Question

Write a balanced chemical equation for the overall cell reaction represented as (a) \(\mathrm{Ag}\left|\mathrm{Ag}^{+} \| \mathrm{Sn}^{4+}, \mathrm{Sn}^{2+}\right| \mathrm{Pt}\). (b) \(\mathrm{Al}\left|\mathrm{Al}^{3+} \| \mathrm{Cu}^{2+}\right| \mathrm{Cu}\). (c) \(\mathrm{Pt}\left|\mathrm{Fe}^{2+}, \mathrm{Fe}^{3+} \| \mathrm{MnO}_{4}^{-}, \mathrm{Mn}^{2+}\right| \mathrm{Pt}\).

Short answer

Question: Write balanced chemical equations for the overall cell reactions represented by the given cell notations. a) Ag|Ag+ || Sn4+, Sn2+| Pt b) Al|Al3+ || Cu2+| Cu c) Pt|Fe2+, Fe3+ || MnO4-, Mn2+| Pt Answer: a) 2Ag + Sn4+ → 2Ag+ + Sn2+ b) 2Al + 3Cu2+ → 2Al3+ + 3Cu c) 5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn2+ + 4H2O

Step-by-step solution

Identify the half-reactions

In this cell notation, we have: - Oxidation half-reaction: \(\mathrm{Ag} \rightarrow \mathrm{Ag}^{+}\) - Reduction half-reaction: \(\mathrm{Sn}^{4+} \rightarrow \mathrm{Sn}^{2+}\)

Balance the half-reactions

Oxidation half-reaction: \(\mathrm{Ag} \rightarrow \mathrm{Ag}^{+} + e^-\) Reduction half-reaction: \(\mathrm{Sn}^{4+} + 2e^- \rightarrow \mathrm{Sn}^{2+}\)

Combine the half-reactions

Multiply the oxidation half-reaction by 2 to balance the electrons: \(2(\mathrm{Ag} \rightarrow \mathrm{Ag}^{+} + e^-)\) gives \(2\mathrm{Ag} \rightarrow 2\mathrm{Ag}^{+} + 2e^-\) Now combine the balanced half-reactions and cancel the electrons: \(2\mathrm{Ag} \rightarrow 2\mathrm{Ag}^{+} + 2e^-\) \(\mathrm{Sn}^{4+} + 2e^- \rightarrow \mathrm{Sn}^{2+}\) Balanced chemical equation: \(2\mathrm{Ag} + \mathrm{Sn}^{4+} \rightarrow 2\mathrm{Ag}^{+} + \mathrm{Sn}^{2+}\) (b) \(\mathrm{Al}\left|\mathrm{Al}^{3+} \| \mathrm{Cu}^{2+}\right| \mathrm{Cu}\)

Identify the half-reactions

In this cell notation, we have: - Oxidation half-reaction: \(\mathrm{Al} \rightarrow \mathrm{Al}^{3+}\) - Reduction half-reaction: \(\mathrm{Cu}^{2+} \rightarrow \mathrm{Cu}\)

Balance the half-reactions

Oxidation half-reaction: \(\mathrm{Al} \rightarrow \mathrm{Al}^{3+} + 3e^-\) Reduction half-reaction: \(\mathrm{Cu}^{2+} + 2e^- \rightarrow \mathrm{Cu}\)

Combine the half-reactions

Multiply the oxidation half-reaction by 2 and the reduction half-reaction by 3 to balance the electrons: \(2(\mathrm{Al} \rightarrow \mathrm{Al}^{3+} + 3e^-)\) \(3(\mathrm{Cu}^{2+} + 2e^- \rightarrow \mathrm{Cu})\) This gives: \(2\mathrm{Al} \rightarrow 2\mathrm{Al}^{3+} + 6e^-\) \(3\mathrm{Cu}^{2+} + 6e^- \rightarrow 3\mathrm{Cu}\) Now combine the balanced half-reactions and cancel the electrons: \(2\mathrm{Al} \rightarrow 2\mathrm{Al}^{3+} + 6e^-\) \(3\mathrm{Cu}^{2+} + 6e^- \rightarrow 3\mathrm{Cu}\) Balanced chemical equation: \(2\mathrm{Al} + 3\mathrm{Cu}^{2+} \rightarrow 2\mathrm{Al}^{3+} + 3\mathrm{Cu}\) (c) \(\mathrm{Pt}\left|\mathrm{Fe}^{2+}, \mathrm{Fe}^{3+} \| \mathrm{MnO}_{4}^{-}, \mathrm{Mn}^{2+}\right| \mathrm{Pt}\)

Identify the half-reactions

In this cell notation, we have: - Oxidation half-reaction: \(\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+}\) - Reduction half-reaction: \(\mathrm{MnO}_{4}^{-} \rightarrow \mathrm{Mn}^{2+}\)

