Manganese ions `(Mn^(2+))` can be oxidised by Persulphate ions `S_(2)O_(8)^(2-)` according to the following half-equations,
`S_(2)O_(8)^(2-)+2e^(-) to 2SO_(4)^(2-)`
`Mn^(2+)+4H_(2)O to MnO_(4)^(-)+4H_(2)O to MnO_(4)^(-)+8H^(+)+5e^(-)`
How many moles of `S_(2)O_(8)^(2-)` are required to oxidise 1mole of `Mn^(2+)`?

Options

A. `2.5`
B. `2.0`
C. `11.0`
D. `0.4`

Answer

To determine how many moles of persulfate ions `(S2O8^(2-))` are required to oxidize 1 mole of manganese ions `(Mn^(2+))`, we will follow these steps:

### Step 1: Write the half-reactions
The half-reactions provided are:
1. Reduction of persulfate:
\[ S2O8^{2-} + 2e^- \rightarrow 2SO4^{2-} \]
2. Oxidation of manganese:
\[ Mn^{2+} + 4H2O \rightarrow MnO4^{-} + 8H^{+} + 5e^- \]

### Step 2: Balance the electrons
In the first half-reaction, 2 electrons are involved, while in the second half-reaction, 5 electrons are involved. To balance the number of electrons transferred, we need to find a common multiple. The least common multiple of 2 and 5 is 10.

### Step 3: Multiply the half-reactions
To balance the electrons:
- Multiply the first half-reaction by 5:
\[ 5(S2O8^{2-} + 2e^- \rightarrow 2SO4^{2-}) \implies 5S2O8^{2-} + 10e^- \rightarrow 10SO4^{2-} \]
- Multiply the second half-reaction by 2:
\[ 2(Mn^{2+} + 4H2O \rightarrow MnO4^{-} + 8H^{+} + 5e^-) \implies 2Mn^{2+} + 8H2O \rightarrow 2MnO4^{-} + 16H^{+} + 10e^- \]

### Step 4: Add the balanced half-reactions
Now we can add the two half-reactions:
\[ 5S2O8^{2-} + 10e^- + 2Mn^{2+} + 8H2O \rightarrow 10SO4^{2-} + 2MnO4^{-} + 16H^{+} + 10e^- \]
The electrons cancel out, giving us:
\[ 5S2O8^{2-} + 2Mn^{2+} + 8H2O \rightarrow 10SO4^{2-} + 2MnO4^{-} + 16H^{+} \]

### Step 5: Determine the moles of `S2O8^(2-)` required for `1 mole of Mn^(2+)`
From the balanced equation, we see that 5 moles of `S2O8^(2-)` are required to oxidize 2 moles of `Mn^(2+)`. Therefore, to find out how many moles of `S2O8^(2-)` are required to oxidize 1 mole of `Mn^(2+)`, we can set up a ratio:
\[ \frac{5 \text{ moles } S2O8^{2-}}{2 \text{ moles } Mn^{2+}} = x \text{ moles } S2O8^{2-} \text{ for } 1 \text{ mole } Mn^{2+} \]
Solving for \(x\):
\[ x = \frac{5}{2} = 2.5 \text{ moles of } S2O8^{2-} \]

### Final Answer
Thus, **2.5 moles of `S2O8^(2-)` are required to oxidize 1 mole of `Mn^(2+)`.**

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