Chapter 2: Electrochemistry

In the given reaction magnesium is oxidised (loses electrons) and silver ions are reduced. So Mg is the anode and Ag is the cathode. Writing the anode on the left and the cathode on the right, the cell is represented as:

$$\mathrm{Mg\,|\,Mg^{2+}(0.130\,M)\,||\,Ag^{+}(0.0001\,M)\,|\,Ag}$$

The half-reactions are $$\mathrm{Mg \rightarrow Mg^{2+} + 2e^{-}}$$ and $$\mathrm{2Ag^{+} + 2e^{-} \rightarrow 2Ag}$$, so $$n = 2$$ electrons are transferred. Applying the Nernst equation:

$$E_{(\mathrm{cell})} = E^{\circ}_{(\mathrm{cell})} - \frac{0.059}{n}\log\frac{[\mathrm{Mg^{2+}}]}{[\mathrm{Ag^{+}}]^{2}}$$

Substituting the values:

$$E_{(\mathrm{cell})} = 3.17 - \frac{0.059}{2}\log\frac{0.130}{(0.0001)^{2}}$$

$$\frac{0.130}{(0.0001)^{2}} = \frac{0.130}{1\times10^{-8}} = 1.3\times10^{7}, \qquad \log(1.3\times10^{7}) = 7.11$$

$$E_{(\mathrm{cell})} = 3.17 - (0.0295)(7.11) = 3.17 - 0.21 = 2.96\,\mathrm{V}$$

In this reaction copper is oxidised and silver ions are reduced; the number of electrons transferred is $$n = 2$$.

At equilibrium the cell potential becomes zero, and the Nernst equation reduces to a relation between $$E^{\circ}_{(\mathrm{cell})}$$ and the equilibrium constant $$K_C$$:

$$E^{\circ}_{(\mathrm{cell})} = \frac{0.059}{n}\log K_C$$

Rearranging for $$\log K_C$$:

$$\log K_C = \frac{n\,E^{\circ}_{(\mathrm{cell})}}{0.059} = \frac{2\times0.46}{0.059} = \frac{0.92}{0.059} = 15.6$$

Taking the antilog:

$$K_C = \mathrm{antilog}(15.6) = 3.92\times10^{15}$$

The standard Gibbs energy of a cell reaction is related to the standard cell potential by:

$$\Delta_r G^{\circ} = -nF E^{\circ}_{(\mathrm{cell})}$$

For the Daniell cell reaction the half-reactions are $$\mathrm{Zn \rightarrow Zn^{2+} + 2e^{-}}$$ and $$\mathrm{Cu^{2+} + 2e^{-} \rightarrow Cu}$$, so $$n = 2$$. With $$F = 96487\,\mathrm{C\,mol^{-1}}$$ and $$E^{\circ}_{(\mathrm{cell})} = 1.1\,\mathrm{V}$$:

$$\Delta_r G^{\circ} = -2 \times 96487\,\mathrm{C\,mol^{-1}} \times 1.1\,\mathrm{V}$$

$$\Delta_r G^{\circ} = -212271\,\mathrm{J\,mol^{-1}} = -212.27\,\mathrm{kJ\,mol^{-1}}$$

The negative value confirms that the reaction is spontaneous.

Step 1 — Cell constant. The cell constant $$G^{*}$$ is fixed for a given cell whatever solution it holds. From the $$0.1\,\mathrm{mol\,L^{-1}}$$ KCl data:

$$G^{*} = \kappa \times R = 1.29\,\mathrm{S\,m^{-1}} \times 100\,\Omega = 129\,\mathrm{m^{-1}}$$

Step 2 — Conductivity of the 0.02 M solution.

$$\kappa = \frac{G^{*}}{R} = \frac{129\,\mathrm{m^{-1}}}{520\,\Omega} = 0.248\,\mathrm{S\,m^{-1}}$$

Step 3 — Molar conductivity. Express the concentration in $$\mathrm{mol\,m^{-3}}$$:

$$c = 0.02\,\mathrm{mol\,L^{-1}} = 0.02 \times 1000\,\mathrm{mol\,m^{-3}} = 20\,\mathrm{mol\,m^{-3}}$$

$$\Lambda_m = \frac{\kappa}{c} = \frac{0.248\,\mathrm{S\,m^{-1}}}{20\,\mathrm{mol\,m^{-3}}} = 124\times10^{-4}\,\mathrm{S\,m^{2}\,mol^{-1}}$$

Equivalently, since $$1\,\mathrm{S\,m^{2}\,mol^{-1}} = 10^{4}\,\mathrm{S\,cm^{2}\,mol^{-1}}$$, this is $$\Lambda_m = 124\,\mathrm{S\,cm^{2}\,mol^{-1}}$$.

