The measured voltage for the reaction with the indicated concentration is 1.50 V. Calculate `E^(@)`.
`Cr(s)+3Ag^(+)(aq,0.10M)to3Ag(s)+Cr^(3+)(aq,0.30 M)`

Options

A. 1.35
B. 1.4
C. 1.65
D. 1.55

Answer

To calculate the standard electrode potential \( E^\circ \) for the given reaction, we will use the Nernst equation. The reaction provided is:

\[ \text{Cr(s)} + 3 \text{Ag}^+(aq, 0.10M) \rightarrow 3 \text{Ag(s)} + \text{Cr}^{3+}(aq, 0.30M) \]

### Step-by-Step Solution:

1. **Identify the Reaction Components:**
- The oxidation half-reaction involves chromium:
\[ \text{Cr(s)} \rightarrow \text{Cr}^{3+}(aq) + 3e^- \]
- The reduction half-reaction involves silver:
\[ 3\text{Ag}^+(aq) + 3e^- \rightarrow 3\text{Ag(s)} \]

2. **Determine the Number of Electrons Transferred (n):**
- From the oxidation of chromium, 3 electrons are transferred. Thus, \( n = 3 \).

3. **Write the Nernst Equation:**
The Nernst equation is given by:
\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log Q \]
where \( Q \) is the reaction quotient.

4. **Calculate the Reaction Quotient (Q):**
The reaction quotient \( Q \) is given by:
\[ Q = \frac{[\text{Cr}^{3+}]}{[\text{Ag}^+]^3} \]
Substituting the given concentrations:
\[ Q = \frac{0.30}{(0.10)^3} = \frac{0.30}{0.001} = 300 \]

5. **Substitute Values into the Nernst Equation:**
Given that \( E_{cell} = 1.50 \, V \):
\[ 1.50 = E^\circ_{cell} - \frac{0.0591}{3} \log(300) \]

6. **Calculate \( \log(300) \):**
Using a calculator:
\[ \log(300) \approx 2.477 \]

7. **Substitute \( \log(300) \) into the Equation:**
\[ 1.50 = E^\circ_{cell} - \frac{0.0591}{3} \times 2.477 \]

8. **Calculate \( \frac{0.0591}{3} \times 2.477 \):**
\[ \frac{0.0591}{3} \approx 0.0197 \]
\[ 0.0197 \times 2.477 \approx 0.0487 \]

9. **Rearranging to Find \( E^\circ_{cell} \):**
\[ E^\circ_{cell} = 1.50 + 0.0487 \]
\[ E^\circ_{cell} \approx 1.5487 \, V \]

10. **Round the Result:**
Rounding to two decimal places, we get:
\[ E^\circ_{cell} \approx 1.55 \, V \]

### Final Answer:
The standard electrode potential \( E^\circ \) for the reaction is approximately **1.55 V**.