A 1.0 M solution of `Cd^(2+)` is added to excess iron and the system is allowed to reach equillibrium. What is the concentration of `Cd^(2+)` ?
`Cd^(2+)(aq)+Fe(s)toCd(s)+Fe^(2+)(aq), E^(@)=0.037`

Options

A. 0.195
B. 0.097
C. 0.053
D. 0.145

Answer

To solve the problem, we need to analyze the electrochemical reaction and determine the concentration of \( \text{Cd}^{2+} \) at equilibrium.

### Step 1: Write the half-reaction and identify the standard reduction potential.
The half-reaction given is:
\[ \text{Cd}^{2+}(aq) + 2e^- \rightarrow \text{Cd}(s) \]
The standard reduction potential \( E^\circ \) for this reaction is given as \( 0.037 \, \text{V} \).

### Step 2: Write the oxidation half-reaction for iron.
The oxidation half-reaction for iron is:
\[ \text{Fe}(s) \rightarrow \text{Fe}^{2+}(aq) + 2e^- \]
The standard reduction potential for the reduction of \( \text{Fe}^{2+} \) to \( \text{Fe} \) is about \( -0.44 \, \text{V} \).

### Step 3: Determine the cell potential.
To find the cell potential \( E^\circ_{\text{cell}} \), we use the formula:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
Here, \( E^\circ_{\text{cathode}} \) is for the reduction of \( \text{Cd}^{2+} \) and \( E^\circ_{\text{anode}} \) is for the oxidation of \( \text{Fe} \).

Substituting the values:
\[ E^\circ_{\text{cell}} = 0.037 \, \text{V} - (-0.44 \, \text{V}) = 0.037 \, \text{V} + 0.44 \, \text{V} = 0.477 \, \text{V} \]

### Step 4: Use the Nernst equation to find the concentration of \( \text{Cd}^{2+} \) at equilibrium.
The Nernst equation is given by:
\[ E = E^\circ - \frac{RT}{nF} \ln Q \]
At equilibrium, \( E = 0 \). Therefore, we can rearrange the equation to solve for \( Q \):
\[ 0 = 0.477 - \frac{RT}{nF} \ln Q \]
Rearranging gives:
\[ \frac{RT}{nF} \ln Q = 0.477 \]
Where:
- \( R = 8.314 \, \text{J/(mol K)} \)
- \( T = 298 \, \text{K} \)
- \( n = 2 \) (number of electrons transferred)
- \( F = 96485 \, \text{C/mol} \)

### Step 5: Calculate \( Q \) and find the concentration of \( \text{Cd}^{2+} \).
Substituting the values into the equation:
\[ \frac{(8.314)(298)}{(2)(96485)} \ln Q = 0.477 \]
Calculating the left side:
\[ \frac{(8.314)(298)}{(2)(96485)} \approx 0.0041 \]
Now substituting back:
\[ 0.0041 \ln Q = 0.477 \]
\[ \ln Q = \frac{0.477}{0.0041} \approx 116.5 \]
Taking the exponential:
\[ Q \approx e^{116.5} \]
Since \( Q = \frac{[\text{Fe}^{2+}]}{[\text{Cd}^{2+}]} \) and assuming that the concentration of \( \text{Fe}^{2+} \) is negligible compared to the initial concentration of \( \text{Cd}^{2+} \):
\[ [\text{Cd}^{2+}] \approx \frac{1.0}{e^{116.5}} \]
This concentration is extremely low, effectively approaching zero.

### Conclusion:
The concentration of \( \text{Cd}^{2+} \) at equilibrium is approximately \( 0 \, \text{M} \).