Problem 38 - Chemistry The Science in Context

Question

Voltaic cells based on the following pairs of half-reactions are constructed. For each pair, write a balanced equation for the cell reaction, and identify which half-reaction takes place at each anode and cathode. a. \(\mathrm{Cd}^{2+}(a q)+2 \mathrm{e}^{-} \rightarrow \mathrm{Cd}(s)\) \(\mathrm{Ag}^{+}(a q)+\mathrm{e}^{-} \rightarrow \mathrm{Ag}(s)\) b. \(\mathrm{AgBr}(s)+\mathrm{e}^{-} \rightarrow \mathrm{Ag}(s)+\mathrm{Br}^{-}(a q)\) \(\mathrm{MnO}_{2}(s)+4 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(\ell)\) c. \(\mathrm{PtCl}_{4}^{2-}(a q)+2 \mathrm{e}^{-} \rightarrow \mathrm{Pt}(s)+4 \mathrm{Cl}^{-}(a q)\) \(\mathrm{AgCl}(s)+\mathrm{e}^{-} \rightarrow \mathrm{Ag}(s)+\mathrm{Cl}^{-}(a q)\)

Step-by-Step Solution

Answer

a. Anode: \(\mathrm{Cd}^{2+}(a q)+2 \mathrm{e}^{-} \rightarrow \mathrm{Cd}(s)\) Cathode: \(\mathrm{Ag}^{+}(a q)+\mathrm{e}^{-} \rightarrow \mathrm{Ag}(s)\) b. Anode: \(2\mathrm{AgBr}(s)+2\mathrm{e}^{-} \rightarrow 2\mathrm{Ag}(s)+2\mathrm{Br}^{-}(a q)\) Cathode: \(\mathrm{MnO}_{2}(s)+4 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(\ell)\) c. Anode: \(2\mathrm{AgCl}(s)+2\mathrm{e}^{-} \rightarrow 2\mathrm{Ag}(s)+2\mathrm{Cl}^{-}(a q)\) Cathode: \(\mathrm{PtCl}_{4}^{2-}(a q)+2\mathrm{e}^{-} \rightarrow \mathrm{Pt}(s)+4 \mathrm{Cl}^{-}(a q)\)

Step 1: Balancing electrons in the two half-reactions

Multiply the second half-reaction by 2 to balance the electrons: 1. \(\mathrm{Cd}^{2+}(a q)+2 \mathrm{e}^{-} \rightarrow \mathrm{Cd}(s)\) 2. \(2\mathrm{Ag}^{+}(a q)+2\mathrm{e}^{-} \rightarrow 2\mathrm{Ag}(s)\)

Step 2: Adding half-reactions to obtain the cell reaction

Combine the two balanced half-reactions: \(\mathrm{Cd}^{2+}(a q)+2\mathrm{Ag}^{+}(a q) \rightarrow \mathrm{Cd}(s)+2\mathrm{Ag}(s)\)

Step 3: Identify anode and cathode half-reactions

Since oxidation occurs at the anode, \(\mathrm{Cd}^{2+}(a q)+2 \mathrm{e}^{-} \rightarrow \mathrm{Cd}(s)\) is the anode reaction. Therefore, the second reaction, \(\mathrm{Ag}^{+}(a q)+\mathrm{e}^{-} \rightarrow \mathrm{Ag}(s)\), occurs at the cathode. b. Half-reactions: 1. \(\mathrm{AgBr}(s)+\mathrm{e}^{-} \rightarrow \mathrm{Ag}(s)+\mathrm{Br}^{-}(a q)\) 2. \(\mathrm{MnO}_{2}(s)+4 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(\ell)\)

Step 1: Balancing electrons in the two half-reactions

Multiply the first half-reaction by 2 to balance the electrons: 1. \(2\mathrm{AgBr}(s)+2\mathrm{e}^{-} \rightarrow 2\mathrm{Ag}(s)+2\mathrm{Br}^{-}(a q)\) 2. \(\mathrm{MnO}_{2}(s)+4 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(\ell)\)

Step 2: Adding half-reactions to obtain the cell reaction

Combine the two balanced half-reactions: \(2\mathrm{AgBr}(s)+\mathrm{MnO}_{2}(s)+4 \mathrm{H}^{+}(a q) \rightarrow 2\mathrm{Ag}(s)+2\mathrm{Br}^{-}(a q)+\mathrm{Mn}^{2+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(\ell)\)

