COURSE TITLE: ELECTROCHEMISTRY
COURSE CODE: CHM 406
DURATION: 1 Hour
Instructions
ATTEMPT ANY ALL QUESTIONS
Question Question 1(a)(i): Differentiate between Primary Cell and Secondary Cell (with examples and equations).
Question Question 1(a)(ii): Differentiate between Electrode Concentration Cell and Electrolytic Concentration Cell.
Question Question 1(b)(i): Describe the hydrogen-oxygen fuel cell with a labeled diagram and electrode reactions.
Question Question 1(b)(ii): Calculate the reversible emf of the hydrogen-oxygen fuel cell at 25 °C.
Question Question 1(c)(i): What is liquid junction potential and how can it be suppressed?
Question Question 1(c)(ii): Write the half-cell reaction and calculate the electrode potential for Zn in 0.1 M ZnSO₄ (20% dissociation).
Question Question 1(d)(i): Describe the construction of a Standard Hydrogen Electrode (SHE).
Question Question 1(d)(ii): Calculate the equilibrium constant
Question Question 2(a): State expressions for mean ionic terms and chemical potential.
Question Question 2(b)(i): Determine moles of Mg²⁺ and Cl⁻ migrated after 15 minutes.
Question Question 2(b)(ii): Calculate drift speed of Mg²⁺ and Cl⁻ ions.
Question Question 2(b)(iii): Compare ion flux across a 1 cm² plane in 1 second.
Question Question 2(c): Relate transport number to ionic mobility and show
Question Question 2(d): Five differences between electrolytic and galvanic cells.
Answers
Answer Question 1(a)(i):
Primary Cell: Non-rechargeable, irreversible reactions. Example: Zinc-Carbon Cell. Reactions:
Primary Cell: Non-rechargeable, irreversible reactions. Example: Zinc-Carbon Cell. Reactions:
- Anode: \(\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-\)
- Cathode: \(2\text{MnO}_2 + 2\text{H}_2\text{O} + 2e^- \rightarrow 2\text{MnO(OH)} + 2\text{OH}^-\)
- Overall: \(\text{Zn} + 2\text{MnO}_2 + \text{H}_2\text{O} \rightarrow \text{ZnO} + 2\text{MnO(OH)}\)
Secondary Cell: Rechargeable, reversible reactions. Example: Lead-Acid Battery. Discharge Reactions:
- Anode: \(\text{Pb} + \text{SO}_4^{2-} \rightarrow \text{PbSO}_4 + 2e^-\)
- Cathode: \(\text{PbO}_2 + 4\text{H}^+ + \text{SO}_4^{2-} + 2e^- \rightarrow \text{PbSO}_4 + 2\text{H}_2\text{O}\)
- Overall: \(\text{Pb} + \text{PbO}_2 + 2\text{H}_2\text{SO}_4 \rightarrow 2\text{PbSO}_4 + 2\text{H}_2\text{O}\)
Answer Question 1(a)(ii):
Electrode Concentration Cell:
Electrode Concentration Cell:
- Same electrolyte, different concentrations. Example: \(\text{Cu} \,|\, \text{Cu}^{2+}(0.1\,\text{M}) \,||\, \text{Cu}^{2+}(1.0\,\text{M}) \,|\, \text{Cu}\) Reaction: \(\text{Cu}^{2+}(1.0\,\text{M}) \rightarrow \text{Cu}^{2+}(0.1\,\text{M})\)
Electrolytic Concentration Cell:
- Same electrodes, different electrolyte concentrations. Example: \(\text{Pt} \,|\, \text{H}_2(1\,\text{atm}) \,|\, \text{HCl}(0.1\,\text{M}) \,||\, \text{HCl}(1.0\,\text{M}) \,|\, \text{H}_2(1\,\text{atm}) \,|\, \text{Pt}\) Reaction: \(\text{H}^+(1.0\,\text{M}) \rightarrow \text{H}^+(0.1\,\text{M})\)
Answer Question 1(b)(i):
- Diagram:
- Anode: Porous Pt electrode (H₂ inlet).
- Cathode: Porous Pt electrode (O₂ inlet).
- Electrolyte: Aqueous KOH.
- External Load: Connects anode and cathode.
