Chapter 2: Electrochemistry (Physical Chemistry)

Set up a galvanic cell with a magnesium electrode dipped in 1 M $\mathrm{Mg}{}^{2+}$ solution as one half-cell and the standard hydrogen electrode (SHE, $E^{\circ }=0$) as the other:

$$\mathrm{Mg}(\mathrm{s})|\mathrm{Mg}{}^{2+}(1\mathrm{M})||\mathrm{H}{}^{+}(1\mathrm{M})|\mathrm{H}{}_2(\mathrm{g},1\mathrm{bar})|\mathrm{Pt}(\mathrm{s})$$

Measure the emf with a voltmeter. Magnesium is the anode (it is oxidised), so $E_{\text{cell}}^{\circ }=E_{\text{SHE}}^{\circ }-E_{\text{Mg}}^{\circ }=0-E_{\text{Mg}}^{\circ }$. The measured emf is 2.37 V, so $E^{\circ }(\mathrm{Mg}{}^{2+}/\mathrm{Mg})=-2.37$ V.

Combine Mg | Mg²⁺(1 M) with the SHE, measure the emf; since Mg is the anode, E° = −(emf) = −2.37 V.

No. $E^{\circ }(\mathrm{Zn}{}^{2+}/\mathrm{Zn})=-0.76$ V is lower than $E^{\circ }(\mathrm{Cu}{}^{2+}/\mathrm{Cu})=+0.34$ V, so zinc is a stronger reducing agent and displaces copper:

$$\mathrm{Zn}(\mathrm{s})+\mathrm{CuSO}{}_4(\mathrm{aq})\to \mathrm{ZnSO}{}_4(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})$$

The pot would be eaten away and copper would deposit.

No; zinc reacts with CuSO₄ and displaces copper.

$\mathrm{Fe}{}^{3+}+\mathrm{e}{}^{-}\to \mathrm{Fe}{}^{2+}$ has $E^{\circ }=0.77$ V. Any species with a higher reduction potential can oxidise $\mathrm{Fe}{}^{2+}$ to $\mathrm{Fe}{}^{3+}$, for example:

• $\mathrm{Cl}{}_2$ ($E^{\circ }=1.36$ V)
• $\mathrm{Br}{}_2$ ($E^{\circ }=1.09$ V)
• acidified $\mathrm{KMnO}{}_4$ ($E^{\circ }=1.51$ V) or acidified $\mathrm{K}{}_2\mathrm{Cr}{}_2\mathrm{O}{}_7$ (1.33 V), and also $\mathrm{F}{}_2$, $\mathrm{H}{}_2\mathrm{O}{}_2$, $\mathrm{O}{}_2$ in acid

For example Cl₂, Br₂ and acidified KMnO₄ (all have E° above 0.77 V).

For $2\mathrm{H}{}^{+}+2\mathrm{e}{}^{-}\to \mathrm{H}{}_2$ (1 bar): $E=-\frac{0.059}{2}\log \frac{1}{[\mathrm{H}{}^{+}{]}^2}=-0.059\text{pH}$.

$E=-0.059\times 10=-0.59$ V.

−0.59 V.

$n=2$.

$$E_{\text{cell}}=1.05-\frac{0.059}{2}\log \frac{[\mathrm{Ni}{}^{2+}]}{[\mathrm{Ag}{}^{+}{]}^2}=1.05-0.0295\log \frac{0.160}{(0.002{)}^2}$$

$\frac{0.160}{4\times 10^{-6}}=4\times 10^4$, $\log (4\times 10^4)=4.60$.

$E_{\text{cell}}=1.05-0.0295\times 4.60=1.05-0.136\approx 0.91$ V.

About 0.91 V.

