Write the cell notation for the voltaic cell that incorporates each of the following redox reactions: (a) \(\mathrm{Al}(s)+\mathrm{Cr}^{3+}(a q) \longrightarrow \mathrm{Al}^{3+}(a q)+\mathrm{Cr}(s)\) (b) \(\mathrm{Cu}^{2+}(a q)+\mathrm{SO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) $$\mathrm{Cu}(s)+\mathrm{SO}_{4}^{2-}(a q)+4 \mathrm{H}^{+}(a q)$$

Redox reactions, or oxidation-reduction reactions, are chemical processes where there is a transfer of electrons between two species. This transfer changes the oxidation states of the chemicals involved. Redox reactions always consist of two half-reactions: one for oxidation and one for reduction.

The substance that loses electrons undergoes oxidation, while the substance that gains electrons undergoes reduction. It's important to remember that oxidation and reduction occur simultaneously. For example, in the reaction \(\text{Al}(s) + \text{Cr}^{3+}(aq) \rightarrow \text{Al}^{3+}(aq) + \text{Cr}(s)\), aluminum (Al) is oxidized and chromium ion \(\text{Cr}^{3+}\) is reduced.

Recognizing oxidation and reduction processes in these reactions is crucial for understanding cell notation and electrochemistry.

The oxidation half-reaction is where a substance loses electrons. This leads to an increase in the oxidation state of the substance.

In the provided exercise, for reaction (a), aluminum (Al) loses three electrons to form aluminum ion (\text{Al}^{3+}). The oxidation half-reaction is: \(\text{Al}(s) \rightarrow \text{Al}^{3+}(aq) + 3e^-\).

Similarly, in reaction (b), sulfur dioxide (\text{SO}_2) is oxidized when it combines with water to produce sulfate ions (\text{SO}_4^{2-}), protons (\text{H}^+), and electrons. The oxidation half-reaction here is: \(\text{SO}_2(g) + 2\text{H}_2O(l) \rightarrow \text{SO}_4^{2-}(aq) + 4\text{H}^+(aq) + 2e^-\).

Understanding oxidation is key to breaking down redox reactions into manageable parts.

In the reduction half-reaction, a substance gains electrons, which leads to a decrease in its oxidation state.

Looking at reaction (a) again, chromium ion (\text{Cr}^{3+}) gains three electrons to form chromium metal (Cr). The reduction half-reaction here is: \(\text{Cr}^{3+}(aq) + 3e^- \rightarrow \text{Cr}(s)\).

In reaction (b), copper ion (\text{Cu}^{2+}) gains two electrons to form copper metal (Cu). The reduction half-reaction is: \(\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)\).

By breaking down the overall redox reaction into these half-reactions, we can better understand the electron transfer process that powers voltaic cells.

Cell notation is a shorthand way to represent the components of a voltaic cell and how they are connected. It follows this format: \(\text{Anode} | \text{Anode's ion} || \text{Cathode's ion} | \text{Cathode}\).

For the given redox reactions, we convert the half-reactions to cell notation.

For reaction (a):

• Oxidation at the anode: \(\text{Al}(s) \rightarrow \text{Al}^{3+}(aq)\)
• Reduction at the cathode: \(\text{Cr}^{3+}(aq) \rightarrow \text{Cr}(s)\)
• Complete cell notation: \(\text{Al}(s) | \text{Al}^{3+}(aq) || \text{Cr}^{3+}(aq) | \text{Cr}(s)\)

For reaction (b):

• Oxidation at the anode: \(\text{SO}_2(g), \text{H}_2O(l) | \text{SO}_4^{2-}(aq), \text{H}^+(aq)\)
• Reduction at the cathode: \(\text{Cu}^{2+}(aq) | \text{Cu}(s)\)
• Complete cell notation: \(\text{SO}_2(g), \text{H}_2O(l) | \text{SO}_4^{2-}(aq), \text{H}^+(aq) || \text{Cu}^{2+}(aq) | \text{Cu}(s)\)

In cell notation, double vertical lines \(||\) separate the two half-cells and single vertical lines \(|\) separate different phases or species within each half-cell. This format provides a clear and concise way to communicate the redox reactions taking place in a voltaic cell.