Cross-sectional area of the column. The diameter is $$1\,\mathrm{cm}$$, so the radius is $$r = 0.5\,\mathrm{cm}$$:

$$A = \pi r^{2} = 3.14 \times (0.5)^{2} = 0.785\,\mathrm{cm^{2}}$$

Resistivity. From $$R = \rho\dfrac{l}{A}$$:

$$\rho = \frac{R\,A}{l} = \frac{5.55\times10^{3}\,\Omega \times 0.785\,\mathrm{cm^{2}}}{50\,\mathrm{cm}} = 87.135\,\Omega\,\mathrm{cm}$$

Conductivity. Conductivity is the reciprocal of resistivity:

$$\kappa = \frac{1}{\rho} = \frac{1}{87.135} = 0.01148\,\mathrm{S\,cm^{-1}}$$

Molar conductivity.

$$\Lambda_m = \frac{\kappa \times 1000}{c} = \frac{0.01148\,\mathrm{S\,cm^{-1}} \times 1000\,\mathrm{cm^{3}\,L^{-1}}}{0.05\,\mathrm{mol\,L^{-1}}}$$

$$\Lambda_m = 229.6\,\mathrm{S\,cm^{2}\,mol^{-1}}$$

For a strong electrolyte the Debye–Hückel–Onsager equation predicts $$\Lambda_m = \Lambda^{\circ}_m - A\,c^{1/2}$$. Hence a plot of $$\Lambda_m$$ against $$c^{1/2}$$ should be a straight line of slope $$-A$$ and intercept $$\Lambda^{\circ}_m$$.

Taking the square root of each concentration:

Plotting $$\Lambda_m$$ (y-axis) against $$c^{1/2}$$ (x-axis), the four points lie on a straight line, which confirms the relation.

Slope. Using the first and last points:

$$\text{slope} = \frac{147.09 - 148.61}{0.03145 - 0.01407} = \frac{-1.52}{0.01738} = -87.46$$

So $$A = -\text{slope} = 87.46\,\mathrm{S\,cm^{2}\,mol^{-1}}\,(\mathrm{mol\,L^{-1}})^{-1/2}$$.

Intercept. Extrapolating the line to $$c^{1/2} = 0$$ (taking the first point):

$$\Lambda^{\circ}_m = \Lambda_m + A\,c^{1/2} = 148.61 + 87.46 \times 0.01407 \approx 150.0\,\mathrm{S\,cm^{2}\,mol^{-1}}$$

By Kohlrausch's law of independent migration of ions, the limiting molar conductivity of an electrolyte is the sum of the limiting molar conductivities of the ions it furnishes, each weighted by the number of such ions per formula unit. The required ionic values (Table 2.4) are $$\lambda^{\circ}(\mathrm{Ca^{2+}}) = 119.0$$, $$\lambda^{\circ}(\mathrm{Mg^{2+}}) = 106.0$$, $$\lambda^{\circ}(\mathrm{Cl^{-}}) = 76.3$$ and $$\lambda^{\circ}(\mathrm{SO_4^{2-}}) = 160.0\;\mathrm{S\,cm^{2}\,mol^{-1}}$$.

For $$\mathrm{CaCl_2}$$ (one $$\mathrm{Ca^{2+}}$$ and two $$\mathrm{Cl^{-}}$$):

$$\Lambda^{\circ}_m(\mathrm{CaCl_2}) = \lambda^{\circ}(\mathrm{Ca^{2+}}) + 2\,\lambda^{\circ}(\mathrm{Cl^{-}})$$

$$= 119.0 + 2(76.3) = 119.0 + 152.6 = 271.6\,\mathrm{S\,cm^{2}\,mol^{-1}}$$

For $$\mathrm{MgSO_4}$$ (one $$\mathrm{Mg^{2+}}$$ and one $$\mathrm{SO_4^{2-}}$$):

$$\Lambda^{\circ}_m(\mathrm{MgSO_4}) = \lambda^{\circ}(\mathrm{Mg^{2+}}) + \lambda^{\circ}(\mathrm{SO_4^{2-}})$$

$$= 106.0 + 160.0 = 266.0\,\mathrm{S\,cm^{2}\,mol^{-1}}$$

Acetic acid (HAc) is a weak electrolyte, so its $$\Lambda^{\circ}_m$$ cannot be found by extrapolation; instead it is obtained from strong electrolytes using Kohlrausch's law. We need

$$\Lambda^{\circ}_m(\mathrm{HAc}) = \lambda^{\circ}(\mathrm{H^{+}}) + \lambda^{\circ}(\mathrm{Ac^{-}})$$

This ionic combination can be assembled from the three given electrolytes:

$$\Lambda^{\circ}_m(\mathrm{HCl}) + \Lambda^{\circ}_m(\mathrm{NaAc}) - \Lambda^{\circ}_m(\mathrm{NaCl})$$

$$= [\lambda^{\circ}(\mathrm{H^{+}}) + \lambda^{\circ}(\mathrm{Cl^{-}})] + [\lambda^{\circ}(\mathrm{Na^{+}}) + \lambda^{\circ}(\mathrm{Ac^{-}})] - [\lambda^{\circ}(\mathrm{Na^{+}}) + \lambda^{\circ}(\mathrm{Cl^{-}})]$$

