`Cu^(2+)+2e^(-)toCu,E^(@)=+0.34V,Ag^(+)+1e^(-)toAg,E^(@)=+0.80V`
(i) Construct a galvanic cell using the above data.
(ii) For what concentration of `Ag^(+)` ions will the emf of the cell be zero at `25^(@)C`, if the concentration of `Cu^(2+)` is 0.01 M? (log 3.919=0.593).

Explanation

To solve the given problem step-by-step, we will follow the instructions provided in the question.

### Step 1: Identify the Anode and Cathode
The standard reduction potentials are given as:
- \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}, \quad E^\circ = +0.34 \, \text{V} \)
- \( \text{Ag}^+ + e^- \rightarrow \text{Ag}, \quad E^\circ = +0.80 \, \text{V} \)

The half-reaction with the higher reduction potential will act as the cathode, while the one with the lower reduction potential will act as the anode. Here, since \( E^\circ_{\text{Ag}^+} > E^\circ_{\text{Cu}^{2+}} \), we have:
- **Anode**: \( \text{Cu} \) (oxidation)
- **Cathode**: \( \text{Ag}^+ \) (reduction)

### Step 2: Write the Cell Reaction
The oxidation reaction at the anode is:
\[ \text{Cu} \rightarrow \text{Cu}^{2+} + 2e^- \]

The reduction reaction at the cathode is:
\[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \]

To balance the electrons, we multiply the silver reaction by 2:
\[ 2\text{Ag}^+ + 2e^- \rightarrow 2\text{Ag} \]

Now, we can combine the two half-reactions:
\[ \text{Cu} + 2\text{Ag}^+ \rightarrow \text{Cu}^{2+} + 2\text{Ag} \]

### Step 3: Calculate the Standard Cell Potential
The standard cell potential \( E^\circ_{\text{cell}} \) is calculated as:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
\[ E^\circ_{\text{cell}} = 0.80 \, \text{V} - 0.34 \, \text{V} = 0.46 \, \text{V} \]

### Step 4: Use the Nernst Equation
The Nernst equation is given by:
\[ E = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log \left( \frac{[\text{Cu}^{2+}]}{[\text{Ag}^+]^2} \right) \]

Where:
- \( n = 2 \) (number of electrons transferred)
- \( [\text{Cu}^{2+}] = 0.01 \, \text{M} \)

We want to find the concentration of \( \text{Ag}^+ \) when \( E = 0 \):
\[ 0 = 0.46 - \frac{0.0591}{2} \log \left( \frac{0.01}{[\text{Ag}^+]^2} \right) \]

### Step 5: Rearranging the Equation
Rearranging gives:
\[ \frac{0.0591}{2} \log \left( \frac{0.01}{[\text{Ag}^+]^2} \right) = 0.46 \]

### Step 6: Solve for \([\text{Ag}^+]\)
Multiplying both sides by \( \frac{2}{0.0591} \):
\[ \log \left( \frac{0.01}{[\text{Ag}^+]^2} \right) = \frac{0.46 \times 2}{0.0591} \]

Calculating the right side:
\[ \log \left( \frac{0.01}{[\text{Ag}^+]^2} \right) = 15.59 \]

Taking the antilogarithm:
\[ \frac{0.01}{[\text{Ag}^+]^2} = 10^{15.59} \]
\[ [\text{Ag}^+]^2 = \frac{0.01}{10^{15.59}} \]
\[ [\text{Ag}^+]^2 = 5.0 \times 10^{-17} \]
\[ [\text{Ag}^+] = 5.0 \times 10^{-9} \, \text{M} \]

### Final Answer
The concentration of \( \text{Ag}^+ \) ions for the emf of the cell to be zero at \( 25^\circ C \) is:
\[ [\text{Ag}^+] = 5.0 \times 10^{-9} \, \text{M} \]