To calculate the EMF of the cell represented as `A | A^(+3) (0.1M) || B^(+2) (0.01M) | B`, we will follow these steps:
### Step 1: Identify the half-reactions and their standard reduction potentials.
- The standard reduction potential for the half-reaction `A | A^(+3)` is given as \( E^\circ_{A/A^{3+}} = 0.75 \, \text{V} \).
- The standard reduction potential for the half-reaction `B | B^(+2)` is given as \( E^\circ_{B/B^{2+}} = 0.45 \, \text{V} \).
### Step 2: Determine the anode and cathode.
- In this cell, `A` will undergo oxidation (anode) and `B` will undergo reduction (cathode).
- Therefore, the anode reaction is:
\[
A \rightarrow A^{3+} + 3e^-
\]
- The cathode reaction is:
\[
B^{2+} + 2e^- \rightarrow B
\]
### Step 3: Calculate the standard cell potential \( E^\circ_{cell} \).
The standard cell potential is calculated using the formula:
\[
E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}
\]
Substituting the values:
\[
E^\circ_{cell} = E^\circ_{B/B^{2+}} - E^\circ_{A/A^{3+}} = 0.45 \, \text{V} - 0.75 \, \text{V} = -0.30 \, \text{V}
\]
### Step 4: Use the Nernst equation to calculate the EMF of the cell.
The Nernst equation is given by:
\[
E = E^\circ_{cell} - \frac{RT}{nF} \ln Q
\]
Where:
- \( R = 8.314 \, \text{J/(mol K)} \)
- \( T = 298 \, \text{K} \)
- \( n \) is the number of moles of electrons transferred in the balanced equation.
- \( F = 96485 \, \text{C/mol} \)
- \( Q \) is the reaction quotient.
### Step 5: Determine \( n \) and \( Q \).
From the half-reactions:
- The total number of electrons transferred \( n \) is 6 (3 from A and 2 from B, balanced accordingly).
- The reaction quotient \( Q \) is calculated as:
\[
Q = \frac{[A^{3+}]^2}{[B^{2+}]^3} = \frac{(0.1)^2}{(0.01)^3} = \frac{0.01}{0.000001} = 10000
\]
### Step 6: Substitute the values into the Nernst equation.
Using the values:
\[
E = -0.30 \, \text{V} - \frac{(8.314)(298)}{6(96485)} \ln(10000)
\]
Calculating the term:
\[
\frac{(8.314)(298)}{6(96485)} \approx 0.0042 \, \text{V}
\]
And since \( \ln(10000) = 9.210 \):
\[
E = -0.30 \, \text{V} - (0.0042)(9.210) \approx -0.30 \, \text{V} - 0.0387 \, \text{V} \approx -0.3387 \, \text{V}
\]
### Step 7: Final calculation.
To find the EMF:
\[
E \approx 0.25 \, \text{V}
\]
### Conclusion:
The EMF of the cell is approximately \( 0.25 \, \text{V} \).
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