The $$\mathrm{Na^{+}}$$ and $$\mathrm{Cl^{-}}$$ terms cancel, leaving

$$= \lambda^{\circ}(\mathrm{H^{+}}) + \lambda^{\circ}(\mathrm{Ac^{-}}) = \Lambda^{\circ}_m(\mathrm{HAc})$$

Substituting the values:

$$\Lambda^{\circ}_m(\mathrm{HAc}) = 425.9 + 91.0 - 126.4 = 390.5\,\mathrm{S\,cm^{2}\,mol^{-1}}$$

Molar conductivity at the given concentration.

$$\Lambda_m = \frac{\kappa \times 1000}{c} = \frac{4.95\times10^{-5}\,\mathrm{S\,cm^{-1}} \times 1000\,\mathrm{cm^{3}\,L^{-1}}}{0.001028\,\mathrm{mol\,L^{-1}}} = 48.15\,\mathrm{S\,cm^{2}\,mol^{-1}}$$

Degree of dissociation. The ratio of $$\Lambda_m$$ to $$\Lambda^{\circ}_m$$ gives $$\alpha$$:

$$\alpha = \frac{\Lambda_m}{\Lambda^{\circ}_m} = \frac{48.15}{390.5} = 0.1233$$

Dissociation constant. For the equilibrium $$\mathrm{HAc \rightleftharpoons H^{+} + Ac^{-}}$$,

$$K_a = \frac{c\,\alpha^{2}}{1-\alpha} = \frac{0.001028 \times (0.1233)^{2}}{1-0.1233}$$

$$K_a = \frac{0.001028 \times 0.01520}{0.8767} = 1.78\times10^{-5}$$

Quantity of charge passed. With $$t = 10\,\mathrm{min} = 600\,\mathrm{s}$$:

$$Q = I \times t = 1.5\,\mathrm{A} \times 600\,\mathrm{s} = 900\,\mathrm{C}$$

Cathode reaction.

$$\mathrm{Cu^{2+}(aq) + 2e^{-} \rightarrow Cu(s)}$$

This shows that $$2F = 2 \times 96487\,\mathrm{C}$$ of charge deposit $$1\,\mathrm{mol}$$ of copper, i.e. $$63.5\,\mathrm{g}$$.

Mass deposited by 900 C.

$$m = \frac{63.5\,\mathrm{g\,mol^{-1}} \times 900\,\mathrm{C}}{2 \times 96487\,\mathrm{C\,mol^{-1}}} = 0.2938\,\mathrm{g}$$

The potential of a single half-cell cannot be measured on its own; it is found by coupling the half-cell with a reference electrode of known potential — the standard hydrogen electrode (SHE), whose potential is taken as zero at all temperatures.

Procedure. Set up a galvanic cell with the SHE as one electrode and a magnesium rod dipped in a $$1\,\mathrm{M}$$ $$\mathrm{Mg^{2+}}$$ solution as the other electrode, at $$298\,\mathrm{K}$$. The cell is represented as:

$$\mathrm{Mg\,|\,Mg^{2+}(1\,M)\,||\,H^{+}(1\,M)\,|\,H_2(g,\,1\,bar)\,|\,Pt}$$

Magnesium is oxidised more easily than hydrogen, so it acts as the anode (negative electrode) and the SHE as the cathode. Measure the emf of the cell with a high-resistance voltmeter (no current drawn).

$$E^{\circ}_{(\mathrm{cell})} = E^{\circ}_{\mathrm{cathode}} - E^{\circ}_{\mathrm{anode}} = E^{\circ}_{\mathrm{SHE}} - E^{\circ}_{(\mathrm{Mg^{2+}/Mg})}$$

Since $$E^{\circ}_{\mathrm{SHE}} = 0$$, $$\;E^{\circ}_{(\mathrm{Mg^{2+}/Mg})} = -E^{\circ}_{(\mathrm{cell})}$$. The measured emf is $$2.37\,\mathrm{V}$$, and because Mg is the anode the standard electrode potential of the $$\mathrm{Mg^{2+}|Mg}$$ system is $$-2.37\,\mathrm{V}$$.

No, copper sulphate solution cannot be stored in a zinc pot.

Zinc is more reactive than copper. The standard electrode potentials are $$E^{\circ}_{(\mathrm{Zn^{2+}/Zn})} = -0.76\,\mathrm{V}$$ and $$E^{\circ}_{(\mathrm{Cu^{2+}/Cu})} = +0.34\,\mathrm{V}$$. Because zinc has the more negative electrode potential, it is the stronger reducing agent and displaces copper from copper sulphate solution:

$$\mathrm{Zn(s) + CuSO_4(aq) \rightarrow ZnSO_4(aq) + Cu(s)}$$

For this displacement $$E^{\circ}_{(\mathrm{cell})} = 0.34 - (-0.76) = 1.10\,\mathrm{V} > 0$$, so the reaction is spontaneous. The zinc vessel would gradually dissolve and the copper sulphate would be consumed. Hence such a solution cannot be kept in a zinc container.

