The standard reduction potential of `Cu^(2+)|Cu` and `Ag^(o+)|Ag` electrodes are `0.337` and `0.799V`, respectively. Construct a galvanic cell using these electrodes so that its standard `EMF` is positive. For what concentration of `Ag^(o+)` will the `EMF` of the cell , at `25^(@)C`, be zero if the concentration of `Cu^(2+)` is `0.01M`?

Explanation

To solve the problem step by step, we will follow the outlined procedure to construct a galvanic cell and find the concentration of \( \text{Ag}^+ \) ions when the EMF of the cell is zero.

### Step 1: Identify the electrodes and their standard reduction potentials
We have the following standard reduction potentials:
- \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \) with \( E^\circ = 0.337 \, \text{V} \)
- \( \text{Ag}^+ + e^- \rightarrow \text{Ag} \) with \( E^\circ = 0.799 \, \text{V} \)

### Step 2: Construct the galvanic cell
To ensure a positive EMF, we choose:
- Anode: Copper (\( \text{Cu} \))
- Cathode: Silver (\( \text{Ag} \))

### Step 3: Calculate the standard EMF of the cell
Using the formula for EMF:
\[ E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} \]
Substituting the values:
\[ E_{\text{cell}} = 0.799 \, \text{V} - 0.337 \, \text{V} = 0.462 \, \text{V} \]
Since \( E_{\text{cell}} \) is positive, the cell is viable.

### Step 4: Use the Nernst equation to find the concentration of \( \text{Ag}^+ \) when EMF is zero
The Nernst equation is given by:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln Q \]
At \( 25^\circ C \) (298 K), the equation simplifies to:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n} \log Q \]
Where:
- \( n = 2 \) (number of electrons transferred)
- \( Q = \frac{[\text{Cu}^{2+}]}{[\text{Ag}^+]^2} \)

Given that \( E_{\text{cell}} = 0 \) when we want to find the concentration of \( \text{Ag}^+ \):
\[ 0 = 0.462 - \frac{0.059}{2} \log \left( \frac{0.01}{[\text{Ag}^+]^2} \right) \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ \frac{0.059}{2} \log \left( \frac{0.01}{[\text{Ag}^+]^2} \right) = 0.462 \]
\[ \log \left( \frac{0.01}{[\text{Ag}^+]^2} \right) = \frac{0.462 \times 2}{0.059} \]
Calculating the right side:
\[ \log \left( \frac{0.01}{[\text{Ag}^+]^2} \right) = 15.685 \]

### Step 6: Solve for \( [\text{Ag}^+] \)
Taking the antilog:
\[ \frac{0.01}{[\text{Ag}^+]^2} = 10^{15.685} \]
\[ [\text{Ag}^+]^2 = \frac{0.01}{10^{15.685}} \]
Calculating:
\[ [\text{Ag}^+]^2 = 1.523 \times 10^{-17} \]
Taking the square root gives:
\[ [\text{Ag}^+] = 1.523 \times 10^{-9} \, \text{M} \]

### Final Answer
The concentration of \( \text{Ag}^+ \) ions when the EMF of the cell is zero, given that the concentration of \( \text{Cu}^{2+} \) is \( 0.01 \, \text{M} \), is:
\[ [\text{Ag}^+] = 1.523 \times 10^{-9} \, \text{M} \]