Balance the following equations by the ion electron method:
a. `MnO_(4)^(Ө) + Cl^(Ө) + H^(o+) rarr Mn^(2+) + H_(2)O + Cl_(2)`
b. `Cr_(2)O_(7)^(2-) + I^(Ө) + H^(o+) rarr Cr^(3+) + H_(2)O + I_(2)`

Explanation

To balance the given redox reactions using the ion-electron method, we will follow these steps for each equation.

### Part a: Balancing `MnO4^- + Cl^- + H^+ → Mn^2+ + H2O + Cl2`

#### Step 1: Identify the half-reactions
1. **Oxidation half-reaction**:
\[ 2Cl^- \rightarrow Cl_2 \]
2. **Reduction half-reaction**:
\[ MnO_4^- \rightarrow Mn^{2+} \]

#### Step 2: Balance the oxidation half-reaction
- Balance chlorine atoms:
\[ 2Cl^- \rightarrow Cl_2 \]
- Balance charge by adding 2 electrons:
\[ 2Cl^- \rightarrow Cl_2 + 2e^- \]

#### Step 3: Balance the reduction half-reaction
- Identify oxidation states: Mn in MnO4^- is +7 and in Mn^2+ is +2. The change is 5 electrons.
- Write the half-reaction:
\[ MnO_4^- + 5e^- \rightarrow Mn^{2+} \]
- Balance oxygen by adding water:
\[ MnO_4^- + 5e^- + 4H_2O \rightarrow Mn^{2+} + 8H^+ \]

#### Step 4: Equalize the number of electrons
- The oxidation half-reaction has 2 electrons, and the reduction half-reaction has 5 electrons. To equalize, multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2:
1. **Oxidation half-reaction**:
\[ 10Cl^- \rightarrow 5Cl_2 + 10e^- \]
2. **Reduction half-reaction**:
\[ 2MnO_4^- + 10e^- + 8H^+ \rightarrow 2Mn^{2+} + 4H_2O \]

#### Step 5: Combine the half-reactions
- Add both half-reactions:
\[ 10Cl^- + 2MnO_4^- + 8H^+ \rightarrow 5Cl_2 + 2Mn^{2+} + 4H_2O \]

### Final Balanced Equation for Part a:
\[ 10Cl^- + 2MnO_4^- + 8H^+ \rightarrow 5Cl_2 + 2Mn^{2+} + 4H_2O \]

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### Part b: Balancing `Cr2O7^2- + I^- + H^+ → Cr^3+ + H2O + I2`

#### Step 1: Identify the half-reactions
1. **Oxidation half-reaction**:
\[ 2I^- \rightarrow I_2 \]
2. **Reduction half-reaction**:
\[ Cr_2O_7^{2-} \rightarrow Cr^{3+} \]

#### Step 2: Balance the oxidation half-reaction
- Balance iodine atoms:
\[ 2I^- \rightarrow I_2 \]
- Balance charge by adding 2 electrons:
\[ 2I^- \rightarrow I_2 + 2e^- \]

#### Step 3: Balance the reduction half-reaction
- Identify oxidation states: Cr in Cr2O7^2- is +6 and in Cr^3+ is +3. The change is 6 electrons for 2 Cr.
- Write the half-reaction:
\[ Cr_2O_7^{2-} + 6e^- \rightarrow 2Cr^{3+} \]
- Balance oxygen by adding water:
\[ Cr_2O_7^{2-} + 6e^- + 7H_2O \rightarrow 2Cr^{3+} + 14H^+ \]

#### Step 4: Equalize the number of electrons
- The oxidation half-reaction has 2 electrons, and the reduction half-reaction has 6 electrons. To equalize, multiply the oxidation half-reaction by 3 and the reduction half-reaction by 1:
1. **Oxidation half-reaction**:
\[ 6I^- \rightarrow 3I_2 + 6e^- \]
2. **Reduction half-reaction**:
\[ Cr_2O_7^{2-} + 6e^- + 7H_2O \rightarrow 2Cr^{3+} + 14H^+ \]

#### Step 5: Combine the half-reactions
- Add both half-reactions:
\[ Cr_2O_7^{2-} + 6I^- + 7H_2O \rightarrow 3I_2 + 2Cr^{3+} + 14H^+ \]

### Final Balanced Equation for Part b:
\[ Cr_2O_7^{2-} + 6I^- + 7H^+ \rightarrow 3I_2 + 2Cr^{3+} + 7H_2O \]

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