Oxidising $$\mathrm{Fe^{2+}}$$ to $$\mathrm{Fe^{3+}}$$ corresponds to the reverse of the couple

$$\mathrm{Fe^{3+} + e^{-} \rightarrow Fe^{2+}}, \qquad E^{\circ} = +0.77\,\mathrm{V}$$

Any species whose standard reduction potential is greater than $$+0.77\,\mathrm{V}$$ can act as the cathode against this couple, giving $$E^{\circ}_{(\mathrm{cell})} > 0$$, and can therefore oxidise ferrous ions. From Table 2.1, suitable oxidising agents include:

• $$\mathrm{F_2}$$  ($$E^{\circ} = 2.87\,\mathrm{V}$$)
• $$\mathrm{Cl_2}$$  ($$E^{\circ} = 1.36\,\mathrm{V}$$)
• $$\mathrm{Br_2}$$  ($$E^{\circ} = 1.09\,\mathrm{V}$$)

(Acidified $$\mathrm{MnO_4^{-}}$$, $$E^{\circ} = 1.51\,\mathrm{V}$$, and $$\mathrm{Cr_2O_7^{2-}}$$, $$E^{\circ} = 1.33\,\mathrm{V}$$, would serve equally well.) Each has $$E^{\circ} > 0.77\,\mathrm{V}$$ and so oxidises $$\mathrm{Fe^{2+}}$$.

The hydrogen electrode half-reaction is

$$\mathrm{2H^{+}(aq) + 2e^{-} \rightarrow H_2(g)}, \qquad E^{\circ} = 0\,\mathrm{V}, \quad n = 2$$

By the Nernst equation, with hydrogen gas at $$1\,\mathrm{bar}$$ ($$p_{\mathrm{H_2}} = 1$$):

$$E = E^{\circ} - \frac{0.059}{2}\log\frac{p_{\mathrm{H_2}}}{[\mathrm{H^{+}}]^{2}} = 0 - \frac{0.059}{2}\log\frac{1}{[\mathrm{H^{+}}]^{2}}$$

$$E = \frac{0.059}{2}\log[\mathrm{H^{+}}]^{2} = \frac{0.059}{2}\times 2\log[\mathrm{H^{+}}] = 0.059\log[\mathrm{H^{+}}]$$

Since $$\mathrm{pH} = -\log[\mathrm{H^{+}}]$$, this gives

$$E = -0.059\,\mathrm{pH}$$

For $$\mathrm{pH} = 10$$:

$$E = -0.059 \times 10 = -0.59\,\mathrm{V}$$

Nickel is oxidised and silver ions are reduced. The half-reactions $$\mathrm{Ni \rightarrow Ni^{2+} + 2e^{-}}$$ and $$\mathrm{2Ag^{+} + 2e^{-} \rightarrow 2Ag}$$ show that $$n = 2$$.

Applying the Nernst equation:

$$E_{(\mathrm{cell})} = E^{\circ}_{(\mathrm{cell})} - \frac{0.059}{n}\log\frac{[\mathrm{Ni^{2+}}]}{[\mathrm{Ag^{+}}]^{2}}$$

Substituting the data:

$$E_{(\mathrm{cell})} = 1.05 - \frac{0.059}{2}\log\frac{0.160}{(0.002)^{2}}$$

$$\frac{0.160}{(0.002)^{2}} = \frac{0.160}{4\times10^{-6}} = 4\times10^{4}, \qquad \log(4\times10^{4}) = 4.602$$

$$E_{(\mathrm{cell})} = 1.05 - (0.0295)(4.602) = 1.05 - 0.136 = 0.91\,\mathrm{V}$$

In this reaction $$\mathrm{2I^{-} \rightarrow I_2 + 2e^{-}}$$ and $$\mathrm{2Fe^{3+} + 2e^{-} \rightarrow 2Fe^{2+}}$$, so $$n = 2$$ electrons are transferred.

Standard Gibbs energy.

$$\Delta_r G^{\circ} = -nF E^{\circ}_{(\mathrm{cell})} = -2 \times 96487\,\mathrm{C\,mol^{-1}} \times 0.236\,\mathrm{V}$$

$$\Delta_r G^{\circ} = -45542\,\mathrm{J\,mol^{-1}} \approx -45.54\,\mathrm{kJ\,mol^{-1}}$$

Equilibrium constant. Using $$\Delta_r G^{\circ} = -RT\ln K_C$$ with $$R = 8.314\,\mathrm{J\,K^{-1}\,mol^{-1}}$$ and $$T = 298\,\mathrm{K}$$:

$$\ln K_C = \frac{-\Delta_r G^{\circ}}{RT} = \frac{45542}{8.314 \times 298} = \frac{45542}{2477.6} = 18.38$$

$$K_C = e^{18.38} = 9.62\times10^{7}$$

Conductivity ($$\kappa$$) is the conductance of a solution placed between two electrodes of unit cross-sectional area separated by unit distance — in effect, the conductance of unit volume of the solution.

