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 follow the steps of identifying the anode and cathode reactions, along with the overall cell reaction.

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

1. **Identify the Anode and Cathode**:
- The anode is where oxidation occurs, and the cathode is where reduction occurs.
- In this cell, the left side (Pt|Fe^(2+),Fe^(3+)) is the anode and the right side (MnO_(4)^(-),Mn^(2+),H^(+)|Pt) is the cathode.

2. **Anode Reaction**:
- The oxidation reaction at the anode involves the conversion of Fe^(2+) to Fe^(3+) with the loss of one electron:
\[ \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^- \]

3. **Cathode Reaction**:
- The reduction reaction at the cathode involves the conversion of MnO4^(-) to Mn^(2+) in an acidic medium. The half-reaction is:
\[ \text{MnO}_4^{-} + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \]

4. **Balancing the Overall Reaction**:
- To balance the overall cell reaction, we need to ensure that the number of electrons lost in the oxidation reaction equals the number of electrons gained in the reduction reaction.
- The oxidation reaction produces 1 electron, while the reduction reaction consumes 5 electrons. Therefore, we multiply the oxidation reaction by 5:
\[ 5\text{Fe}^{2+} \rightarrow 5\text{Fe}^{3+} + 5e^- \]
- 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 Cell 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 for `Pt,Cl_(2)|Cl^(-)(aq)||Ag^(+)(aq)|Ag`

1. **Identify the Anode and Cathode**:
- The left side (Pt,Cl_(2)|Cl^(-)(aq)) is the anode and the right side (Ag^(+)(aq)|Ag) is the cathode.

2. **Anode Reaction**:
- The oxidation reaction at the anode involves the conversion of Cl^(-) to Cl2 with the loss of electrons:
\[ 2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^- \]

3. **Cathode Reaction**:
- The reduction reaction at the cathode involves the conversion of Ag^(+) to Ag:
\[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \]

4. **Balancing the Overall Reaction**:
- The oxidation reaction produces 2 electrons, while the reduction reaction consumes 1 electron. Therefore, we multiply the reduction reaction by 2:
\[ 2\text{Ag}^+ + 2e^- \rightarrow 2\text{Ag} \]
- Now we can combine the two half-reactions:
\[ 2\text{Cl}^- + 2\text{Ag}^+ \rightarrow \text{Cl}_2 + 2\text{Ag} \]

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

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