$n=2$.

$${\Delta }_r{G}^{\circ }=-nF{E}^{\circ }=-2\times 96487\times 0.236\approx -45.54{\text{kJ mol}}^{-1}$$

$$\log K=\frac{-{\Delta }_r{G}^{\circ }}{2.303RT}=\frac{45540}{2.303\times 8.314\times 298}\approx 7.98\Rightarrow K\approx 9.6\times 10^7$$

${\Delta }_r{G}^{\circ }$ ≈ −45.54 kJ mol⁻¹; K ≈ $9.6\times 10^7$.

Conductivity is the conductance of 1 cm³ of solution. On dilution, the number of ions present in each cm³ decreases (even though a weak electrolyte ionises more). With fewer ions to carry current per unit volume, the conductivity falls.

Fewer ions are present per unit volume after dilution.

By Kohlrausch's law of independent migration of ions, use the limiting molar conductivities of strong electrolytes, which can be measured:

$${\Lambda }_m^{\circ }(\mathrm{H}{}_2\mathrm{O})={\lambda }^{\circ }(\mathrm{H}{}^{+})+{\lambda }^{\circ }(\mathrm{OH}{}^{-})={\Lambda }_m^{\circ }(\mathrm{HCl})+{\Lambda }_m^{\circ }(\mathrm{NaOH})-{\Lambda }_m^{\circ }(\mathrm{NaCl})$$

${\Lambda }_m^{\circ }({\text{H}}_2\text{O})={\Lambda }_m^{\circ }(\text{HCl})+{\Lambda }_m^{\circ }(\text{NaOH})-{\Lambda }_m^{\circ }(\text{NaCl})$ (Kohlrausch's law).

${\Lambda }_m^{\circ }=349.6+54.6=404.2$ S cm² mol⁻¹.

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

$$K_a=\frac{c{\alpha }^2}{1-\alpha }=\frac{0.025\times (0.114{)}^2}{0.886}\approx 3.67\times 10^{-4}{\text{mol L}}^{-1}$$

α ≈ 0.114; $K_a$ ≈ $3.67\times 10^{-4}$ mol L⁻¹.

$Q=It=0.5\times 2\times 3600=3600$ C.

Number of electrons $=\frac{3600}{1.602\times 10^{-19}}\approx 2.25\times 10^{22}$.

About $2.25\times 10^{22}$ electrons.

Highly reactive metals with very negative electrode potentials, which cannot be reduced by carbon: sodium, potassium, lithium, calcium, magnesium and aluminium.

Na, K, Li, Ca, Mg, Al.

6 mol of electrons are needed: $Q=6F=6\times 96487\approx 5.79\times 10^5$ C.

$5.79\times 10^5$ C (6 F).

On recharging, an external current reverses the discharge reaction. The $\mathrm{PbSO}{}_4$ on both plates is converted back:

• At the negative plate (now the cathode): $\mathrm{PbSO}{}_4(\mathrm{s})+2\mathrm{e}{}^{-}\to \mathrm{Pb}(\mathrm{s})+\mathrm{SO}{}_4{}^{2-}(\mathrm{aq})$
• At the positive plate (now the anode): $\mathrm{PbSO}{}_4(\mathrm{s})+2\mathrm{H}{}_2\mathrm{O}(\mathrm{l})\to \mathrm{PbO}{}_2(\mathrm{s})+\mathrm{SO}{}_4{}^{2-}(\mathrm{aq})+4\mathrm{H}{}^{+}(\mathrm{aq})+2\mathrm{e}{}^{-}$

Overall: $2\mathrm{PbSO}{}_4(\mathrm{s})+2\mathrm{H}{}_2\mathrm{O}(\mathrm{l})\to \mathrm{Pb}(\mathrm{s})+\mathrm{PbO}{}_2(\mathrm{s})+2\mathrm{H}{}_2\mathrm{SO}{}_4(\mathrm{aq})$

Lead, lead dioxide and sulphuric acid are regenerated, so the acid becomes more concentrated again.

$2{\text{PbSO}}_4+2{\text{H}}_2\text{O}\to \text{Pb}+{\text{PbO}}_2+2{\text{H}}_2{\text{SO}}_4$ (the discharge reaction reversed).

