To calculate the cell voltage for the electrochemical cell formed by the half-cells \( X//X^{2+}(0.1M) \) and \( Y//Y^{2+}(1.0M) \), we will follow these steps:
### Step 1: Identify the half-reactions and their standard reduction potentials
We need to know the standard reduction potentials for the half-reactions involved. Let's assume:
- The standard reduction potential for the reaction \( X^{2+} + 2e^- \rightarrow X \) is \( E^\circ_{X} = 0.25 \, V \).
- The standard reduction potential for the reaction \( Y^{2+} + 2e^- \rightarrow Y \) is \( E^\circ_{Y} = 0.34 \, V \).
### Step 2: Determine the anode and cathode
In an electrochemical cell:
- The half-cell with the higher reduction potential acts as the cathode (where reduction occurs).
- The half-cell with the lower reduction potential acts as the anode (where oxidation occurs).
Here, since \( E^\circ_{Y} (0.34 \, V) > E^\circ_{X} (0.25 \, V) \):
- \( Y \) is the cathode.
- \( X \) is the anode.
### Step 3: Calculate the standard cell potential \( E^\circ_{cell} \)
The standard cell potential can be calculated using the formula:
\[
E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}
\]
Substituting the values:
\[
E^\circ_{cell} = E^\circ_{Y} - E^\circ_{X} = 0.34 \, V - 0.25 \, V = 0.09 \, V
\]
### Step 4: Apply the Nernst equation
The Nernst equation is given by:
\[
E_{cell} = E^\circ_{cell} - \frac{0.059}{n} \log \left( \frac{[X^{2+}]}{[Y^{2+}]} \right)
\]
Where:
- \( n \) is the number of moles of electrons transferred (in this case, \( n = 2 \)).
- \( [X^{2+}] = 0.1 \, M \)
- \( [Y^{2+}] = 1.0 \, M \)
Substituting the values into the Nernst equation:
\[
E_{cell} = 0.09 \, V - \frac{0.059}{2} \log \left( \frac{0.1}{1.0} \right)
\]
### Step 5: Calculate the logarithm
Calculating the logarithm:
\[
\log \left( \frac{0.1}{1.0} \right) = \log(0.1) = -1
\]
### Step 6: Substitute the logarithm back into the equation
Now substituting back:
\[
E_{cell} = 0.09 \, V - \frac{0.059}{2} \times (-1)
\]
\[
E_{cell} = 0.09 \, V + \frac{0.059}{2}
\]
\[
E_{cell} = 0.09 \, V + 0.0295 \, V = 0.1195 \, V
\]
### Final Result
The cell voltage \( E_{cell} \) is approximately \( 0.12 \, V \).
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