To solve the problem of matching the standard reduction potentials from List-I with the values in List-II, we will follow these steps:
### Step 1: Calculate \( E^\circ \) for \( \text{Fe}^{3+} \) to \( \text{Fe} \)
We know from the data:
- \( E^\circ (\text{Fe}^{3+}, \text{Fe}^{2+}) = +0.77 \, \text{V} \)
- \( E^\circ (\text{Fe}^{2+}, \text{Fe}) = -0.44 \, \text{V} \)
Using the formula for combining standard reduction potentials:
\[
E^\circ (\text{Fe}^{3+}, \text{Fe}) = E^\circ (\text{Fe}^{3+}, \text{Fe}^{2+}) + E^\circ (\text{Fe}^{2+}, \text{Fe})
\]
Substituting the values:
\[
E^\circ (\text{Fe}^{3+}, \text{Fe}) = 0.77 \, \text{V} + (-0.44 \, \text{V}) = 0.33 \, \text{V}
\]
### Step 2: Calculate \( E^\circ \) for the reaction \( 4 \text{H}_2\text{O} \rightleftharpoons 4 \text{H}^+ + 4 \text{OH}^- \)
Using the given potentials:
- \( E^\circ (\text{O}_2 + 4\text{H}^+ + 4e^- \rightarrow 2\text{H}_2\text{O}) = +1.23 \, \text{V} \)
- \( E^\circ (\text{O}_2 + 2\text{H}_2\text{O} + 4e^- \rightarrow 4\text{OH}^-) = +0.40 \, \text{V} \)
We can find the potential for the reaction by subtracting:
\[
E^\circ = E^\circ (\text{OH}^-) - E^\circ (\text{H}_2\text{O})
\]
Calculating:
\[
E^\circ = 0.40 \, \text{V} - 1.23 \, \text{V} = -0.83 \, \text{V}
\]
### Step 3: Calculate \( E^\circ \) for \( \text{Cu}^{2+} + \text{Cu} \rightleftharpoons 2\text{Cu}^{+} \)
Using the potentials:
- \( E^\circ (\text{Cu}^{2+}, \text{Cu}) = +0.34 \, \text{V} \)
- \( E^\circ (\text{Cu}^{+}, \text{Cu}) = +0.52 \, \text{V} \)
The relation is:
\[
E^\circ (\text{Cu}^{2+}, \text{Cu}^{+}) = E^\circ (\text{Cu}^{2+}, \text{Cu}) - E^\circ (\text{Cu}^{+}, \text{Cu})
\]
Calculating:
\[
E^\circ = 0.34 \, \text{V} - 0.52 \, \text{V} = -0.18 \, \text{V}
\]
### Step 4: Calculate \( E^\circ \) for \( \text{Cr}^{3+} \) to \( \text{Cr}^{2+} \)
Using the potentials:
- \( E^\circ (\text{Cr}^{3+}, \text{Cr}) = -0.74 \, \text{V} \)
- \( E^\circ (\text{Cr}^{2+}, \text{Cr}) = -0.91 \, \text{V} \)
The relation is:
\[
E^\circ (\text{Cr}^{3+}, \text{Cr}^{2+}) = E^\circ (\text{Cr}^{3+}, \text{Cr}) - E^\circ (\text{Cr}^{2+}, \text{Cr})
\]
Calculating:
\[
E^\circ = -0.74 \, \text{V} - (-0.91 \, \text{V}) = 0.17 \, \text{V}
\]
### Summary of Results
1. \( E^\circ (\text{Fe}^{3+}, \text{Fe}) = 0.33 \, \text{V} \) (not matching any)
2. \( E^\circ (4 \text{H}_2\text{O} \rightleftharpoons 4 \text{H}^+ + 4 \text{OH}^-) = -0.83 \, \text{V} \) matches with (4)
3. \( E^\circ (\text{Cu}^{2+} + \text{Cu} \rightleftharpoons 2\text{Cu}^{+}) = -0.18 \, \text{V} \) matches with (1)
4. \( E^\circ (\text{Cr}^{3+}, \text{Cr}^{2+}) = 0.17 \, \text{V} \) matches with (2)
### Final Matching
- \( P \) matches with (1)
- \( Q \) matches with (4)
- \( R \) matches with (2)
- \( S \) matches with (3)