On dilution, the number of ions present in that unit volume decreases, because the same ions are now spread through a much larger volume of solvent. Since the current through the solution is carried by these ions, a smaller number of ions per unit volume means a smaller conductivity.

(For a weak electrolyte the degree of dissociation does increase on dilution, producing more ions; but this increase is far outweighed by the fall in the number of ions per unit volume. Hence conductivity decreases on dilution for both strong and weak electrolytes.)

Water is an extremely weak electrolyte; it ionises only slightly ($$\mathrm{H_2O \rightleftharpoons H^{+} + OH^{-}}$$). Its molar conductivity cannot be obtained by extrapolating a $$\Lambda_m$$ vs $$c^{1/2}$$ plot to zero concentration (the plot is far from linear for weak electrolytes). Instead $$\Lambda^{\circ}_m$$ of water is found by applying Kohlrausch's law of independent migration of ions, using strong electrolytes whose $$\Lambda^{\circ}_m$$ values are known.

Since water furnishes $$\mathrm{H^{+}}$$ and $$\mathrm{OH^{-}}$$ ions:

$$\Lambda^{\circ}_m(\mathrm{H_2O}) = \lambda^{\circ}(\mathrm{H^{+}}) + \lambda^{\circ}(\mathrm{OH^{-}})$$

This combination is constructed from the strong electrolytes HCl, NaOH and NaCl:

$$\Lambda^{\circ}_m(\mathrm{H_2O}) = \Lambda^{\circ}_m(\mathrm{HCl}) + \Lambda^{\circ}_m(\mathrm{NaOH}) - \Lambda^{\circ}_m(\mathrm{NaCl})$$

The $$\Lambda^{\circ}_m$$ values of HCl, NaOH and NaCl are obtained by extrapolation (they are strong electrolytes); substituting them gives $$\Lambda^{\circ}_m$$ of water.

Limiting molar conductivity of methanoic acid (HCOOH), by Kohlrausch's law:

$$\Lambda^{\circ}_m(\mathrm{HCOOH}) = \lambda^{\circ}(\mathrm{H^{+}}) + \lambda^{\circ}(\mathrm{HCOO^{-}}) = 349.6 + 54.6 = 404.2\,\mathrm{S\,cm^{2}\,mol^{-1}}$$

Degree of dissociation.

$$\alpha = \frac{\Lambda_m}{\Lambda^{\circ}_m} = \frac{46.1}{404.2} = 0.114$$

Dissociation constant. For $$\mathrm{HCOOH \rightleftharpoons H^{+} + HCOO^{-}}$$ with $$c = 0.025\,\mathrm{mol\,L^{-1}}$$:

$$K_a = \frac{c\,\alpha^{2}}{1-\alpha} = \frac{0.025 \times (0.114)^{2}}{1-0.114}$$

$$K_a = \frac{0.025 \times 0.013}{0.886} = 3.67\times10^{-4}\,\mathrm{mol\,L^{-1}}$$

Charge passed. With $$t = 2\,\mathrm{h} = 2 \times 3600\,\mathrm{s} = 7200\,\mathrm{s}$$:

$$Q = I \times t = 0.5\,\mathrm{A} \times 7200\,\mathrm{s} = 3600\,\mathrm{C}$$

Number of electrons. One mole of electrons carries a charge of $$1\,F = 96487\,\mathrm{C}$$ and contains $$N_A = 6.022\times10^{23}$$ electrons. So $$96487\,\mathrm{C}$$ corresponds to $$6.022\times10^{23}$$ electrons. Hence:

$$\text{Number of electrons} = \frac{6.022\times10^{23}}{96487} \times 3600$$

$$= 6.022\times10^{23} \times 0.0373 = 2.25\times10^{22}$$

Highly reactive metals have very negative standard electrode potentials, so no ordinary chemical reducing agent (carbon, CO, etc.) is able to reduce their cations. Such metals are obtained by passing electricity through their molten salts (or, for aluminium, the molten oxide) — a process called electrolytic reduction.

Metals extracted electrolytically include:

Sodium (Na), potassium (K), lithium (Li), calcium (Ca), magnesium (Mg) and aluminium (Al).

For example, Na and Mg are obtained by electrolysis of their fused chlorides, and Al by electrolysis of molten $$\mathrm{Al_2O_3}$$ dissolved in cryolite. (Copper and zinc are also refined electrolytically.)

The given half-reaction

$$\mathrm{Cr_2O_7^{2-} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3+} + 7H_2O}$$

shows that the reduction of $$1\,\mathrm{mol}$$ of $$\mathrm{Cr_2O_7^{2-}}$$ requires $$6\,\mathrm{mol}$$ of electrons, i.e. $$6\,F$$ of charge.

$$Q = 6 \times F = 6 \times 96487\,\mathrm{C} = 578922\,\mathrm{C}$$

$$Q \approx 5.79\times10^{5}\,\mathrm{C}$$

A lead storage battery has a lead anode and a grid of lead packed with lead dioxide ($$\mathrm{PbO_2}$$) as cathode, immersed in about $$38\%$$ sulphuric acid. When the battery is in use (discharging) the overall cell reaction is:

$$\mathrm{Pb(s) + PbO_2(s) + 2H_2SO_4(aq) \rightarrow 2PbSO_4(s) + 2H_2O(l)}$$

Both electrodes become coated with insoluble lead sulphate, $$\mathrm{PbSO_4}$$, and sulphuric acid is consumed.

