Calculate the emf of the cell
`Fe(s)+2H^(+)(1M)toFe^(+2)(0.001M)+H_(2)//(g),1 atm)`
`(Given : E_(Fe^(+2)//Fe)=-0.44V)`

Options

A. 0.215 V
B. 0.38 V
C. `-0.48 V`
D. 0.5285 V

Answer

To calculate the electromotive force (emf) of the cell given by the reaction:

\[ \text{Fe(s)} + 2\text{H}^+(1M) \rightarrow \text{Fe}^{2+}(0.001M) + \text{H}_2(g), 1 \text{ atm} \]

we will use the Nernst equation:

\[ E = E^\circ - \frac{0.059}{n} \log \left( \frac{[\text{Ox}]}{[\text{Red}]^2} \right) \]

### Step 1: Identify the standard electrode potentials
The standard electrode potential for the half-reaction involving \( \text{Fe}^{2+}/\text{Fe} \) is given as:

\[ E^\circ_{\text{Fe}^{2+}/\text{Fe}} = -0.44 \text{ V} \]

The standard electrode potential for the reduction of hydrogen ions is:

\[ E^\circ_{\text{H}^+/H_2} = 0 \text{ V} \]

### Step 2: Determine the overall standard cell potential
The overall standard cell potential \( E^\circ_{\text{cell}} \) can be calculated using:

\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]

In this case, the cathode is where the reduction occurs (hydrogen) and the anode is where the oxidation occurs (iron):

\[ E^\circ_{\text{cell}} = E^\circ_{\text{H}^+/H_2} - E^\circ_{\text{Fe}^{2+}/Fe} = 0 - (-0.44) = 0.44 \text{ V} \]

### Step 3: Identify the number of electrons transferred
In this reaction, 2 electrons are transferred (from Fe to H\(^+\)):

\[ n = 2 \]

### Step 4: Substitute concentrations into the Nernst equation
We need to substitute the concentrations into the Nernst equation. The concentration of \( \text{Fe}^{2+} \) is \( 0.001 \, \text{M} \) and the concentration of \( \text{H}^+ \) is \( 1 \, \text{M} \). The pressure of \( \text{H}_2 \) is \( 1 \, \text{atm} \).

The Nernst equation becomes:

\[ E = E^\circ_{\text{cell}} - \frac{0.059}{2} \log \left( \frac{[\text{Fe}^{2+}]}{[\text{H}^+]^2} \right) \]

Substituting the values:

\[ E = 0.44 - \frac{0.059}{2} \log \left( \frac{0.001}{(1)^2} \right) \]

### Step 5: Calculate the logarithm
Calculating the logarithm:

\[ \log(0.001) = -3 \]

### Step 6: Substitute the logarithm back into the equation
Now we substitute back into the Nernst equation:

\[ E = 0.44 - \frac{0.059}{2} \times (-3) \]

### Step 7: Simplify the equation
Calculating the term:

\[ E = 0.44 + \frac{0.059 \times 3}{2} = 0.44 + 0.0885 \]

### Step 8: Final calculation
Now, adding these values together:

\[ E = 0.44 + 0.0885 = 0.5285 \text{ V} \]

### Final Answer
The emf of the cell is approximately:

\[ \text{EMF} = 0.5285 \text{ V} \]

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