Given that `E_(Fe^(2+)//Fe)^(.)=-0.44V,E_(Fe^(3+)//Fe^(2+))^(@)=0.77V` if `Fe^(2+),Fe^(3+)` and `Fe` solid are kept together then

Options

A. the concentration of `Fe^(3+)` increases
B. the concentration of `Fe^(3+)` decreases
C. the mass of Fe increases
D. the concentration of `Fe^(2+)` decreases

Answer

To solve the problem, we need to analyze the given reduction potentials and determine the reactions that will occur when Fe²⁺, Fe³⁺, and solid Fe are kept together.

### Step-by-Step Solution:

1. **Identify the Reduction Potentials:**
- The reduction potential of Fe²⁺ to Fe is given as \( E_{Fe^{2+}/Fe} = -0.44 \, V \).
- The reduction potential of Fe³⁺ to Fe²⁺ is given as \( E_{Fe^{3+}/Fe^{2+}} = 0.77 \, V \).

2. **Determine Which Species Will Be Reduced and Oxidized:**
- A higher reduction potential indicates a greater tendency to gain electrons (be reduced). Here, Fe³⁺ has a higher reduction potential than Fe²⁺.
- Therefore, Fe³⁺ will be reduced to Fe²⁺, and Fe will be oxidized to Fe²⁺.

3. **Write the Half-Reactions:**
- **Oxidation (Anode):**
\[ Fe \rightarrow Fe^{2+} + 2e^{-} \]
- **Reduction (Cathode):**
\[ Fe^{3+} + e^{-} \rightarrow Fe^{2+} \]

4. **Balance the Electrons:**
- To balance the number of electrons transferred, we need to multiply the reduction half-reaction by 2:
\[ 2Fe^{3+} + 2e^{-} \rightarrow 2Fe^{2+} \]
- Now, the overall balanced reaction becomes:
\[ Fe + 2Fe^{3+} \rightarrow 2Fe^{2+} \]

5. **Analyze the Changes in Concentration:**
- In the reaction, Fe is being oxidized to form Fe²⁺, and Fe³⁺ is being reduced to Fe²⁺.
- As Fe³⁺ is consumed, its concentration decreases.
- The concentration of Fe²⁺ increases as it is produced from both the oxidation of Fe and the reduction of Fe³⁺.

6. **Conclusion:**
- Therefore, the concentration of Fe³⁺ decreases, and the concentration of Fe²⁺ increases. The mass of solid Fe will decrease as it is converted to Fe²⁺.

### Final Answer:
The concentration of Fe³⁺ decreases.