Fill the missing products in the following reactions.
(i) `2MnI_4^(-)+16 H^(+)+10I^(-)to2Mn^(2+)+8H_2O+"_________"`.
(ii) `2MnO_4^(-)+H-2O+I^(-)to2MnO_2+2OH^(-)+"__________" `.

Options

A. (i) HI, (ii) `I_2`
B. (ii)I2, (ii) IO_3^(-)`
C. `(iii)I_2,(ii) I_2`
D. `(iv) IO_3^(-), (ii) I_2`

Answer

To fill in the missing products in the given reactions, let's analyze each reaction step by step.

### Reaction (i):
Given reaction:
\[ 2 \text{MnI}_4^{-} + 16 \text{H}^{+} + 10 \text{I}^{-} \rightarrow 2 \text{Mn}^{2+} + 8 \text{H}_2\text{O} + \_\_\_\_\_\_\_\_ \]

**Step 1: Identify the oxidation states.**
- In \(\text{MnI}_4^{-}\), manganese (Mn) is in the +7 oxidation state.
- In \(\text{Mn}^{2+}\), manganese is in the +2 oxidation state.

**Step 2: Determine the change in oxidation states.**
- Mn is reduced from +7 to +2.
- The iodide ions (\(I^{-}\)) are oxidized to molecular iodine (\(I_2\)).

**Step 3: Balance the iodine atoms.**
- 10 \(I^{-}\) ions will produce 5 \(I_2\) molecules.

**Step 4: Write the complete balanced equation.**
The complete balanced equation is:
\[ 2 \text{MnI}_4^{-} + 16 \text{H}^{+} + 10 \text{I}^{-} \rightarrow 2 \text{Mn}^{2+} + 8 \text{H}_2\text{O} + 5 \text{I}_2 \]

### Missing product for Reaction (i):
**Answer:** \(5 \text{I}_2\)

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### Reaction (ii):
Given reaction:
\[ 2 \text{MnO}_4^{-} + \text{H}_2\text{O} + \text{I}^{-} \rightarrow 2 \text{MnO}_2 + 2 \text{OH}^{-} + \_\_\_\_\_\_\_\_ \]

**Step 1: Identify the oxidation states.**
- In \(\text{MnO}_4^{-}\), manganese (Mn) is in the +7 oxidation state.
- In \(\text{MnO}_2\), manganese is in the +4 oxidation state.

**Step 2: Determine the change in oxidation states.**
- Mn is reduced from +7 to +4.
- The iodide ion (\(I^{-}\)) is oxidized to iodate (\(IO_3^{-}\)).

**Step 3: Balance the iodine atoms.**
- The \(I^{-}\) ion will produce \(IO_3^{-}\).

**Step 4: Write the complete balanced equation.**
The complete balanced equation is:
\[ 2 \text{MnO}_4^{-} + \text{H}_2\text{O} + 2 \text{I}^{-} \rightarrow 2 \text{MnO}_2 + 2 \text{OH}^{-} + 2 \text{IO}_3^{-} \]

### Missing product for Reaction (ii):
**Answer:** \(2 \text{IO}_3^{-}\)

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### Summary of Answers:
1. For Reaction (i): \(5 \text{I}_2\)
2. For Reaction (ii): \(2 \text{IO}_3^{-}\)

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