| Converts Electrical Energy into Chemical Energy. | Converts Chemical Energy into Electrical Energy. |
| Non-spontaneous reaction is driven by an external battery. | Spontaneous chemical reaction generates electricity. |
| Anode is Positive (+), Cathode is Negative (-). | Anode is Negative (-), Cathode is Positive (+). |
| Both electrodes are in the same container. | Electrodes are in separate half-cells connected by a salt bridge. |
Note: In BOTH cells, Oxidation always occurs at the Anode, and Reduction always occurs at the Cathode. (Trick to remember: An Ox, Red Cat).
5. Faraday's Laws of Electrolysis
These laws govern the quantitative aspects of electrolysis in an Electrolytic cell.
First Law
Statement: The mass of any substance deposited or liberated at an electrode is directly proportional to the quantity of electricity ($Q$) passed through the electrolyte.
$$W \propto Q \implies W = Z \cdot Q$$Since $Q = I \times t$ (Current $\times$ time),
$$W = Z \cdot I \cdot t$$
Where $Z$ is Electrochemical Equivalent (ECE). $Z = \frac{\text{Molar Mass}(M)}{n \times F}$.
($n$ = moles of electrons required, $F$ = Faraday's constant = $96500 \text{ C/mol}$).
Second Law
Statement: When the same quantity of electricity is passed through different electrolytes connected in series, the masses of substances liberated at the respective electrodes are directly proportional to their equivalent masses.
$$\frac{W_1}{W_2} = \frac{E_1}{E_2}$$6. Galvanic or Voltaic Cells
A classic example is the Daniell Cell, which uses zinc and copper electrodes.
- Anode (-): Zinc rod dipped in $ZnSO_4$. (Oxidation: $Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-$)
- Cathode (+): Copper rod dipped in $CuSO_4$. (Reduction: $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$)
Salt Bridge
A U-shaped glass tube containing a saturated solution of an inert electrolyte (like $KCl, NH_4NO_3, KNO_3$) in agar-agar gel.
Functions of Salt Bridge:- Completes the electrical circuit by allowing the flow of ions.
- Maintains the electrical neutrality of both the half-cell solutions.
- Prevents the mixing of the two electrolytic solutions.
Cell Notation (IUPAC Convention)
The standard way to write a cell representation:
Example for Daniell Cell: $Zn(s) | Zn^{2+}(1M) || Cu^{2+}(1M) | Cu(s)$
7. Thermodynamics of Galvanic Cells
Electrical work done by a galvanic cell is equal to the decrease in its Gibbs free energy ($-\Delta G$).
$$\text{Electrical Work} = \text{Charge} \times \text{Cell Potential} = nF \times E_{cell}$$Therefore, $$\Delta G = -nFE_{cell}$$
Under standard conditions ($1 \text{ M}$, $1 \text{ atm}$, $298 \text{ K}$):
$$\Delta G^\circ = -nFE^\circ_{cell}$$
- For a spontaneous cell reaction, $\Delta G$ must be negative, which means $E_{cell}$ must be positive.
- Relationship with Equilibrium Constant ($K$): $\Delta G^\circ = -RT \ln K = -nFE^\circ_{cell}$.
Therefore, $\ln K = \frac{nFE^\circ_{cell}}{RT}$ or $E^\circ_{cell} = \frac{2.303 RT}{nF} \log_{10} K$.
8. Nernst Equation
The Nernst equation gives the relationship between the electrode potential (or cell potential) and the concentration of the electrolytic solutions.
For a general cell reaction: $aA + bB \rightarrow cC + dD$
$$E_{cell} = E^\circ_{cell} - \frac{RT}{nF} \ln \frac{[C]^c [D]^d}{[A]^a [B]^b}$$At 298 K ($25^\circ\text{C}$), substituting values of $R, T,$ and $F$, and converting to base-10 log:
$$E_{cell} = E^\circ_{cell} - \frac{0.0592}{n} \log_{10} \frac{[\text{Products}]}{[\text{Reactants}]}$$
Note: The concentration of pure solids and pure liquids is taken as 1.