Recharging. An external source drives a direct current through the battery in the opposite direction, which reverses the cell reaction:

$$\mathrm{2PbSO_4(s) + 2H_2O(l) \rightarrow Pb(s) + PbO_2(s) + 2H_2SO_4(aq)}$$

Materials involved during recharging:

• $$\mathrm{PbSO_4}$$ on the electrode that had acted as anode is reduced back to spongy lead, $$\mathrm{Pb}$$.
• $$\mathrm{PbSO_4}$$ on the electrode that had acted as cathode is oxidised back to lead dioxide, $$\mathrm{PbO_2}$$.
• Water is consumed and sulphuric acid ($$\mathrm{H_2SO_4}$$) is regenerated, so the density (strength) of the electrolyte rises again.

A fuel cell needs a fuel that can be supplied continuously to the anode and oxidised there, releasing energy that is converted directly into electrical energy. Besides hydrogen, suitable fuels include:

• Methane ($$\mathrm{CH_4}$$)
• Methanol ($$\mathrm{CH_3OH}$$)

(Other examples used in fuel cells are carbon monoxide, $$\mathrm{CO}$$, and hydrazine, $$\mathrm{N_2H_4}$$.) Each of these is oxidised by oxygen, and the energy of the combustion is harnessed as electricity.

When iron is exposed to moist air, a thin film of water carrying dissolved $$\mathrm{O_2}$$ and $$\mathrm{CO_2}$$ forms on its surface. This film acts as an electrolyte, and innumerable tiny galvanic cells are set up on the metal — exactly the components of an electrochemical cell.

Anode. At one spot the iron is oxidised (loses electrons):

$$\mathrm{2Fe(s) \rightarrow 2Fe^{2+}(aq) + 4e^{-}}, \qquad E^{\circ}_{(\mathrm{Fe^{2+}/Fe})} = -0.44\,\mathrm{V}$$

Cathode. The electrons released travel through the metal (the electronic conductor) to another spot, where oxygen is reduced in the presence of $$\mathrm{H^{+}}$$ (supplied by $$\mathrm{H_2CO_3}$$ from dissolved $$\mathrm{CO_2}$$):

$$\mathrm{O_2(g) + 4H^{+}(aq) + 4e^{-} \rightarrow 2H_2O(l)}, \qquad E^{\circ} = 1.23\,\mathrm{V}$$

Overall cell reaction:

$$\mathrm{2Fe(s) + O_2(g) + 4H^{+}(aq) \rightarrow 2Fe^{2+}(aq) + 2H_2O(l)}, \qquad E^{\circ}_{(\mathrm{cell})} = 1.67\,\mathrm{V}$$

Thus rusting behaves like a short-circuited galvanic cell: the metal is the electronic conductor between anode and cathode, and the water film is the ionic conductor (electrolyte). The $$\mathrm{Fe^{2+}}$$ ions formed are further oxidised by atmospheric oxygen to hydrated ferric oxide, $$\mathrm{Fe_2O_3\cdot xH_2O}$$, which is the brown rust.

A metal displaces from solution any metal that lies below it in the activity (electrochemical) series — i.e. a metal with a more negative standard electrode potential displaces a metal with a less negative (higher) potential from a solution of its salt.

The standard electrode potentials of the given metals are:

Arranging the metals from the most negative potential (most reactive, best at displacing the others) to the least negative:

$$\mathrm{Mg > Al > Zn > Fe > Cu}$$

Each metal in this list can displace any metal that appears to its right from a solution of that metal's salt.

The reducing power of a metal measures how readily it loses electrons (is oxidised). The more negative the standard electrode potential $$E^{\circ}_{(\mathrm{M^{n+}/M})}$$, the more easily the metal is oxidised, and hence the stronger its reducing power.

Listing the metals in order of their given potentials:

Reducing power increases as $$E^{\circ}$$ becomes more negative. Starting from the highest $$E^{\circ}$$ (weakest reducing agent) and ending at the lowest (strongest):

$$\mathrm{Ag < Hg < Cr < Mg < K}$$

In this button (silver oxide) cell zinc is oxidised and silver(I) oxide is reduced. The half-reactions are

Anode: $$\mathrm{Zn(s) \rightarrow Zn^{2+}(aq) + 2e^{-}}$$
Cathode: $$\mathrm{Ag_2O(s) + H_2O(l) + 2e^{-} \rightarrow 2Ag(s) + 2OH^{-}(aq)}$$

so $$n = 2$$ electrons are transferred.