Step 3: Identify anode and cathode half-reactions

Since oxidation occurs at the anode, \(2\mathrm{AgBr}(s)+2\mathrm{e}^{-} \rightarrow 2\mathrm{Ag}(s)+2\mathrm{Br}^{-}(a q)\) is the anode reaction. Therefore, the second reaction, \(\mathrm{MnO}_{2}(s)+4 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(\ell)\), occurs at the cathode. c. Half-reactions: 1. \(\mathrm{PtCl}_{4}^{2-}(a q)+2\mathrm{e}^{-} \rightarrow \mathrm{Pt}(s)+4 \mathrm{Cl}^{-}(a q)\) 2. \(\mathrm{AgCl}(s)+\mathrm{e}^{-} \rightarrow \mathrm{Ag}(s)+\mathrm{Cl}^{-}(a q)\)

Step 1: Balancing electrons in the two half-reactions

Multiply the second half-reaction by 2 to balance the electrons: 1. \(\mathrm{PtCl}_{4}^{2-}(a q)+2\mathrm{e}^{-} \rightarrow \mathrm{Pt}(s)+4 \mathrm{Cl}^{-}(a q)\) 2. \(2\mathrm{AgCl}(s)+2\mathrm{e}^{-} \rightarrow 2\mathrm{Ag}(s)+2\mathrm{Cl}^{-}(a q)\)

Step 2: Adding half-reactions to obtain the cell reaction

Combine the two balanced half-reactions: \(\mathrm{PtCl}_{4}^{2-}(a q)+2\mathrm{AgCl}(s) \rightarrow \mathrm{Pt}(s)+2\mathrm{Ag}(s)+6 \mathrm{Cl}^{-}(a q)\)

Step 3: Identify anode and cathode half-reactions

Since oxidation occurs at the anode, \(2\mathrm{AgCl}(s)+2\mathrm{e}^{-} \rightarrow 2\mathrm{Ag}(s)+2\mathrm{Cl}^{-}(a q)\) is the anode reaction. Therefore, the first reaction, \(\mathrm{PtCl}_{4}^{2-}(a q)+2\mathrm{e}^{-} \rightarrow \mathrm{Pt}(s)+4 \mathrm{Cl}^{-}(a q)\), occurs at the cathode.

Key Concepts

Half-Reactions Balancing

Understanding the concept of half-reactions is crucial for grasping how voltaic cells work. A half-reaction is a part of the overall redox reaction that occurs in a voltaic cell and involves either oxidation or reduction, but not both. Each half-reaction can be thought of as a separate chemical process occurring at the electrodes.
To balance half-reactions, it is essential to ensure that the total number of electrons lost in the oxidation half-reaction equals the total number of electrons gained in the reduction half-reaction. This balance is necessary for the conservation of charge and mass.
Here is how you balance them:

• Look at the electron transfer in both half-reactions.
• If one half-reaction involves a different number of electrons than the other, multiply the equations by the appropriate number to balance the electrons.

Balancing allows the half-reactions to seamlessly combine into a complete redox reaction, which represents the total chemical changes occurring during the operation of the cell.

Anode and Cathode Identification

Anodes and cathodes are fundamental components where the half-reactions of oxidation and reduction occur, thereby facilitating the flow of electrons within a voltaic cell. Understanding the roles of these electrodes can help you predict the direction of electron flow and the type of reactions taking place.
To identify the anode and the cathode, you need to determine where oxidation and reduction happen:

• The anode is the electrode where oxidation occurs. During this process, species lose electrons. It is often represented by an increase in oxidation state.
• The cathode is the electrode where reduction takes place. Here, species gain electrons, resulting in a decrease in oxidation state.

Remember the mnemonic "An Ox, Red Cat": **An**ode is for **Ox**idation, and **Red**uction occurs at the **Cat**hode. By correctly identifying the anode and cathode, you can predict the direction in which electrons flow, from the anode (losing electrons) to the cathode (gaining electrons). This is crucial for understanding and designing electrochemical cells.

Cell Reaction Equations

Cell reaction equations represent the entire chemical reaction that takes place within a voltaic cell. By understanding how to derive these from half-reactions, you'll get insights into the overall electron transfer process.
The process to write cell reaction equations involves:

• Balancing each half-reaction so that the electrons lost and gained are equal.
• Combining the two balanced half-reactions to cancel out the electrons and reveal the net cell reaction.

When combined, the balanced half-reactions should reflect no net change in electron count, as they will perfectly offset each other.
This net cell reaction not only showcases the transfer of electrons but also describes the chemical transformation from reactants to products. Understanding and writing accurate cell reaction equations allows for more powerful insights into the energy changes and efficiency of the voltaic cell.