JEE Chemistry Electrochemistry Notes Set 06

Electrolytic Conduction

Electrolytes (acids, bases, salts) conduct current through the movement of ions; strong electrolytes ionise completely and weak electrolytes partly. Non-electrolytes (sugar, urea, glucose, ethanol) give no ions. Metals conduct through free electrons, and their conductance falls on heating, whereas the conductance of an electrolyte rises on heating because ionisation and ion mobility increase.
Resistance \( R = \rho\,l/a \); conductance \( G = 1/R \) (unit ohm\(^{-1}\) or siemens). Specific conductivity \( \kappa = G\times\dfrac{l}{a} = G\times \)cell constant. Equivalent conductance \( \Lambda_{eq} = \dfrac{1000\,\kappa}{N} \); molar conductance \( \Lambda_m = \dfrac{1000\,\kappa}{M} \).
Kohlrausch's law: at infinite dilution each ion contributes independently, \( \Lambda^\infty = \lambda^\infty_+ + \lambda^\infty_- \). Degree of dissociation of a weak electrolyte \( \alpha = \Lambda/\Lambda^\infty \). Conductance of strong electrolytes falls slowly with concentration (Debye-Huckel-Onsager); weak electrolytes show a steep rise on dilution.

Electrolysis and Faraday's Laws

First law: the mass deposited is proportional to the charge passed, \( w = ZQ = ZIt \), where Z is the electrochemical equivalent. Second law: for the same charge, the masses deposited are proportional to the equivalent weights.
One faraday = 96500 C is the charge of one mole of electrons and liberates one gram equivalent of any substance. \( w = \dfrac{E\,I\,t}{96500} \). For n electrons per ion, moles deposited = \( \dfrac{Q}{nF} \).
Products: molten NaCl gives Na at the cathode and Cl\(_2\) at the anode; aqueous NaCl gives H\(_2\), Cl\(_2\) and NaOH; dilute H\(_2\)SO\(_4\) with platinum electrodes gives H\(_2\) at the cathode and O\(_2\) at the anode; aqueous CuSO\(_4\) with copper electrodes dissolves copper at the anode and deposits it at the cathode (refining). Uses: electroplating, electrorefining, extraction of reactive metals (Na, Al, Ca) and electrotyping.

Galvanic Cells and Electrode Potential

A galvanic cell converts chemical energy into electrical energy. Oxidation occurs at the anode (negative pole) and reduction at the cathode (positive pole); the salt bridge completes the circuit and removes the liquid junction potential. Cell notation: Zn | Zn\(^{2+}\) || Cu\(^{2+}\) | Cu.
Standard electrode potential is measured against the standard hydrogen electrode (SHE, 0.00 V, 1 atm H\(_2\), 1 M H\(^+\)). \( E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \). A metal with a more negative potential is a stronger reducing agent and displaces metals of higher potential from their salt solutions; metals above hydrogen in the series liberate hydrogen from dilute acids.
Alkali metals such as Li, K and Ca have the most negative potentials and are the strongest reducing agents; fluorine has the highest positive value and is the strongest oxidising agent.

Nernst Equation and Equilibrium

For \( M^{n+} + ne^- \rightarrow M \): \( E = E^\circ - \dfrac{0.0591}{n}\log\dfrac{1}{[M^{n+}]} \) at 298 K. For a cell: \( E_{cell} = E^\circ_{cell} - \dfrac{0.0591}{n}\log Q \).
\( \Delta G^\circ = -nFE^\circ \) and \( \Delta G^\circ = -2.303\,RT\log K \), so \( \log K = \dfrac{nE^\circ}{0.0591} \). At equilibrium \( E_{cell} = 0 \). A spontaneous cell has \( E>0 \), \( \Delta G<0 \) and \( K>1 \).

Commercial Cells

Dry (Leclanche) cell: zinc anode, graphite cathode, paste of NH\(_4\)Cl, ZnCl\(_2\) and MnO\(_2\); about 1.5 V; primary cell.
Lead storage battery: anode Pb, cathode PbO\(_2\), electrolyte about 38% H\(_2\)SO\(_4\); on discharge both plates become PbSO\(_4\) and the acid is used up; on charging the acid is regenerated. About 2 V per cell.
Fuel cell: H\(_2\) and O\(_2\) react at porous electrodes in KOH; overall \( 2H_2 + O_2 \rightarrow 2H_2O \), efficiency about 70% (far higher than a heat engine), continuous supply of fuel.

Worked Examples

Example 1. How long must a current of 2 A pass through CuSO\(_4\) solution to deposit 3.175 g of copper? (Cu = 63.5)
Answer:
Equivalent weight of Cu = 31.75, so \( Q = \dfrac{3.175\times96500}{31.75} = 9650 \) C.
\( t = Q/I = 9650/2 = 4825 \) s.

Example 2. Find \( E^\circ_{cell} \) for Zn | Zn\(^{2+}\) || Ag\(^+\) | Ag given \( E^\circ(Zn^{2+}/Zn) = -0.76 \) V and \( E^\circ(Ag^+/Ag) = +0.80 \) V.
Answer:
\( E^\circ_{cell} = 0.80 - (-0.76) = 1.56 \) V.