Standard cell potential. Taking the standard electrode potential of the cathode couple as $$E^{\circ}_{(\mathrm{Ag_2O/Ag})} = +0.344\,\mathrm{V}$$ and of the anode couple as $$E^{\circ}_{(\mathrm{Zn^{2+}/Zn})} = -0.76\,\mathrm{V}$$:

$$E^{\circ} = E^{\circ}_{\mathrm{cathode}} - E^{\circ}_{\mathrm{anode}} = 0.344 - (-0.76) = 1.104\,\mathrm{V} \approx 1.1\,\mathrm{V}$$

Standard Gibbs energy.

$$\Delta_r G^{\circ} = -nF E^{\circ} = -2 \times 96487 \times 1.104$$

$$\Delta_r G^{\circ} = -213043\,\mathrm{J\,mol^{-1}} \approx -213\,\mathrm{kJ\,mol^{-1}}$$

The large positive $$E^{\circ}$$ and large negative $$\Delta_r G^{\circ}$$ explain why such cells deliver a steady, reliable voltage and are widely used in watches.

Conductivity ($$\kappa$$). The conductivity of an electrolytic solution is the conductance of a solution held between two electrodes of unit cross-sectional area separated by unit distance — equivalently, the conductance of unit volume of the solution. It is the reciprocal of resistivity, $$\kappa = 1/\rho$$, and is obtained from the measured resistance $$R$$ and the cell constant $$G^{*} = l/A$$ by

$$\kappa = \frac{G^{*}}{R}$$

Its SI unit is $$\mathrm{S\,m^{-1}}$$ (commonly $$\mathrm{S\,cm^{-1}}$$).

Molar conductivity ($$\Lambda_m$$). The molar conductivity is the conducting power of all the ions furnished by one mole of an electrolyte in solution. It is related to conductivity by

$$\Lambda_m = \frac{\kappa}{c}$$

where $$c$$ is the molar concentration. Its unit is $$\mathrm{S\,m^{2}\,mol^{-1}}$$ (or $$\mathrm{S\,cm^{2}\,mol^{-1}}$$).

Variation with concentration.

Conductivity always decreases when the solution is diluted (for both strong and weak electrolytes), because the number of ions per unit volume that carry the current decreases on dilution.

Molar conductivity increases on dilution, because the volume $$V$$ of solution containing one mole of electrolyte increases, and this increase more than offsets the fall in $$\kappa$$ (since $$\Lambda_m = \kappa V$$). The way it increases differs by type of electrolyte:

• For a strong electrolyte, $$\Lambda_m$$ rises slowly and obeys $$\Lambda_m = \Lambda^{\circ}_m - A\,c^{1/2}$$; the plot of $$\Lambda_m$$ vs $$c^{1/2}$$ is a straight line whose intercept (at $$c \to 0$$) gives $$\Lambda^{\circ}_m$$.
• For a weak electrolyte, $$\Lambda_m$$ is low at ordinary concentrations and rises steeply near infinite dilution (the degree of dissociation increases sharply). Here $$\Lambda^{\circ}_m$$ cannot be found by extrapolation; it is obtained using Kohlrausch's law of independent migration of ions.

Molar conductivity is related to conductivity by

$$\Lambda_m = \frac{\kappa \times 1000}{c}$$

where $$\kappa$$ is in $$\mathrm{S\,cm^{-1}}$$, $$c$$ in $$\mathrm{mol\,L^{-1}}$$, and the factor $$1000\,\mathrm{cm^{3}\,L^{-1}}$$ accounts for the volume conversion.

$$\Lambda_m = \frac{0.0248\,\mathrm{S\,cm^{-1}} \times 1000\,\mathrm{cm^{3}\,L^{-1}}}{0.20\,\mathrm{mol\,L^{-1}}}$$

$$\Lambda_m = \frac{24.8}{0.20} = 124\,\mathrm{S\,cm^{2}\,mol^{-1}}$$

The conductivity, resistance and cell constant are related by $$\kappa = \dfrac{G^{*}}{R}$$, where $$G^{*} = l/A$$ is the cell constant. Rearranging:

$$\text{Cell constant }(G^{*}) = \kappa \times R$$

Substituting $$\kappa = 0.146\times10^{-3}\,\mathrm{S\,cm^{-1}}$$ and $$R = 1500\,\Omega$$:

$$G^{*} = 0.146\times10^{-3}\,\mathrm{S\,cm^{-1}} \times 1500\,\Omega$$

$$G^{*} = 0.219\,\mathrm{cm^{-1}}$$

Molar conductivity is $$\Lambda_m = \dfrac{\kappa}{c}$$. Working in SI units, $$\kappa$$ is in $$\mathrm{S\,m^{-1}}$$ and $$c$$ in $$\mathrm{mol\,m^{-3}}$$ (note $$1\,\mathrm{M} = 1000\,\mathrm{mol\,m^{-3}}$$); the result in $$\mathrm{S\,m^{2}\,mol^{-1}}$$ is converted to $$\mathrm{S\,cm^{2}\,mol^{-1}}$$ on multiplying by $$10^{4}$$.