Methane and methanol (ethanol and propane are also used).

Methane and methanol.

On a wet iron surface, small regions act as tiny electrochemical cells:

• Anode (where iron is oxidised): $2\mathrm{Fe}(\mathrm{s})\to 2\mathrm{Fe}{}^{2+}+4\mathrm{e}{}^{-}$, $E^{\circ }=-0.44$ V
• Cathode (another spot, where oxygen is reduced in the presence of $\mathrm{H}{}^{+}$ from dissolved $\mathrm{CO}{}_2$): $\mathrm{O}{}_2(\mathrm{g})+4\mathrm{H}{}^{+}+4\mathrm{e}{}^{-}\to 2\mathrm{H}{}_2\mathrm{O}(\mathrm{l})$, $E^{\circ }=1.23$ V

Overall: $2\mathrm{Fe}+\mathrm{O}{}_2+4\mathrm{H}{}^{+}\to 2\mathrm{Fe}{}^{2+}+2\mathrm{H}{}_2\mathrm{O}$, $E_{\text{cell}}^{\circ }=1.67$ V. The water film is the electrolyte and the metal carries the electrons. The $\mathrm{Fe}{}^{2+}$ is further oxidised by air to hydrated ferric oxide, $\mathrm{Fe}{}_2\mathrm{O}{}_3\cdot x\mathrm{H}{}_2\mathrm{O}$ (rust).

Iron is oxidised at anodic spots, oxygen is reduced at cathodic spots, and the water film acts as the electrolyte; Fe²⁺ is then oxidised to Fe₂O₃·xH₂O.

A metal with a lower (more negative) electrode potential displaces those below it. Order: Mg > Al > Zn > Fe > Cu (Mg displaces all the others; Cu displaces none).

Mg, Al, Zn, Fe, Cu.

The lower the reduction potential, the stronger the reducing agent:

Ag < Hg < Cr < Mg < K

Ag < Hg < Cr < Mg < K.

$$\mathrm{Zn}(\mathrm{s})|\mathrm{Zn}{}^{2+}(\mathrm{aq})||\mathrm{Ag}{}^{+}(\mathrm{aq})|\mathrm{Ag}(\mathrm{s})$$

(i) The zinc electrode (anode) is negative.

(ii) In the external wire, electrons carry the current (from Zn to Ag). Inside the cell and the salt bridge, ions carry it.

(iii) Anode: $\mathrm{Zn}(\mathrm{s})\to \mathrm{Zn}{}^{2+}(\mathrm{aq})+2\mathrm{e}{}^{-}$; cathode: $\mathrm{Ag}{}^{+}(\mathrm{aq})+\mathrm{e}{}^{-}\to \mathrm{Ag}(\mathrm{s})$.

Zn | Zn²⁺ || Ag⁺ | Ag; (i) Zn is negative; (ii) electrons in the wire, ions in the solution; (iii) Zn → Zn²⁺ + 2e⁻ and Ag⁺ + e⁻ → Ag.

(i) $E^{\circ }(\mathrm{Cd}{}^{2+}/\mathrm{Cd})=-0.40$ V, $E^{\circ }(\mathrm{Cr}{}^{3+}/\mathrm{Cr})=-0.74$ V; $n=6$.

$E_{\text{cell}}^{\circ }=-0.40-(-0.74)=0.34$ V.

${\Delta }_r{G}^{\circ }=-6\times 96487\times 0.34\approx -196.8$ kJ mol⁻¹.

$\log K=\frac{196800}{2.303\times 8.314\times 298}\approx 34.5$, so $K\approx 3.2\times 10^{34}$.

(ii) $E^{\circ }(\mathrm{Ag}{}^{+}/\mathrm{Ag})=0.80$ V, $E^{\circ }(\mathrm{Fe}{}^{3+}/\mathrm{Fe}{}^{2+})=0.77$ V; $n=1$.

