Calculate `E_(cell)` of the reaction
`{:(Mg(s),+2Ag^(+)to,Mg^(+2)+,2Ag(s),),(,(0.0001M),(0.100M),,):}`
`If E_(cell)^(0)=3.17V`

Options

A. `-2.96 V`
B. `+2.96 V`
C. 3.38 V
D. `-3.38 V`

Answer

To calculate the cell potential \( E_{\text{cell}} \) for the given reaction:

\[ \text{Mg(s)} + 2\text{Ag}^+ \rightarrow \text{Mg}^{2+} + 2\text{Ag(s)} \]

with the standard cell potential \( E_{\text{cell}}^{\circ} = 3.17 \, \text{V} \), we will use the Nernst equation:

\[ E_{\text{cell}} = E_{\text{cell}}^{\circ} - \frac{0.0591}{n} \log Q \]

where:
- \( n \) is the number of moles of electrons transferred in the reaction.
- \( Q \) is the reaction quotient.

### Step 1: Identify the values for \( n \) and \( Q \)

In this reaction:
- Magnesium (Mg) loses 2 electrons to become \( \text{Mg}^{2+} \), so \( n = 2 \).
- The reaction quotient \( Q \) is given by the formula:

\[ Q = \frac{[\text{Mg}^{2+}]}{[\text{Ag}^+]^2} \]

Given:
- \( [\text{Mg}^{2+}] = 0.1 \, \text{M} \)
- \( [\text{Ag}^+] = 0.0001 \, \text{M} \)

Thus, we can calculate \( Q \):

\[ Q = \frac{0.1}{(0.0001)^2} = \frac{0.1}{0.00000001} = 10^8 \]

### Step 2: Substitute values into the Nernst equation

Now substituting the values into the Nernst equation:

\[ E_{\text{cell}} = 3.17 \, \text{V} - \frac{0.0591}{2} \log(10^8) \]

### Step 3: Calculate \( \log(10^8) \)

We know that:

\[ \log(10^8) = 8 \]

### Step 4: Substitute and simplify

Now substituting \( \log(10^8) \) back into the equation:

\[ E_{\text{cell}} = 3.17 \, \text{V} - \frac{0.0591}{2} \cdot 8 \]

Calculating \( \frac{0.0591}{2} \cdot 8 \):

\[ \frac{0.0591}{2} = 0.02955 \]
\[ 0.02955 \cdot 8 = 0.2364 \]

### Step 5: Final calculation

Now substituting this back into the equation:

\[ E_{\text{cell}} = 3.17 \, \text{V} - 0.2364 = 2.9336 \, \text{V} \]

### Step 6: Round to appropriate significant figures

Rounding to four significant figures, we get:

\[ E_{\text{cell}} \approx 2.934 \, \text{V} \]

### Final Answer

The cell potential \( E_{\text{cell}} \) is approximately \( 2.934 \, \text{V} \).

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