What must be concentration of `Ag^(+)` in an aqueous solution containing `Cu^(2+) =1.0 M` so that bot the metals can be
deposited on the cathode simultaneously. Given that
`E_(Cu//Cu^(2+))^(0)=0.34V` and `E_(Ag^(+)//Ag)^(0)=0.812V,T=298K`

Options

A. nearly `10^(-19)M`
B. `10^(-12)M`
C. `10^(-8)M`
D. nearly `10^(-16)M`

Explanation

To find the concentration of \( \text{Ag}^+ \) in an aqueous solution containing \( \text{Cu}^{2+} = 1.0 \, \text{M} \) so that both metals can be deposited on the cathode simultaneously, we can follow these steps:

### Step 1: Write the half-reactions
The half-reactions for the reduction of copper and silver are:
- For copper:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \quad (E^\circ = 0.34 \, \text{V}) \]
- For silver:
\[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \quad (E^\circ = 0.812 \, \text{V}) \]

### Step 2: Set up the Nernst equation
The Nernst equation relates the standard electrode potential to the concentrations of the ions:
\[ E = E^\circ - \frac{0.059}{n} \log \frac{1}{[\text{oxidized form}]} \]
For copper:
\[ E_{\text{Cu}} = 0.34 - \frac{0.059}{2} \log \frac{1}{[\text{Cu}^{2+}]} \]
For silver:
\[ E_{\text{Ag}} = 0.812 - 0.059 \log \frac{1}{[\text{Ag}^+]} \]

### Step 3: Set the cell potentials equal
Since both metals are deposited simultaneously, we can set the potentials equal:
\[ E_{\text{Cu}} = E_{\text{Ag}} \]
Substituting the Nernst equations:
\[ 0.34 - \frac{0.059}{2} \log \frac{1}{1.0} = 0.812 - 0.059 \log \frac{1}{[\text{Ag}^+]} \]

### Step 4: Simplify the equation
Since \( \log(1) = 0 \):
\[ 0.34 - 0 = 0.812 - 0.059 \log \frac{1}{[\text{Ag}^+]} \]
This simplifies to:
\[ 0.34 = 0.812 - 0.059 \log \frac{1}{[\text{Ag}^+]} \]

### Step 5: Rearrange to find \( \log [\text{Ag}^+] \)
Rearranging gives:
\[ 0.059 \log \frac{1}{[\text{Ag}^+]} = 0.812 - 0.34 \]
\[ 0.059 \log \frac{1}{[\text{Ag}^+]} = 0.472 \]
Dividing both sides by 0.059:
\[ \log \frac{1}{[\text{Ag}^+]} = \frac{0.472}{0.059} \approx 8 \]

### Step 6: Solve for \( [\text{Ag}^+] \)
Taking the antilogarithm:
\[ \frac{1}{[\text{Ag}^+]} = 10^8 \]
Thus:
\[ [\text{Ag}^+] = 10^{-8} \, \text{M} \]

### Final Answer
The concentration of \( \text{Ag}^+ \) must be \( 10^{-8} \, \text{M} \).

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