$E_{\text{cell}}^{\circ }=0.80-0.77=0.03$ V.

${\Delta }_r{G}^{\circ }=-1\times 96487\times 0.03\approx -2.89$ kJ mol⁻¹.

$\log K=\frac{2894}{5706}\approx 0.507$, so $K\approx 3.2$.

(i) 0.34 V, −196.8 kJ mol⁻¹, K ≈ $3.2\times 10^{34}$. (ii) 0.03 V, −2.89 kJ mol⁻¹, K ≈ 3.2.

(i) $\mathrm{Mg}+\mathrm{Cu}{}^{2+}\to \mathrm{Mg}{}^{2+}+\mathrm{Cu}$, $n=2$, $E^{\circ }=0.34-(-2.37)=2.71$ V.

$E=E^{\circ }-\frac{0.059}{2}\log \frac{[\mathrm{Mg}{}^{2+}]}{[\mathrm{Cu}{}^{2+}]}=2.71-0.0295\log \frac{0.001}{0.0001}=2.71-0.0295=2.68$ V.

(ii) $\mathrm{Fe}+2\mathrm{H}{}^{+}\to \mathrm{Fe}{}^{2+}+\mathrm{H}{}_2$, $n=2$, $E^{\circ }=0-(-0.44)=0.44$ V.

$E=0.44-0.0295\log \frac{[\mathrm{Fe}{}^{2+}]}{[\mathrm{H}{}^{+}{]}^2}=0.44-0.0295\log (0.001)=0.44+0.089=0.53$ V.

(iii) $\mathrm{Sn}+2\mathrm{H}{}^{+}\to \mathrm{Sn}{}^{2+}+\mathrm{H}{}_2$, $n=2$, $E^{\circ }=0-(-0.14)=0.14$ V.

$E=0.14-0.0295\log \frac{0.050}{(0.020{)}^2}=0.14-0.0295\log 125=0.14-0.062=0.08$ V.

(iv) $2\mathrm{Br}{}^{-}+2\mathrm{H}{}^{+}\to \mathrm{Br}{}_2+\mathrm{H}{}_2$, $n=2$, $E^{\circ }=0-1.09=-1.09$ V.

$E=-1.09-0.0295\log \frac{1}{[\mathrm{Br}{}^{-}{]}^2[\mathrm{H}{}^{+}{]}^2}=-1.09-0.0295\log \frac{1}{(0.010{)}^2(0.030{)}^2}=-1.09-0.0295\times 7.05\approx -1.30$ V.

(The negative emf means this cell reaction is not spontaneous as written.)

(i) 2.68 V. (ii) 0.53 V. (iii) 0.08 V. (iv) −1.30 V.

Zinc is oxidised ($E^{\circ }=-0.76$ V) and silver is reduced ($E^{\circ }=0.80$ V for $\mathrm{Ag}{}^{+}/\mathrm{Ag}$, as used in the NCERT answer); $n=2$.

$E^{\circ }=0.80-(-0.76)=1.56$ V.

${\Delta }_r{G}^{\circ }=-nF{E}^{\circ }=-2\times 96487\times 1.56\approx -3.01\times 10^5$ J mol⁻¹ $=-301$ kJ mol⁻¹.

E° = 1.56 V; ${\Delta }_r{G}^{\circ }$ ≈ −301 kJ mol⁻¹.

Conductivity (κ): the conductance of a solution placed between two electrodes of 1 cm² area, 1 cm apart, i.e. of 1 cm³ of solution. Unit S cm⁻¹ (or S m⁻¹).

Molar conductivity (Λ_m): the conductance of the volume of solution that contains 1 mole of electrolyte, kept between electrodes 1 cm apart. ${\Lambda }_m=\frac{\kappa \times 1000}{c}$ (S cm² mol⁻¹).

