JEE Chemistry Electrochemistry Notes Set 04

Nernst Equation

The reaction free energy depends on composition: \( \Delta G = \Delta G^\circ + RT\ln Q \). Since \( \Delta G = -nFE \), this gives, at 25 \( ^\circ \)C,

\[ E = E^\circ - \frac{0.0591}{n}\log Q \qquad Q = \frac{[\text{products}]}{[\text{reactants}]} \]

1. Electrode (reduction): \( E = E^\circ + \frac{0.0591}{n}\log[\text{M}^{n+}] \), so the potential rises with ion concentration.
2. Hydrogen electrode at 1 atm: \( E_{red} = -0.0591\,\text{pH} \) and \( E_{ox} = +0.0591\,\text{pH} \). For general pressure, \( E = -\frac{0.0591}{2}\log\frac{p_{H_2}}{[H^+]^2} \).
3. Daniell cell: \( E = 1.10 - \frac{0.0591}{2}\log\frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \). Raising [Cu\( ^{2+} \)] or lowering [Zn\( ^{2+} \)] increases the emf.
4. Concentration cell: both electrodes are identical, so \( E^\circ = 0 \). The more concentrated half cell is the cathode. For hydrogen electrodes at 1 atm, \( E = 0.0591\log(C_{cathode}/C_{anode}) \).
5. Feasibility: a reaction is spontaneous if \( \Delta G < 0 \), that is E is positive.
6. Equilibrium: at equilibrium E = 0 and Q = K, so \( E^\circ = \frac{0.0591}{n}\log K \) and \( \Delta G^\circ = -RT\ln K = -2.303RT\log K \).

Worked example. Calculate E for Zn | Zn\( ^{2+} \)(0.001 M) || Cu\( ^{2+} \)(0.1 M) | Cu with \( E^\circ = 1.1 \) V. Here n = 2 and \( E = 1.1 - 0.02955\log(10^{-3}/10^{-1}) = 1.1 + 0.0591 = 1.159 \) V.

Worked example. For Fe\( ^{2+} \) + Ce\( ^{4+} \) \( \rightleftharpoons \) Fe\( ^{3+} \) + Ce\( ^{3+} \), \( E^\circ = 1.44 - 0.68 = 0.76 \) V and n = 1, so \( \log K = 0.76/0.06 \approx 12.67 \) and \( K \approx 4.6 \times 10^{12} \).

Special Cells

1. A reversible cell can be driven backwards by an external emf. If the opposing emf equals that of the Daniell cell (1.1 V), there is no reaction and no current. If it is larger, the cell reaction is reversed: zinc is deposited and copper dissolves.
2. Primary cells cannot be recharged, secondary cells can.

Electrolysis: Qualitative Aspects

Electrolysis uses electrical energy to force a non-spontaneous reaction, in an electrolytic cell. The cathode is negative and takes the reduction, while the anode is positive and takes the oxidation (the opposite polarity to a galvanic cell). Among cations, the one with the higher reduction potential is discharged first at the cathode. Among anions, the one with the lower reduction potential (higher oxidation potential) is discharged first at the anode. Nitrate, phosphate and sulphate ions are not discharged from aqueous solution in practice.

In aqueous NaCl, water is easier to reduce than Na\( ^+ \), so hydrogen forms at the cathode. At the anode water is easier to oxidise than Cl\( ^- \) by potential alone, but Cl\( _2 \) is formed from concentrated brine because of the high overpotential of oxygen evolution. Molten lead bromide needs an applied potential above 1.22 V.

Faraday's Laws

1. First law: the mass deposited is proportional to the charge passed, \( W = Z I t \), where Z is the electrochemical equivalent (mass per coulomb).
2. Second law: for the same charge through different electrolytes in series, the masses deposited are in the ratio of their equivalent weights.
3. One faraday, 96500 C, is the charge of one mole of electrons and liberates one gram equivalent of any substance.
4. Combined form: \( \dfrac{W}{E} = \dfrac{It}{96500} \), where E is the equivalent weight.

Mole relations at an electrode: for \( \text{M}^{n+} + ne^- \rightarrow \text{M} \), n faradays deposit 1 mol. So 1 F deposits 1 mol of Ag, 1/2 mol of Cu or 1/3 mol of Al. In moles, equal charges deposit Ag : Cu : Al in the ratio 6 : 3 : 2. One mole of H\( _2 \) needs 2 F and one mole of O\( _2 \) needs 4 F.

Worked example. Passing 0.4 F through AgNO\( _3 \), CuSO\( _4 \) and FeCl\( _3 \) in series deposits Ag = 0.4 \( \times \) 108 = 43.2 g, Cu = 0.4 \( \times \) 63.5/2 = 12.7 g and Fe = 0.4 \( \times \) 56/3 = 7.47 g.

Worked example. A current of 100 A through molten NaCl for 5 hours: charge = 100 \( \times \) 18000 = 1.8 \( \times \) 10\( ^6 \) C, which is 18.65 F. Each mole of Cl\( _2 \) needs 2 F, so 9.33 mol are formed, which is 209 L at STP.

Commercial Cells

A lead battery discharges as a galvanic cell and is charged as an electrolytic cell. A car battery has six 2 V cells in series, giving 12 V.

Corrosion and Its Prevention

Rusting is an electrochemical process. At one spot on the iron, Fe is oxidised to Fe\( ^{2+} \) (anode, \( E^\circ = -0.44 \) V). Electrons travel through the metal to another spot where oxygen is reduced, \( \text{O}_2 + 4\text{H}^+ + 4e^- \rightarrow 2\text{H}_2\text{O} \) (cathode, \( E^\circ = 1.23 \) V). The Fe\( ^{2+} \) is further oxidised by air to hydrated Fe\( _2 \)O\( _3 \), which is rust. Salt water and impurities speed it up.

1. Coatings of paint, grease or metal keep air and water away.
2. Galvanising (zinc coating): zinc is more reactive than iron, so it is oxidised first and protects the iron even where the coating is scratched. A tin coating protects only while it is intact, because once broken the iron becomes the anode and rusts faster.
3. Cathodic protection: a block of a more reactive metal (Mg or Zn) is connected to buried iron pipes. It acts as the sacrificial anode and is replaced from time to time.