Calculate the voltage `E` of the cell
`Ag(s)| AgIO_(3)(s)|Ag^(+) (x,M), HIO_(3) (0.300M)||Zn^(2+) (0.175M)|Zn(s)`
If `K_(SP) = 3.02 xx 10^(-8)` for `AgIO_(3)(s)` and `K_(a) = 0.162` for `HIO_(3), E^(@) (Zn^(+2)//Zn) =- 0.76 V, E^(@) (Ag//Ag^(+)) =- 0.8V`

Options

A. `-2.188V`
B. `-0.188V`
C. `-1.188V`
D. `-3.188V`

Explanation

To calculate the voltage \( E \) of the cell represented as

\[ \text{Ag(s)| AgIO}_3\text{(s)|Ag}^+(x,M), \text{HIO}_3(0.300M)||\text{Zn}^{2+}(0.175M)|\text{Zn(s)} \]

we will follow these steps:

### Step 1: Calculate the concentration of \( \text{H}^+ \) and \( \text{IO}_3^- \) from \( \text{HIO}_3 \)

The dissociation of \( \text{HIO}_3 \) can be represented as:
\[ \text{HIO}_3 \rightleftharpoons \text{H}^+ + \text{IO}_3^- \]

Given \( K_a = 0.162 \) and the initial concentration of \( \text{HIO}_3 \) is \( 0.300 \, M \), we set up the equilibrium expression:
\[ K_a = \frac{[\text{H}^+][\text{IO}_3^-]}{[\text{HIO}_3]} \]

Let \( x \) be the concentration of \( \text{H}^+ \) and \( \text{IO}_3^- \) at equilibrium. Then:
\[ K_a = \frac{x^2}{0.300 - x} \]

Substituting the value of \( K_a \):
\[ 0.162 = \frac{x^2}{0.300 - x} \]

This leads to the quadratic equation:
\[ x^2 + 0.162x - 0.0486 = 0 \]

Solving this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{-0.162 \pm \sqrt{(0.162)^2 + 4 \times 1 \times 0.0486}}{2 \times 1} \]

Calculating the discriminant and solving gives:
\[ x \approx 0.1538 \, M \]

Thus, the concentrations are:
\[ [\text{H}^+] = 0.1538 \, M, \quad [\text{IO}_3^-] = 0.1538 \, M \]

### Step 2: Calculate the concentration of \( \text{Ag}^+ \) using \( K_{sp} \)

The dissociation of \( \text{AgIO}_3 \) is:
\[ \text{AgIO}_3 \rightleftharpoons \text{Ag}^+ + \text{IO}_3^- \]

The solubility product \( K_{sp} \) is given by:
\[ K_{sp} = [\text{Ag}^+][\text{IO}_3^-] \]

Substituting the known values:
\[ 3.02 \times 10^{-8} = [\text{Ag}^+](0.1538) \]

Solving for \( [\text{Ag}^+] \):
\[ [\text{Ag}^+] = \frac{3.02 \times 10^{-8}}{0.1538} \approx 1.96 \times 10^{-7} \, M \]

### Step 3: Write the cell reaction and calculate \( E^\circ \)

The cell reaction involves the oxidation of silver and the reduction of zinc:
\[ 2\text{Ag} \rightarrow 2\text{Ag}^+ + 2e^- \quad (\text{oxidation}) \]
\[ \text{Zn}^{2+} + 2e^- \rightarrow \text{Zn} \quad (\text{reduction}) \]

The standard cell potential \( E^\circ \) is calculated as:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = (-0.76) - (-0.8) = -0.76 + 0.8 = 0.04 \, V \]

### Step 4: Apply the Nernst equation

The Nernst equation is given by:
\[ E = E^\circ - \frac{0.0591}{n} \log \left( \frac{[\text{Ag}^+]^2}{[\text{Zn}^{2+}]}\right) \]

Where \( n = 2 \) (number of electrons transferred). Substituting the values:
\[ E = 0.04 - \frac{0.0591}{2} \log \left( \frac{(1.96 \times 10^{-7})^2}{0.175} \right) \]

Calculating the logarithm:
\[ \log \left( \frac{(1.96 \times 10^{-7})^2}{0.175} \right) = \log \left( \frac{3.84 \times 10^{-14}}{0.175} \right) \approx \log(2.194 \times 10^{-13}) \approx -12.26 \]

Thus:
\[ E = 0.04 - \frac{0.0591}{2} \times (-12.26) \approx 0.04 + 0.362 \]
\[ E \approx 0.402 \, V \]

### Final Answer
The voltage \( E \) of the cell is approximately \( 0.402 \, V \).

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