Complete the given reaction.
`2Mn^(2+)+5ul((i))+8H_2Oto2ul((ii))+10ul((iii))+16H^(+)`

Options

A. `{:("(i)","(ii)","(iii)"),(SO_4^(2-),MnO_4^(-),S_2O_8^(2-)):}`
B. `{:("(i)","(ii)","(iii)"),(S_2O_8^(2-),MnO_4^(-),SO_4^(2-)):}`
C. `{:("(i)","(ii)","(iii)"),(MnO_4^(-),S_2O_8^(2-),SO_4^(2-)):}`
D. `{:("(i)","(ii)","(iii)"),(S_2O_8^(2-),SO_4^(2-),MnO_4^(-)):}`

Answer

To complete the given reaction, we need to identify the species that are being oxidized and reduced. The reaction provided is:

\[ 2 \text{Mn}^{2+} + 5 \_\_\_ + 8 \text{H}_2\text{O} \rightarrow 2 \_\_\_ + 10 \_\_\_ + 16 \text{H}^+ \]

### Step 1: Identify the oxidation states
- Manganese in \(\text{Mn}^{2+}\) has an oxidation state of +2.
- In the products, we need to find the species that corresponds to the oxidation of manganese to a higher oxidation state. The common higher oxidation state for manganese is +7, which corresponds to \(\text{MnO}_4^-\).

### Step 2: Determine the reducing agent
- The reducing agent will be the species that gets oxidized. In this case, we need to find a species that can provide electrons to \(\text{Mn}^{2+}\) to convert it into \(\text{MnO}_4^-\).
- The species that can be oxidized to a lower oxidation state is likely to be a sulfur compound. The compound that has sulfur in a high oxidation state is \(\text{S}_2\text{O}_8^{2-}\) (peroxydisulfate), which has sulfur in the +6 oxidation state.

### Step 3: Write the products
- The product of the oxidation of \(\text{S}_2\text{O}_8^{2-}\) will be \(\text{SO}_4^{2-}\) (sulfate), where sulfur has an oxidation state of +6.
- Therefore, the first blank will be filled with \(\text{S}_2\text{O}_8^{2-}\), the second blank with \(\text{MnO}_4^{-}\), and the third blank with \(\text{SO}_4^{2-}\).

### Final Reaction
Putting it all together, we have:

\[ 2 \text{Mn}^{2+} + 5 \text{S}_2\text{O}_8^{2-} + 8 \text{H}_2\text{O} \rightarrow 2 \text{MnO}_4^{-} + 10 \text{SO}_4^{2-} + 16 \text{H}^+ \]

### Summary of the Blanks
- First blank: \(\text{S}_2\text{O}_8^{2-}\)
- Second blank: \(\text{MnO}_4^{-}\)
- Third blank: \(\text{SO}_4^{2-}\)

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