Write cell reaction of the following cells:
(a). `Pt|Fe^(2+),Fe^(3+)||MnO_(4)^(-),Mn^(2+),H^(+)|Pt`
(b). `Pt,Cl_(2)|Cl^(-)(aq)||Ag^(+)(aq)|Ag`

Answer

To write the cell reactions for the given electrochemical cells, we will analyze each cell separately, identifying the oxidation and reduction processes.

### (a) Cell Reaction: `Pt|Fe^(2+),Fe^(3+)||MnO_(4)^(-),Mn^(2+),H^(+)|Pt`

1. **Identify the Anode and Cathode:**
- **Anode (Left Side):** This is where oxidation occurs. Here, \( \text{Fe}^{2+} \) is oxidized to \( \text{Fe}^{3+} \).
- **Cathode (Right Side):** This is where reduction occurs. Here, \( \text{MnO}_4^{-} \) is reduced to \( \text{Mn}^{2+} \).

2. **Write the Half Reactions:**
- **Oxidation Half-Reaction (Anode):**
\[ \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^{-} \]
- **Reduction Half-Reaction (Cathode):**
To balance the reduction of \( \text{MnO}_4^{-} \) to \( \text{Mn}^{2+} \), we need to consider the acidic medium (H\(^+\)):
\[ \text{MnO}_4^{-} + 8\text{H}^{+} + 5e^{-} \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \]

3. **Balance the Electrons:**
- The oxidation reaction produces 1 electron, while the reduction reaction consumes 5 electrons. To balance, we multiply the oxidation half-reaction by 5:
\[ 5\text{Fe}^{2+} \rightarrow 5\text{Fe}^{3+} + 5e^{-} \]

4. **Combine the Half Reactions:**
- Now we can combine the two half-reactions:
\[ 5\text{Fe}^{2+} + \text{MnO}_4^{-} + 8\text{H}^{+} \rightarrow 5\text{Fe}^{3+} + \text{Mn}^{2+} + 4\text{H}_2\text{O} \]

### Final Reaction for (a):
\[ 5\text{Fe}^{2+} + \text{MnO}_4^{-} + 8\text{H}^{+} \rightarrow 5\text{Fe}^{3+} + \text{Mn}^{2+} + 4\text{H}_2\text{O} \]

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### (b) Cell Reaction: `Pt,Cl_(2)|Cl^(-)(aq)||Ag^(+)(aq)|Ag`

1. **Identify the Anode and Cathode:**
- **Anode (Left Side):** Here, \( \text{Cl}^{-} \) is oxidized to \( \text{Cl}_2 \).
- **Cathode (Right Side):** Here, \( \text{Ag}^{+} \) is reduced to \( \text{Ag} \).

2. **Write the Half Reactions:**
- **Oxidation Half-Reaction (Anode):**
\[ 2\text{Cl}^{-} \rightarrow \text{Cl}_2 + 2e^{-} \]
- **Reduction Half-Reaction (Cathode):**
\[ \text{Ag}^{+} + e^{-} \rightarrow \text{Ag} \]

3. **Balance the Electrons:**
- The oxidation reaction produces 2 electrons, while the reduction reaction consumes 1 electron. To balance, we multiply the reduction half-reaction by 2:
\[ 2\text{Ag}^{+} + 2e^{-} \rightarrow 2\text{Ag} \]

4. **Combine the Half Reactions:**
- Now we can combine the two half-reactions:
\[ 2\text{Cl}^{-} + 2\text{Ag}^{+} \rightarrow \text{Cl}_2 + 2\text{Ag} \]

### Final Reaction for (b):
\[ 2\text{Cl}^{-} + 2\text{Ag}^{+} \rightarrow \text{Cl}_2 + 2\text{Ag} \]

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