- Reactions:
- Anode: \(\text{H}_2 + 2\text{OH}^- \rightarrow 2\text{H}_2\text{O} + 2e^-\)
- Cathode: \(\text{O}_2 + 2\text{H}_2\text{O} + 4e^- \rightarrow 4\text{OH}^-\)
- Overall: \(2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}\)
Answer Question 1(b)(ii):
\[ \begin{aligned} \Delta G^\circ_{\text{H}_2\text{O(l)}} &= -56700\ \text{cal/mol} \\ \Delta G^\circ_{\text{ionization}} &= 19050\ \text{cal/mol} \\ \text{Total } \Delta G^\circ &= -56700 + 19050 = -37650\ \text{cal/mol} \\ &= -37650 \times 4.184 = -157599.6\ \text{J/mol} \\ E &= -\frac{\Delta G^\circ}{nF} = \frac{157599.6}{2 \times 96500} \\ &= \boxed{0.82\ \text{V}} \end{aligned} \]
\[ \begin{aligned} \Delta G^\circ_{\text{H}_2\text{O(l)}} &= -56700\ \text{cal/mol} \\ \Delta G^\circ_{\text{ionization}} &= 19050\ \text{cal/mol} \\ \text{Total } \Delta G^\circ &= -56700 + 19050 = -37650\ \text{cal/mol} \\ &= -37650 \times 4.184 = -157599.6\ \text{J/mol} \\ E &= -\frac{\Delta G^\circ}{nF} = \frac{157599.6}{2 \times 96500} \\ &= \boxed{0.82\ \text{V}} \end{aligned} \]
Answer Question 1(c)(i):
Liquid Junction Potential: Potential difference due to unequal ion mobilities at the interface of two electrolytes. Suppression: Use a salt bridge with inert electrolytes (e.g., KCl) where cation and anion have similar mobilities.
Liquid Junction Potential: Potential difference due to unequal ion mobilities at the interface of two electrolytes. Suppression: Use a salt bridge with inert electrolytes (e.g., KCl) where cation and anion have similar mobilities.
Answer Question 1(c)(ii):
- Reaction: \(\text{Zn}^{2+} + 2e^- \rightarrow \text{Zn(s)}\)
- Effective [Zn²⁺]: \(0.1 \times 0.2 = 0.02\ \text{M}\)
- Nernst Equation:
\[ \begin{aligned} E &= E^\circ + \frac{0.059}{2} \log[\text{Zn}^{2+}] \\ &= -0.76 + 0.0295 \log(0.02) \\ \log(0.02) &\approx -1.698 \\ E &= -0.76 – 0.0295 \times 1.698 \\ &= \boxed{-0.81\ \text{V}} \end{aligned} \]
Answer Question 1(d)(i):
- Construction:
- Platinum electrode coated with Pt black.
- Immersed in 1 M HCl solution.
- H₂ gas bubbled at 1 atm pressure.
- Reaction: \(2\text{H}^+ + 2e^- \rightarrow \text{H}_2(g)\)
Answer Question 1(d)(ii):
\[ \begin{aligned} E^\circ_{\text{cell}} &= E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} – E^\circ_{\text{Sn}^{4+}/\text{Sn}^{2+}} \\ &= 0.771\ \text{V} – 0.15\ \text{V} = 0.621\ \text{V} \\ \ln K &= \frac{nFE^\circ}{RT} = \frac{2 \times 96500 \times 0.621}{8.314 \times 298} \\ &\approx 48.38 \\ K &= e^{48.38} \approx \boxed{1.3 \times 10^{21}} \end{aligned} \]
\[ \begin{aligned} E^\circ_{\text{cell}} &= E^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} – E^\circ_{\text{Sn}^{4+}/\text{Sn}^{2+}} \\ &= 0.771\ \text{V} – 0.15\ \text{V} = 0.621\ \text{V} \\ \ln K &= \frac{nFE^\circ}{RT} = \frac{2 \times 96500 \times 0.621}{8.314 \times 298} \\ &\approx 48.38 \\ K &= e^{48.38} \approx \boxed{1.3 \times 10^{21}} \end{aligned} \]
Answer Question 2(a):
\[ \begin{aligned} \mu_{\pm} &= \frac{\nu_+ \mu_+ + \nu_- \mu_-}{\nu} \\ m_{\pm} &= \left(m_+^{\nu_+} m_-^{\nu_-}\right)^{1/\nu} \\ a_{\pm} &= \left(a_+^{\nu_+} a_-^{\nu_-}\right)^{1/\nu} \\ \gamma_{\pm} &= \left(\gamma_+^{\nu_+} \gamma_-^{\nu_-}\right)^{1/\nu} \\ \mu_{\text{solute}} &= \nu RT \ln a_{\pm} \end{aligned} \]
\[ \begin{aligned} \mu_{\pm} &= \frac{\nu_+ \mu_+ + \nu_- \mu_-}{\nu} \\ m_{\pm} &= \left(m_+^{\nu_+} m_-^{\nu_-}\right)^{1/\nu} \\ a_{\pm} &= \left(a_+^{\nu_+} a_-^{\nu_-}\right)^{1/\nu} \\ \gamma_{\pm} &= \left(\gamma_+^{\nu_+} \gamma_-^{\nu_-}\right)^{1/\nu} \\ \mu_{\text{solute}} &= \nu RT \ln a_{\pm} \end{aligned} \]