Variation with concentration:

• κ decreases on dilution, because the number of ions per cm³ falls.
• Λ_m increases on dilution, because the volume containing 1 mol grows faster than κ falls. - For strong electrolytes it rises slowly and linearly with $\sqrt{c}$: ${\Lambda }_m={\Lambda }_m^{\circ }-A\sqrt{c}$; extrapolating to $c=0$ gives ${\Lambda }_m^{\circ }$. - For weak electrolytes it rises steeply at low concentration, because the degree of dissociation increases sharply; ${\Lambda }_m^{\circ }$ cannot be found by extrapolation (Kohlrausch's law is used).

κ = conductance of 1 cm³ (falls on dilution); Λ_m = κ × 1000/c (rises on dilution: slowly for strong, steeply for weak electrolytes).

$${\Lambda }_m=\frac{\kappa \times 1000}{c}=\frac{0.0248\times 1000}{0.20}=124{\text{S cm}}^2{\text{mol}}^{-1}$$

124 S cm² mol⁻¹.

Cell constant $G^{\ast }=\kappa \times R=0.146\times 10^{-3}\times 1500=0.219$ cm⁻¹.

0.219 cm⁻¹.

In SI units, ${\Lambda }_m=\frac{\kappa }{c}$ with $c$ in mol m⁻³ (1 M = 1000 mol m⁻³). For 0.001 M: ${\Lambda }_m=\frac{1.237\times 10^{-2}}{1}=1.237\times 10^{-2}$ S m² mol⁻¹ $=123.7$ S cm² mol⁻¹.

Plotting ${\Lambda }_m$ (y-axis) against $c^{1/2}$ (x-axis) gives a straight line sloping down (slope about −59 S cm² mol⁻¹ M^−½). Extending the line to $c^{1/2}=0$, the intercept is

$${\Lambda }_m^{\circ }\approx 124{\text{S cm}}^2{\text{mol}}^{-1}$$

Λ_m = 123.7, 118.5, 115.8, 111.1, 106.7 S cm² mol⁻¹; the straight-line plot gives ${\Lambda }_m^{\circ }$ ≈ 124 S cm² mol⁻¹.

$${\Lambda }_m=\frac{7.896\times 10^{-5}\times 1000}{0.00241}\approx 32.76{\text{S cm}}^2{\text{mol}}^{-1}$$

$\alpha =\frac{32.76}{390.5}\approx 0.0839$.

$$K_a=\frac{c{\alpha }^2}{1-\alpha }=\frac{0.00241\times (0.0839{)}^2}{0.916}\approx 1.85\times 10^{-5}{\text{mol L}}^{-1}$$

Λ_m ≈ 32.76 S cm² mol⁻¹; $K_a$ ≈ $1.85\times 10^{-5}$ mol L⁻¹.

• (i) $\mathrm{Al}{}^{3+}+3\mathrm{e}{}^{-}\to \mathrm{Al}$: $3F=3\times 96487\approx 2.89\times 10^5$ C
• (ii) $\mathrm{Cu}{}^{2+}+2\mathrm{e}{}^{-}\to \mathrm{Cu}$: $2F\approx 1.93\times 10^5$ C
• (iii) Mn goes from +7 to +2: $\mathrm{MnO}{}_4{}^{-}+8\mathrm{H}{}^{+}+5\mathrm{e}{}^{-}\to \mathrm{Mn}{}^{2+}+4\mathrm{H}{}_2\mathrm{O}$: $5F\approx 4.82\times 10^5$ C

(i) $2.89\times 10^5$ C. (ii) $1.93\times 10^5$ C. (iii) $4.82\times 10^5$ C.

(i) $\mathrm{Ca}{}^{2+}+2\mathrm{e}{}^{-}\to \mathrm{Ca}$. 20.0 g of Ca $=\frac{20}{40}=0.5$ mol, needing $0.5\times 2=1$ F.

