Electrochemistry - Electrochemical cells

Calculate the $E_{cell}$ for the following cell at $298 K$: $Mg(s) | Mg^{2+}(0.001 M) || Cu^{2+}(0.0001 M) | Cu(s)$. Given $E^\circ_{Mg^{2+}/Mg} = -2.36 V$ and $E^\circ_{Cu^{2+}/Cu} = +0.34 V$.

1. Find $E^\circ_{cell}$: $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.34 V - (-2.36 V) = 2.70 V$.
2. Identify $n$: The reaction is $Mg + Cu^{2+} \rightarrow Mg^{2+} + Cu$, so $n = 2$.
3. Apply Nernst Equation: $E_{cell} = 2.70 - \frac{0.0591}{2} \log \frac{[Mg^{2+}]}{[Cu^{2+}]}$.
4. Substitute values: $E_{cell} = 2.70 - 0.02955 \log \frac{10^{-3}}{10^{-4}} = 2.70 - 0.02955 \log(10) = 2.70 - 0.02955 = 2.67045 V$.

We first determine the standard cell potential using reduction potentials. Then, the Nernst equation is used to adjust for non-standard concentrations. Since the concentration of $Mg^{2+}$ is 10 times that of $Cu^{2+}$, the log term reduces the overall potential.

Calculate the standard Gibbs energy change ($\\Delta G^\circ$) for the reaction: $2Fe^{3+}(aq) + 2I^-(aq) \rightarrow 2Fe^{2+}(aq) + I_2(s)$, given $E^\circ_{cell} = 0.236 V$.

1. Identify $n$: The reaction involves the transfer of $2$ electrons ($2I^- \rightarrow I_2 + 2e^-$), so $n = 2$.
2. Use the formula: $\Delta G^\circ = -nFE^\circ_{cell}$.
3. Substitute values: $\Delta G^\circ = -(2) \times (96500 C/mol) \times (0.236 V)$.
4. Calculate: $\Delta G^\circ = -45548 J/mol = -45.55 kJ/mol$.

Standard Gibbs energy change is directly proportional to the standard cell potential. The negative value indicates that the reaction is thermodynamically spontaneous under standard conditions.

Calculate the equilibrium constant ($K_c$) for the reaction occurring in a cell consisting of Silver and Nickel electrodes at $298 K$. The cell is represented as: $Ni(s) | Ni^{2+}(1 M) || Ag^{+}(1 M) | Ag(s)$. Given $E^\circ_{Ni^{2+}/Ni} = -0.25 V$ and $E^\circ_{Ag^{+}/Ag} = +0.80 V$.

1. Calculate $E^\circ_{cell}$: $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$ $E^\circ_{cell} = 0.80 V - (-0.25 V) = 1.05 V$
2. Determine $n$ (number of electrons): $Ni \rightarrow Ni^{2+} + 2e^-$ and $2Ag^+ + 2e^- \rightarrow 2Ag$, so $n=2$.
3. Use the relation: $\log K_c = \frac{n E^\circ_{cell}}{0.0591}$ $\log K_c = \frac{2 \times 1.05}{0.0591} = \frac{2.10}{0.0591} \approx 35.53$ $K_c = antilog(35.53) \approx 3.38 \times 10^{35}$

The equilibrium constant is calculated from the standard cell potential. A large $K_c$ value indicates that the reaction goes nearly to completion.

For the cell $Al(s) | Al^{3+}(0.01 M) || Fe^{2+}(0.02 M) | Fe(s)$, calculate the cell potential ($E_{cell}$) at $298 K$. Given $E^\circ_{Al^{3+}/Al} = -1.66 V$ and $E^\circ_{Fe^{2+}/Fe} = -0.44 V$.

1. Calculate $E^\circ_{cell}$: $E^\circ_{cell} = -0.44 V - (-1.66 V) = 1.22 V$
2. Write the balanced equation: $2Al(s) + 3Fe^{2+}(aq) \rightarrow 2Al^{3+}(aq) + 3Fe(s)$. Here, $n = 6$.
3. Apply Nernst Equation: $E_{cell} = 1.22 - \frac{0.0591}{6} \log \frac{[Al^{3+}]^2}{[Fe^{2+}]^3}$ $E_{cell} = 1.22 - 0.00985 \log \frac{(10^{-2})^2}{(2 \times 10^{-2})^3}$ $E_{cell} = 1.22 - 0.00985 \log \frac{10^{-4}}{8 \times 10^{-6}}$ $E_{cell} = 1.22 - 0.00985 \log (12.5)$ $E_{cell} = 1.22 - (0.00985 \times 1.0969) \approx 1.22 - 0.0108 = 1.2092 V$

The Nernst equation accounts for the non-standard concentrations of the aluminum and iron ions to find the actual electromotive force of the cell.