`[Fe(H_(2)O)_(2)(C_(2)O_(4))_(2)]^(2-)+MnO_(4)^(-)+H^(+)toFe^(3+)+CO_(2)+Mn^(2+)+H_(2)O`
In this reaction, number of protons involved in the balanced equation are

Options

A. 10
B. 8
C. 6
D. 0.04

Answer

To solve the problem, we need to balance the given redox reaction and determine the number of protons (H⁺ ions) involved in the balanced equation. The reaction is:

\[ [Fe(H_2O)_2(C_2O_4)_2]^{2-} + MnO_4^{-} + H^{+} \rightarrow Fe^{3+} + CO_2 + Mn^{2+} + H_2O \]

### Step 1: Identify the oxidation states
- **Iron (Fe)** in the complex \([Fe(H_2O)_2(C_2O_4)_2]^{2-}\):
- Oxalate ion \((C_2O_4)^{2-}\) has a charge of -2. Each oxalate contributes -2, hence two oxalates contribute -4.
- Water is neutral (0).
- Therefore, the oxidation state of Fe can be calculated as:

\[ Y + 0 + (-4) = -2 \implies Y = +2 \]

- **Manganese (Mn)** in \(MnO_4^{-}\):
- Let the oxidation state of Mn be \(X\):

\[ X + 4(-2) = -1 \implies X = +7 \]

- **Carbon (C)** in oxalate:
- Let the oxidation state of carbon be \(X\):

\[ 2X + 4(-2) = -2 \implies 2X - 8 = -2 \implies 2X = +6 \implies X = +3 \]

### Step 2: Determine the changes in oxidation states
- **Fe** changes from +2 to +3 (oxidation).
- **C** changes from +3 in oxalate to +4 in \(CO_2\) (oxidation).
- **Mn** changes from +7 to +2 (reduction).

### Step 3: Write half-reactions
1. **Oxidation half-reaction**:
- \(Fe^{2+} \rightarrow Fe^{3+} + e^{-}\) (1 electron)
- For 2 moles of \(C_2O_4^{2-}\):
- \(C_2O_4^{2-} \rightarrow 2CO_2 + 2e^{-}\) (2 electrons)
- Total for 2 moles of \(C_2O_4^{2-}\): 4 electrons.

2. **Reduction half-reaction**:
- \(MnO_4^{-} + 8H^{+} + 5e^{-} \rightarrow Mn^{2+} + 4H_2O\)

### Step 4: Balance the overall reaction
- Combine the half-reactions ensuring the number of electrons lost equals the number of electrons gained:

\[ 2[Fe^{2+} + 2C_2O_4^{2-}] + MnO_4^{-} + 8H^{+} \rightarrow 2Fe^{3+} + 4CO_2 + Mn^{2+} + 4H_2O \]

### Step 5: Count the protons (H⁺ ions)
In the balanced equation, we see that **8 protons (H⁺)** are involved.

### Final Answer:
The number of protons involved in the balanced equation is **8**.

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