Answer Question 2(b)(i):
\[ \begin{aligned} Q &= 5\ \text{A} \times 900\ \text{s} = 4500\ \text{C} \\ t_+ &= \frac{u_+}{u_+ + u_-} = \frac{5.5}{5.5 + 2 \times 7.91} = 0.258 \\ \text{Moles of Mg}^{2+} &= \frac{t_+ \times Q}{2F} = \frac{0.258 \times 4500}{2 \times 96500} = \boxed{0.00601\ \text{mol}} \\ \text{Moles of Cl}^- &= \frac{t_- \times Q}{F} = \frac{0.742 \times 4500}{96500} = \boxed{0.0346\ \text{mol}} \end{aligned} \]
\[ \begin{aligned} Q &= 5\ \text{A} \times 900\ \text{s} = 4500\ \text{C} \\ t_+ &= \frac{u_+}{u_+ + u_-} = \frac{5.5}{5.5 + 2 \times 7.91} = 0.258 \\ \text{Moles of Mg}^{2+} &= \frac{t_+ \times Q}{2F} = \frac{0.258 \times 4500}{2 \times 96500} = \boxed{0.00601\ \text{mol}} \\ \text{Moles of Cl}^- &= \frac{t_- \times Q}{F} = \frac{0.742 \times 4500}{96500} = \boxed{0.0346\ \text{mol}} \end{aligned} \]
Answer Question 2(b)(ii):
\[ \begin{aligned} v(\text{Mg}^{2+}) &= u_+ \times E = 5.5 \times 10^{-4}\ \text{cm}^2/\text{V s} \times 8.52\ \text{V/cm} = \boxed{0.00469\ \text{cm/s}} \\ v(\text{Cl}^-) &= u_- \times E = 7.91 \times 10^{-4}\ \text{cm}^2/\text{V s} \times 8.52\ \text{V/cm} = \boxed{0.00674\ \text{cm/s}} \end{aligned} \]
\[ \begin{aligned} v(\text{Mg}^{2+}) &= u_+ \times E = 5.5 \times 10^{-4}\ \text{cm}^2/\text{V s} \times 8.52\ \text{V/cm} = \boxed{0.00469\ \text{cm/s}} \\ v(\text{Cl}^-) &= u_- \times E = 7.91 \times 10^{-4}\ \text{cm}^2/\text{V s} \times 8.52\ \text{V/cm} = \boxed{0.00674\ \text{cm/s}} \end{aligned} \]
Answer Question 2(b)(iii):
\[ \begin{aligned} \text{Current carried by Mg}^{2+} &= t_+ \times I = 0.258 \times 5 = 1.29\ \text{A} \\ \text{Current carried by Cl}^- &= t_- \times I = 0.742 \times 5 = 3.71\ \text{A} \\ \text{Ions/s (Mg}^{2+}) &= \frac{1.29\ \text{A}}{2 \times 1.6 \times 10^{-19}\ \text{C}} \approx 4.03 \times 10^{18} \\ \text{Ions/s (Cl}^-) &= \frac{3.71\ \text{A}}{1.6 \times 10^{-19}\ \text{C}} \approx 2.32 \times 10^{19} \\ \text{Ratio} &= \frac{\text{Cl}^-}{\text{Mg}^{2+}} \approx 5.75 \end{aligned} \]
\[ \begin{aligned} \text{Current carried by Mg}^{2+} &= t_+ \times I = 0.258 \times 5 = 1.29\ \text{A} \\ \text{Current carried by Cl}^- &= t_- \times I = 0.742 \times 5 = 3.71\ \text{A} \\ \text{Ions/s (Mg}^{2+}) &= \frac{1.29\ \text{A}}{2 \times 1.6 \times 10^{-19}\ \text{C}} \approx 4.03 \times 10^{18} \\ \text{Ions/s (Cl}^-) &= \frac{3.71\ \text{A}}{1.6 \times 10^{-19}\ \text{C}} \approx 2.32 \times 10^{19} \\ \text{Ratio} &= \frac{\text{Cl}^-}{\text{Mg}^{2+}} \approx 5.75 \end{aligned} \]
Answer Question 2(c):
\[ \begin{aligned} t_+ &= \frac{u_+}{u_+ + u_-}, \quad t_- = \frac{u_-}{u_+ + u_-} \\ \frac{v_+}{v_-} &= \frac{u_+ \times E}{u_- \times E} = \frac{u_+}{u_-} = \frac{t_+}{t_-} \\ \implies r &= \boxed{\frac{t_+}{t_-}} \end{aligned} \]
\[ \begin{aligned} t_+ &= \frac{u_+}{u_+ + u_-}, \quad t_- = \frac{u_-}{u_+ + u_-} \\ \frac{v_+}{v_-} &= \frac{u_+ \times E}{u_- \times E} = \frac{u_+}{u_-} = \frac{t_+}{t_-} \\ \implies r &= \boxed{\frac{t_+}{t_-}} \end{aligned} \]
Answer Question 2(d):
- Energy: Galvanic cells produce energy; electrolytic cells consume energy.
- Anode Polarity: Galvanic anode is negative; electrolytic anode is positive.
- Spontaneity: Galvanic reactions are spontaneous; electrolytic reactions are non-spontaneous.
- ΔG: Galvanic: ΔG<0; electrolytic: ΔG>0.
- Applications: Galvanic cells are batteries; electrolytic cells are used for electroplating.