9. Commercial Cells (Batteries)
Cells used as sources of electrical energy are called batteries. They are of two types:
- Primary Cells: Cannot be recharged as the cell reaction is not reversible. Example: Dry cell (Leclanche cell).
- Secondary Cells: Can be recharged by passing direct current in the opposite direction. Example: Lead accumulator, Nickel-Cadmium cell.
- Fuel Cells: Devices that convert the energy of combustion of fuels (like $H_2, CH_4$) directly into electrical energy.
Lead Storage Battery (Accumulator)
Widely used in automobiles and inverters. It is a secondary cell.
- Anode: Spongy lead ($Pb$).
- Cathode: Lead grid packed with lead dioxide ($PbO_2$).
- Electrolyte: $38\%$ by mass $H_2SO_4$ solution.
Discharging Reactions (Cell in use):
At Anode: $Pb(s) + SO_4^{2-}(aq) \rightarrow PbSO_4(s) + 2e^-$
At Cathode: $PbO_2(s) + 4H^+(aq) + SO_4^{2-}(aq) + 2e^- \rightarrow PbSO_4(s) + 2H_2O(l)$
Net: $Pb(s) + PbO_2(s) + 2H_2SO_4(aq) \rightarrow 2PbSO_4(s) + 2H_2O(l)$
During recharging, the reactions are exactly reversed, and $H_2SO_4$ is regenerated.
10. Solved Textbook Numericals
Problem 1: Nernst Equation
Calculate the EMF of the following cell at 298 K:
$Mg(s) | Mg^{2+}(0.1 \text{ M}) || Cu^{2+}(0.01 \text{ M}) | Cu(s)$
Given: $E^\circ(Mg^{2+}/Mg) = -2.37 \text{ V}$ and $E^\circ(Cu^{2+}/Cu) = +0.34 \text{ V}$.
Solution:
1. Calculate $E^\circ_{cell}$:
$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.34 - (-2.37) = 2.71 \text{ V}$
2. Cell Reaction:
Anode: $Mg \rightarrow Mg^{2+} + 2e^-$
Cathode: $Cu^{2+} + 2e^- \rightarrow Cu$
Net: $Mg + Cu^{2+} \rightarrow Mg^{2+} + Cu$ (Here, $n = 2$ electrons exchanged).
3. Apply Nernst Equation:
$E_{cell} = E^\circ_{cell} - \frac{0.0592}{n} \log_{10} \frac{[Mg^{2+}]}{[Cu^{2+}]}$
$E_{cell} = 2.71 - \frac{0.0592}{2} \log_{10} \frac{0.1}{0.01}$
$E_{cell} = 2.71 - 0.0296 \log_{10}(10)$
$E_{cell} = 2.71 - 0.0296(1) = 2.6804 \text{ V}$
Answer: The EMF of the cell is 2.6804 V.
Problem 2: Faraday's First Law
A current of 1.5 Amperes was passed through an electrolytic cell containing $AgNO_3$ solution with inert electrodes for 2 hours. What mass of silver was deposited? (Molar mass of Ag = 108 g/mol, $F = 96500 \text{ C/mol}$).
Solution:
Given: $I = 1.5 \text{ A}$, $t = 2 \text{ hours} = 2 \times 60 \times 60 = 7200 \text{ s}$
Reaction at cathode: $Ag^+ + e^- \rightarrow Ag(s)$ (Here, $n = 1$ mole of electrons per mole of Ag).
Total charge $Q = I \times t = 1.5 \times 7200 = 10800 \text{ C}$
Using formula: $W = \frac{M}{n \times F} \times Q$
$W = \frac{108}{1 \times 96500} \times 10800$
$W = \frac{108 \times 108}{965} = \frac{11664}{965} = 12.087 \text{ g}$
Answer: The mass of silver deposited is 12.087 g.
