Given `E^(c-)._(Fe^(2+)|Fe)` and `E^(c-)._(Fe^(3+)|Fe^(2+))` are `-0.44` and `0.77V` respectively.
If `Fe^(2+),Fe^(3+)` and `Fe` blocks are kept together, then

Options

A. `Fe^(3+)` increases
B. `Fe^(3+)` decreases
C. `Fe^(2+), Fe^(3+)` remain unchanged
D. `Fe^(2+)` decreases

Explanation

To solve the problem, we need to analyze the given standard electrode potentials and their implications for the reactions involving the iron species (Fe²⁺, Fe³⁺, and Fe).

### Step-by-Step Solution:

1. **Identify the Given Potentials:**
- The standard electrode potential for the half-reaction \( \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe} \) is given as \( E^\circ = -0.44 \, \text{V} \).
- The standard electrode potential for the half-reaction \( \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \) is given as \( E^\circ = 0.77 \, \text{V} \).

2. **Determine Spontaneity of Reactions:**
- The reaction \( \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe} \) is non-spontaneous because the potential is negative.
- The reaction \( \text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+} \) is spontaneous because the potential is positive.

3. **Analyze the System:**
- When \( \text{Fe}^{2+} \), \( \text{Fe}^{3+} \), and solid \( \text{Fe} \) are kept together, the spontaneous reaction will be the reduction of \( \text{Fe}^{3+} \) to \( \text{Fe}^{2+} \).
- This means \( \text{Fe}^{3+} \) will be reduced, leading to an increase in the concentration of \( \text{Fe}^{2+} \).

4. **Conclude the Changes in Concentrations:**
- As \( \text{Fe}^{3+} \) is reduced to \( \text{Fe}^{2+} \), the concentration of \( \text{Fe}^{3+} \) will decrease.
- The concentration of \( \text{Fe}^{2+} \) will increase due to the conversion from \( \text{Fe}^{3+} \).

5. **Final Answer:**
- Therefore, the correct conclusion is that the concentration of \( \text{Fe}^{3+} \) decreases.

### Answer:
The concentration of \( \text{Fe}^{3+} \) decreases.