(ii) $\mathrm{Al}{}^{3+}+3\mathrm{e}{}^{-}\to \mathrm{Al}$. 40.0 g of Al $=\frac{40}{27}=1.48$ mol, needing $1.48\times 3\approx 4.44$ F.

(i) 1 F. (ii) About 4.44 F.

(i) $\mathrm{H}{}_2\mathrm{O}\to \frac{1}{2}\mathrm{O}{}_2+2\mathrm{H}{}^{+}+2\mathrm{e}{}^{-}$: 2 mol of electrons, $2F\approx 1.93\times 10^5$ C.

(ii) Fe goes from +2 to +3: 1 mol of electrons per mol of FeO, $1F=96487$ C.

(i) $1.93\times 10^5$ C. (ii) 96487 C.

$Q=It=5\times 20\times 60=6000$ C. $\mathrm{Ni}{}^{2+}+2\mathrm{e}{}^{-}\to \mathrm{Ni}$.

Moles of Ni $=\frac{6000}{2\times 96487}=0.0311$ mol; mass $=0.0311\times 58.7\approx 1.82$ g.

About 1.82 g.

Moles of Ag $=\frac{1.45}{108}=0.01343$ mol $=$ moles of electrons (since $\mathrm{Ag}{}^{+}+\mathrm{e}{}^{-}\to \mathrm{Ag}$).

$Q=0.01343\times 96487\approx 1296$ C, so $t=\frac{1296}{1.5}\approx 864$ s (about 14.4 min).

The same charge passes through every cell in series:

• Cu: $\frac{0.01343}{2}\times 63.5\approx 0.43$ g
• Zn: $\frac{0.01343}{2}\times 65.4\approx 0.44$ g

About 864 s (14.4 min); about 0.43 g of Cu and 0.44 g of Zn.

A reaction is feasible if $E_{\text{cell}}^{\circ }=E_{\text{(species reduced)}}^{\circ }-E_{\text{(species oxidised)}}^{\circ }$ is positive.

• (i) $0.77-0.54=+0.23$ V: feasible ($\mathrm{Fe}{}^{3+}$ oxidises $\mathrm{I}{}^{-}$ to $\mathrm{I}{}_2$)
• (ii) $0.80-0.34=+0.46$ V: feasible
• (iii) $0.77-1.09=-0.32$ V: not feasible
• (iv) $0.77-0.80=-0.03$ V: not feasible
• (v) $1.09-0.77=+0.32$ V: feasible

Feasible: (i), (ii), (v). Not feasible: (iii), (iv).

• (i) Cathode: $\mathrm{Ag}{}^{+}+\mathrm{e}{}^{-}\to \mathrm{Ag}$ (silver deposits). Anode: the silver anode dissolves, $\mathrm{Ag}\to \mathrm{Ag}{}^{+}+\mathrm{e}{}^{-}$.
• (ii) Cathode: silver deposits. Anode: water is oxidised, $2\mathrm{H}{}_2\mathrm{O}\to \mathrm{O}{}_2+4\mathrm{H}{}^{+}+4\mathrm{e}{}^{-}$, so oxygen is released (nitrate is not oxidised).
• (iii) Cathode: $2\mathrm{H}{}^{+}+2\mathrm{e}{}^{-}\to \mathrm{H}{}_2$ (hydrogen). Anode: oxygen from water (sulphate is not oxidised).
• (iv) Cathode: $\mathrm{Cu}{}^{2+}+2\mathrm{e}{}^{-}\to \mathrm{Cu}$ (copper). Anode: $2\mathrm{Cl}{}^{-}\to \mathrm{Cl}{}_2+2\mathrm{e}{}^{-}$ (chlorine, because of the overpotential for oxygen).

(i) Ag at the cathode, Ag dissolves at the anode. (ii) Ag at the cathode, O₂ at the anode. (iii) H₂ at the cathode, O₂ at the anode. (iv) Cu at the cathode, Cl₂ at the anode.