11. Board PYQs with Complete Answers
Verified previous year questions from the Maharashtra State Board HSC Chemistry exams.
1 Mark Questions (VSA)
Q1. State Kohlrausch’s law. (Oct 2013, March 2017)
Answer: At infinite dilution, each ion of an electrolyte migrates independently of its co-ion and makes its own definite contribution to the total molar conductivity of the electrolyte.
Q2. Write the Nernst equation for a single electrode. (March 2016, Oct 2021)
Answer: For an electrode reaction $M^{n+} + ne^- \rightarrow M(s)$, the Nernst equation is: $E = E^\circ - \frac{2.303RT}{nF} \log_{10} \frac{1}{[M^{n+}]}$.
Q3. What is a salt bridge? (March 2018, March 2023)
Answer: A salt bridge is a U-shaped glass tube containing a saturated solution of an inert electrolyte (like $KCl$ or $NH_4NO_3$) in agar-agar gel, used to connect the two half-cells of a galvanic cell.
2 Mark Questions (SA-I)
Q4. State Faraday's first law of electrolysis. Give its mathematical expression. (March 2014, Oct 2019)
Statement: The mass of any substance deposited or liberated at an electrode during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte.
Expression: $W = Z \cdot I \cdot t$ (where $W$ is mass, $Z$ is electrochemical equivalent, $I$ is current in amperes, and $t$ is time in seconds).
Q5. Write any two functions of a salt bridge. (March 2015, March 2022)
- It provides an electrical contact between the two half-cells, completing the internal circuit.
- It maintains electrical neutrality in both half-cells by allowing the flow of ions.
3 Mark Questions (SA-II)
Q6. Write the discharging reactions occurring at the anode and cathode in a lead storage battery. (Oct 2014, March 2019, Oct 2020)
During discharging (when the battery provides current), it acts as a galvanic cell.
At Anode (Oxidation): Spongy lead is oxidized to $Pb^{2+}$ which immediately precipitates as $PbSO_4$.
$Pb(s) + SO_4^{2-}(aq) \rightarrow PbSO_4(s) + 2e^-$
At Cathode (Reduction): Lead dioxide is reduced to $Pb^{2+}$ which also precipitates as $PbSO_4$.
$PbO_2(s) + 4H^+(aq) + SO_4^{2-}(aq) + 2e^- \rightarrow PbSO_4(s) + 2H_2O(l)$
Net Cell Reaction:
$Pb(s) + PbO_2(s) + 2H_2SO_4(aq) \rightarrow 2PbSO_4(s) + 2H_2O(l)$
Sulphuric acid is consumed during discharging, decreasing its density.
Q7. The standard cell potential for the reaction $Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$ is 1.10 V at 298 K. Calculate the standard Gibbs free energy change ($\Delta G^\circ$) and the equilibrium constant ($K$) for the reaction. ($F = 96500 \text{ C/mol}, R = 8.314 \text{ J/K}\cdot\text{mol}$). (March 2017, March 2021)
Given: $E^\circ_{cell} = 1.10 \text{ V}$, $n = 2$ (moles of electrons transferred).
Part A: Calculate $\Delta G^\circ$
$\Delta G^\circ = -nFE^\circ_{cell}$
$\Delta G^\circ = -(2) \times (96500) \times (1.10)$
$\Delta G^\circ = -212,300 \text{ Joules} = -212.3 \text{ kJ/mol}$
Part B: Calculate Equilibrium Constant ($K$)
Using the relation: $E^\circ_{cell} = \frac{0.0592}{n} \log_{10} K$ (at 298 K)
$1.10 = \frac{0.0592}{2} \log_{10} K$
$1.10 = 0.0296 \log_{10} K$
$\log_{10} K = \frac{1.10}{0.0296} \approx 37.16$
$K = \text{antilog}(37.16) = 1.44 \times 10^{37}$
Final Answers: $\Delta G^\circ = -212.3 \text{ kJ/mol}$ and $K = 1.44 \times 10^{37}$.
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