Calculate the emf of the cell in mV (atleast first two digits must match with correct answer)
`Ag(s),AgIO_(3)(s)|Ag^(+)(xM),HIO_(3)(1M)||Zn^(+2)(1M)|Zn(s)`
if `K_(sp)=3xx10^(-8)` for `AgIO_(3)` and `K_(a)=(1)/(6)` for `HIO_(3)` and `E_(cell)^(0)` for `2Ag+Zn^(+2)to2Ag^(+)+Zn` is `-1.56V`.
`log3=0.48)Take(PT)/(F)=0.059`) (giving your answer in magnitude only)

Explanation

To calculate the EMF of the cell given the reaction and conditions, we will follow these steps:

### Step 1: Determine the concentration of \( \text{IO}_3^- \)
Given that \( K_a = \frac{1}{6} \) for \( \text{HIO}_3 \), we can express the dissociation of \( \text{HIO}_3 \) as follows:

\[ K_a = \frac{[\text{H}^+][\text{IO}_3^-]}{[\text{HIO}_3]} \]

Assuming the initial concentration of \( \text{HIO}_3 \) is 1 M and letting \( \alpha \) be the degree of dissociation, we have:

\[ K_a = \frac{(1)(\alpha)}{(1 - \alpha)} = \frac{1}{6} \]

This leads to the equation:

\[ \frac{\alpha}{1 - \alpha} = \frac{1}{6} \]

Cross-multiplying gives:

\[ 6\alpha = 1 - \alpha \implies 7\alpha = 1 \implies \alpha = \frac{1}{7} \]

The concentration of \( \text{IO}_3^- \) is:

\[ [\text{IO}_3^-] = 1 \cdot \alpha = \frac{1}{7} \text{ M} \]

### Step 2: Calculate the concentration of \( \text{Ag}^+ \)
Using the solubility product \( K_{sp} \) for \( \text{AgIO}_3 \):

\[ K_{sp} = [\text{Ag}^+][\text{IO}_3^-] = 3 \times 10^{-8} \]

Substituting \( [\text{IO}_3^-] = \frac{1}{7} \):

\[ 3 \times 10^{-8} = [\text{Ag}^+] \cdot \frac{1}{7} \]

Solving for \( [\text{Ag}^+] \):

\[ [\text{Ag}^+] = 3 \times 10^{-8} \cdot 7 = 2.1 \times 10^{-7} \text{ M} \]

### Step 3: Calculate the EMF of the cell
The overall cell reaction is:

\[ 2 \text{Ag}^+ + \text{Zn} \rightarrow 2 \text{Ag} + \text{Zn}^{2+} \]

The standard EMF \( E^\circ_{cell} \) is given as -1.56 V. We can use the Nernst equation to find the EMF under non-standard conditions:

\[ E = E^\circ_{cell} + \frac{0.059}{n} \log \left( \frac{[\text{Ag}^+]^2}{[\text{Zn}^{2+}]} \right) \]

Here, \( n = 2 \) (number of electrons transferred), and \( [\text{Zn}^{2+}] = 1 \text{ M} \):

\[ E = -1.56 + \frac{0.059}{2} \log \left( \frac{(2.1 \times 10^{-7})^2}{1} \right) \]

Calculating the logarithm:

\[ \log \left( (2.1 \times 10^{-7})^2 \right) = 2 \log(2.1 \times 10^{-7}) = 2 \left( \log(2.1) + \log(10^{-7}) \right) \]

Using \( \log(2.1) \approx 0.32 \):

\[ \log(2.1 \times 10^{-7}) \approx 0.32 - 7 = -6.68 \]

Thus:

\[ E = -1.56 + \frac{0.059}{2} \cdot (-6.68) = -1.56 - 0.059 \cdot 3.34 \]

Calculating:

\[ E = -1.56 - 0.197 = -1.757 \text{ V} \]

Converting to mV:

\[ E = -1757 \text{ mV} \]

### Final Answer
The EMF of the cell is approximately **1757 mV** (in magnitude).

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