Electrochemistry - Galvanic cells

Calculate the EMF of the following cell at $298\,K$: $Mg(s) | Mg^{2+}(0.1\,M) || Cu^{2+}(0.01\,M) | Cu(s)$. Given $E^\circ_{Mg^{2+}/Mg} = -2.37\,V$ and $E^\circ_{Cu^{2+}/Cu} = +0.34\,V$.

1. Calculate $E^\circ_{cell}$: $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.34\,V - (-2.37\,V) = 2.71\,V$.
2. Identify $n$: In the reaction $Mg + Cu^{2+} \rightarrow Mg^{2+} + Cu$, the number of electrons transferred is $n = 2$.
3. Apply Nernst Equation: $E_{cell} = E^\circ_{cell} - \frac{0.0591}{2} \log \frac{[Mg^{2+}]}{[Cu^{2+}]}$.
4. Substitute values: $E_{cell} = 2.71 - 0.02955 \log \frac{0.1}{0.01} = 2.71 - 0.02955 \log(10) = 2.71 - 0.02955 = 2.68045\,V$.

First, the standard cell potential is determined. Then, the Nernst equation is used to account for the non-standard concentrations of the ions. Since the concentration of the anode ion is higher than the cathode ion, the cell potential decreases slightly from the standard value.

Write the cell reaction and calculate the standard cell potential ($E^\circ_{cell}$) for the following cell: $Al(s) | Al^{3+}(1M) || Ni^{2+}(1M) | Ni(s)$. Given $E^\circ_{Al^{3+}/Al} = -1.66\,V$ and $E^\circ_{Ni^{2+}/Ni} = -0.25\,V$.

1. Identify electrodes: Anode is $Al$ (lower reduction potential), Cathode is $Ni$.
2. Half-reactions: Anode: $2Al(s) \rightarrow 2Al^{3+}(aq) + 6e^-$ Cathode: $3Ni^{2+}(aq) + 6e^- \rightarrow 3Ni(s)$
3. Overall Reaction: $2Al(s) + 3Ni^{2+}(aq) \rightarrow 2Al^{3+}(aq) + 3Ni(s)$
4. $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$ $E^\circ_{cell} = -0.25\,V - (-1.66\,V)$ $E^\circ_{cell} = +1.41\,V$

The cell potential is positive, confirming the redox reaction is spontaneous under standard conditions. Aluminium acts as the reducing agent.

Determine the equilibrium constant ($K_c$) at $298\,K$ for the reaction occurring in the cell: $Fe(s) | Fe^{2+}(aq) || Ag^+(aq) | Ag(s)$. Given $E^\circ_{Fe^{2+}/Fe} = -0.44\,V$ and $E^\circ_{Ag^+/Ag} = +0.80\,V$.

1. Calculate $E^\circ_{cell}$: $E^\circ_{cell} = E^\circ_{Ag^+/Ag} - E^\circ_{Fe^{2+}/Fe}$ $E^\circ_{cell} = 0.80 - (-0.44) = 1.24\,V$
2. Determine $n$ (electrons transferred): $Fe \rightarrow Fe^{2+} + 2e^-$ $2Ag^+ + 2e^- \rightarrow 2Ag$ So, $n = 2$.
3. Use the formula: $\log K_c = \frac{n E^\circ_{cell}}{0.0591}$ $\log K_c = \frac{2 \times 1.24}{0.0591} \approx 41.96$ $K_c = 10^{41.96} \approx 9.12 \times 10^{41}$

A large $K_c$ value indicates that the reaction goes nearly to completion, which is expected from the high positive $E^\circ_{cell}$.