Balance the half-reactions

Oxidation half-reaction: \(\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+} + e^-\) Reduction half-reaction (balancing method for a reaction in acidic solution): 1. Write the unbalanced reduction half-reaction: \(\mathrm{MnO}_{4}^{-} \rightarrow \mathrm{Mn}^{2+}\) 2. Balance oxygen atoms by adding water molecules: \(\mathrm{MnO}_{4}^{-} \rightarrow \mathrm{Mn}^{2+} + 4\mathrm{H}_{2}\mathrm{O}\) 3. Balance hydrogen atoms by adding protons (\(\mathrm{H}^{+}\) ions): \(\mathrm{MnO}_{4}^{-} + 8\mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+} + 4\mathrm{H}_{2}\mathrm{O}\) 4. Balance charge by adding electrons: \(\mathrm{MnO}_{4}^{-} + 8\mathrm{H}^{+} + 5e^- \rightarrow \mathrm{Mn}^{2+} + 4\mathrm{H}_{2}\mathrm{O}\)

Combine the half-reactions

Multiply the oxidation half-reaction by 5 to balance the electrons: \(5(\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+} + e^-)\) gives \(5\mathrm{Fe}^{2+} \rightarrow 5\mathrm{Fe}^{3+} + 5e^-\) Now combine the balanced half-reactions and cancel the electrons: \(5\mathrm{Fe}^{2+} \rightarrow 5\mathrm{Fe}^{3+} + 5e^-\) \(\mathrm{MnO}_{4}^{-} + 8\mathrm{H}^{+} + 5e^- \rightarrow \mathrm{Mn}^{2+} + 4\mathrm{H}_{2}\mathrm{O}\) Balanced chemical equation: \(5\mathrm{Fe}^{2+} + \mathrm{MnO}_{4}^{-} + 8\mathrm{H}^{+} \rightarrow 5\mathrm{Fe}^{3+} + \mathrm{Mn}^{2+} + 4\mathrm{H}_{2}\mathrm{O}\)

Key concepts

Cell Notation

Cell notation is a systematic way to represent the components and their arrangements within an electrochemical cell. This shorthand description is crucial for understanding the flow of electrons between the electrodes during a redox reaction. For example, in notation (a) \(\mathrm{Ag}\left|\mathrm{Ag}^{+} \| \mathrm{Sn}^{4+},\mathrm{Sn}^{2+}\right| \mathrm{Pt}\), elements to the left of the vertical line are at the anode, where oxidation occurs, while elements to the right are at the cathode, where reduction takes place. The double line signifies the salt bridge or membrane separating both half-cells. Understanding cell notation is fundamental for deciphering the electrochemical process and subsequently writing the balanced chemical equation for the cell reaction.

Half-Reactions

Half-reactions are the bread and butter of balancing redox equations. They split the overall redox reaction into two separate processes: oxidation (loss of electrons) and reduction (gain of electrons). Each half-reaction is balanced individually for atoms and charge. For instance, the oxidation half-reaction for silver is \(\mathrm{Ag} \rightarrow \mathrm{Ag}^{+} + e^-\), and for tin(IV) reduction, it is \(\mathrm{Sn}^{4+} + 2e^- \rightarrow \mathrm{Sn}^{2+}\). Analyzing each half-reaction separately makes it easier to balance the entire cell reaction, ensuring that the same number of electrons are lost in oxidation as are gained in reduction.

Electron Balancing

Electron balancing is a critical skill in constructing a balanced chemical equation for a redox reaction, especially within an electrochemical cell. When balancing half-reactions, it's key to ensure that the number of electrons lost in the oxidation half-reaction equals the number gained in the reduction half-reaction. This parity allows for the cancellation of electrons when the half-reactions are added together. For example, with the aluminum and copper cell represented as \(2\mathrm{Al} → 2\mathrm{Al}^{3+} + 6e^-\) and \(3\mathrm{Cu}^{2+} + 6e^- → 3\mathrm{Cu}\), we multiply the aluminum half-reaction by 2 and the copper half-reaction by 3 to balance out the six electrons exchanged in both half-reactions.

Redox Reactions

Redox reactions are the foundation of electrochemical processes, characterized by the transfer of electrons between species. These reactions involve both reduction and oxidation occurring simultaneously. Redox reactions are not only vital in batteries and electrochemistry but also play a pivotal role in many biological systems and industrial processes. The balanced chemical equation \(2\mathrm{Ag} + \mathrm{Sn}^{4+} → 2\mathrm{Ag}^{+} + \mathrm{Sn}^{2+}\) represents a complete redox reaction obtained by adding the balanced half-reactions. Understanding redox reactions and mastering the art of balancing them is crucial for students studying chemistry, as they provide deep insights into the electron transfer that drives many chemical processes.