`6.537 xx 10^(-2)f ` of metallic `Zn` was added to `100 mL` of saturated solution of `AgCl` . Calculate `log .([Zn^(2+)])/([Ag^(o+)^2])`.
Given `: E^(c-)._(Ag^(o+)|Ag)=0.80V, E^(c-)._(Zn^(2+)|Zn)=-0.763V`.
`K_(sp)` of `AgCl~~10^(-10),` atomic weight of `Zn=65.37`
26.5
13.24
53
106

Options

A. `26.5`
B. `13.24`
C. 53
D. 106

Explanation

To solve the problem, we need to calculate the value of \( \log \left( \frac{[Zn^{2+}]}{[Ag^+]^2} \right) \) after adding zinc to a saturated solution of silver chloride (AgCl). Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the reactions
At the anode, zinc undergoes oxidation:
\[ Zn \rightarrow Zn^{2+} + 2e^- \]
At the cathode, silver ions undergo reduction:
\[ Ag^+ + e^- \rightarrow Ag \]
The overall balanced reaction is:
\[ Zn + 2Ag^+ \rightarrow Zn^{2+} + 2Ag \]

### Step 2: Calculate the standard cell potential (\(E^\circ_{cell}\))
The standard cell potential is calculated using the standard reduction potentials given:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]
Substituting the values:
\[ E^\circ_{cell} = 0.80V - (-0.763V) = 0.80V + 0.763V = 1.563V \]

### Step 3: Calculate the solubility of AgCl
The solubility product (\(K_{sp}\)) of AgCl is given as \(10^{-10}\). The solubility of AgCl can be calculated as follows:
\[ K_{sp} = [Ag^+][Cl^-] \]
Since AgCl is a 1:1 compound, we have:
\[ [Ag^+] = [Cl^-] = s \]
Thus,
\[ K_{sp} = s^2 \implies s = \sqrt{K_{sp}} = \sqrt{10^{-10}} = 10^{-5} \text{ M} \]
So, the concentration of \(Ag^+\) is \(10^{-5} \text{ M}\).

### Step 4: Calculate the molarity of \(Zn^{2+}\)
The amount of zinc added is \(6.537 \times 10^{-2} \text{ g}\) and the volume of the solution is \(100 \text{ mL}\). The molarity can be calculated using the formula:
\[ \text{Molarity} = \frac{\text{mass (g)} \times 1000}{\text{molar mass (g/mol)} \times \text{volume (mL)}} \]
Substituting the values:
\[ \text{Molarity of } Zn^{2+} = \frac{6.537 \times 10^{-2} \times 1000}{65.37 \times 100} = \frac{6.537}{65.37} = 0.01 \text{ M} \]

### Step 5: Apply the Nernst equation
Using the Nernst equation, we can relate the concentrations to the cell potential:
\[ E = E^\circ - \frac{0.059}{n} \log \left( \frac{[Zn^{2+}]}{[Ag^+]^2} \right) \]
Where \(n = 2\) (the number of electrons transferred). Rearranging gives:
\[ \log \left( \frac{[Zn^{2+}]}{[Ag^+]^2} \right) = \frac{n(E^\circ - E)}{0.059} \]
At equilibrium, the potential \(E\) can be considered as 0, leading to:
\[ \log \left( \frac{[Zn^{2+}]}{[Ag^+]^2} \right) = \frac{2 \times 1.563}{0.059} \approx 53 \]

### Final Answer
Thus, the value of \( \log \left( \frac{[Zn^{2+}]}{[Ag^+]^2} \right) \) is approximately **53**.

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