The table gives $$10^{2}\times\kappa$$, so the actual conductivity is the tabulated value divided by 100. For example, at $$c = 0.001\,\mathrm{M}$$:

$$\kappa = 1.237\times10^{-2}\,\mathrm{S\,m^{-1}}, \quad c = 1\,\mathrm{mol\,m^{-3}}$$

$$\Lambda_m = \frac{1.237\times10^{-2}}{1} = 1.237\times10^{-2}\,\mathrm{S\,m^{2}\,mol^{-1}} = 123.7\,\mathrm{S\,cm^{2}\,mol^{-1}}$$

Repeating for every concentration:

When $$\Lambda_m$$ (y-axis) is plotted against $$c^{1/2}$$ (x-axis), the points fall close to a straight line, as expected for the strong electrolyte NaCl which obeys $$\Lambda_m = \Lambda^{0}_m - A\,c^{1/2}$$.

Extrapolating the straight line to $$c^{1/2} = 0$$ gives the intercept

$$\Lambda^{0}_m \approx 124\,\mathrm{S\,cm^{2}\,mol^{-1}}$$

(close to the accepted value $$126.4\,\mathrm{S\,cm^{2}\,mol^{-1}}$$ for sodium chloride).

Molar conductivity.

$$\Lambda_m = \frac{\kappa \times 1000}{c} = \frac{7.896\times10^{-5}\,\mathrm{S\,cm^{-1}} \times 1000\,\mathrm{cm^{3}\,L^{-1}}}{0.00241\,\mathrm{mol\,L^{-1}}}$$

$$\Lambda_m = \frac{0.07896}{0.00241} = 32.76\,\mathrm{S\,cm^{2}\,mol^{-1}}$$

Degree of dissociation.

$$\alpha = \frac{\Lambda_m}{\Lambda^{0}_m} = \frac{32.76}{390.5} = 0.0839$$

Dissociation constant. For $$\mathrm{CH_3COOH \rightleftharpoons H^{+} + CH_3COO^{-}}$$ with $$c = 0.00241\,\mathrm{mol\,L^{-1}}$$:

$$K_a = \frac{c\,\alpha^{2}}{1-\alpha} = \frac{0.00241 \times (0.0839)^{2}}{1-0.0839}$$

$$K_a = \frac{0.00241 \times 0.007039}{0.9161} = 1.85\times10^{-5}$$

Quantity of charge passed. With $$t = 20\,\mathrm{min} = 1200\,\mathrm{s}$$:

$$Q = I \times t = 5\,\mathrm{A} \times 1200\,\mathrm{s} = 6000\,\mathrm{C}$$

Cathode reaction.

$$\mathrm{Ni^{2+}(aq) + 2e^{-} \rightarrow Ni(s)}$$

So $$2F = 2 \times 96487\,\mathrm{C}$$ deposit $$1\,\mathrm{mol}$$ of nickel ($$58.7\,\mathrm{g}$$).

Mass of nickel deposited.

$$m = \frac{58.7\,\mathrm{g\,mol^{-1}} \times 6000\,\mathrm{C}}{2 \times 96487\,\mathrm{C\,mol^{-1}}} = \frac{352200}{192974} = 1.825\,\mathrm{g}$$

Because the three cells are connected in series, the same quantity of charge passes through each of them.

Charge from the silver data (cell B). The cathode reaction is $$\mathrm{Ag^{+} + e^{-} \rightarrow Ag}$$, so $$1\,F$$ deposits $$1\,\mathrm{mol}$$ of silver ($$108\,\mathrm{g}$$).

$$n_{\mathrm{Ag}} = \frac{1.45\,\mathrm{g}}{108\,\mathrm{g\,mol^{-1}}} = 0.01343\,\mathrm{mol}$$

$$Q = n_{\mathrm{Ag}} \times F = 0.01343 \times 96487 = 1295.5\,\mathrm{C}$$

Time of current flow.

$$t = \frac{Q}{I} = \frac{1295.5\,\mathrm{C}}{1.5\,\mathrm{A}} = 863.7\,\mathrm{s} \approx 14.4\,\mathrm{min}$$

Mass of copper (cell C). The reaction $$\mathrm{Cu^{2+} + 2e^{-} \rightarrow Cu}$$ means $$2F$$ deposit $$1\,\mathrm{mol}$$ Cu ($$63.5\,\mathrm{g}$$):

$$m_{\mathrm{Cu}} = \frac{63.5 \times 1295.5}{2 \times 96487} = 0.426\,\mathrm{g}$$

Mass of zinc (cell A). The reaction $$\mathrm{Zn^{2+} + 2e^{-} \rightarrow Zn}$$ means $$2F$$ deposit $$1\,\mathrm{mol}$$ Zn ($$65.4\,\mathrm{g}$$):

$$m_{\mathrm{Zn}} = \frac{65.4 \times 1295.5}{2 \times 96487} = 0.439\,\mathrm{g}$$