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 determine the concentration of `Cd^(2+)` at equilibrium after it reacts with excess iron. Here is a step-by-step solution:

### Step 1: Write the balanced chemical equation
The reaction between cadmium ions and iron can be represented as:
\[ \text{Cd}^{2+}(aq) + \text{Fe}(s) \rightarrow \text{Cd}(s) + \text{Fe}^{2+}(aq) \]

### Step 2: Define initial concentrations
Initially, we have:
- Concentration of `Cd^(2+)` = 1.0 M
- Concentration of `Fe` = Excess (we assume it does not limit the reaction)

### Step 3: Define changes at equilibrium
Let \( x \) be the amount of `Cd^(2+)` that reacts. At equilibrium:
- Concentration of `Cd^(2+)` = \( 1.0 - x \)
- Concentration of `Fe^{2+}` produced = \( x \)

### Step 4: Write the expression for the equilibrium constant \( K \)
The equilibrium constant \( K \) for the reaction can be expressed as:
\[ K = \frac{[\text{Fe}^{2+}]}{[\text{Cd}^{2+}]} = \frac{x}{1.0 - x} \]

### Step 5: Use the Nernst equation to find \( K \)
The standard cell potential \( E^0 \) is given as 0.037 V. At equilibrium, the cell potential \( E \) is 0 V. Using the Nernst equation:
\[ E = E^0 - \frac{0.0591}{n} \log K \]
At equilibrium, \( E = 0 \):
\[ 0 = 0.037 - \frac{0.0591}{n} \log K \]

### Step 6: Determine \( n \)
In this reaction, 2 electrons are transferred (1 for `Cd^(2+)` to `Cd` and 1 for `Fe` to `Fe^{2+}`), so \( n = 2 \).

### Step 7: Solve for \( K \)
Rearranging the equation gives:
\[ \log K = \frac{0.037 \cdot n}{0.0591} \]
Substituting \( n = 2 \):
\[ \log K = \frac{0.037 \cdot 2}{0.0591} \approx 1.25 \]
Now, we find \( K \) by taking the antilog:
\[ K \approx 10^{1.25} \approx 17.78 \]

### Step 8: Substitute \( K \) back into the equilibrium expression
Now we have:
\[ 17.78 = \frac{x}{1.0 - x} \]

### Step 9: Solve for \( x \)
Cross-multiplying gives:
\[ 17.78(1.0 - x) = x \]
\[ 17.78 - 17.78x = x \]
\[ 17.78 = 18.78x \]
\[ x = \frac{17.78}{18.78} \approx 0.947 \]

### Step 10: Find the concentration of `Cd^(2+)` at equilibrium
Now, substituting \( x \) back to find the concentration of `Cd^(2+)`:
\[ [\text{Cd}^{2+}] = 1.0 - x = 1.0 - 0.947 = 0.053 \, \text{M} \]

### Final Answer
The concentration of `Cd^(2+)` at equilibrium is **0.053 M**.

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