Fin the reduction potential of AsO4^3- /AsO2^- in a solution when 18 mL of 0.1N solution of NaI is added to 20mL Na3AsO4 solution at pH =5 .The stanerd reduction potential of AsO4^-3/AsO2^-=-0.70.

To find the reduction potential of the arsenate/arsenite (\( AsO_4^{3-} / AsO_2^- \)) couple under the given conditions, we use the **Nernst equation**:

\[
E = E^\circ - \frac{0.0591}{n} \log \frac{[ \text{oxidized form} ]}{[ \text{reduced form} ]}
\]

### Given Data:
- **Standard reduction potential:** \( E^\circ = -0.70V \)
- **Added NaI solution:** \( 18 mL \) of \( 0.1N \)
- **Volume of \( Na_3AsO_4 \) solution:** \( 20mL \)
- **pH:** \( 5 \)
- **Reaction:**
\[
AsO_4^{3-} + 4H^+ + 2e^- \rightarrow AsO_2^- + 2H_2O
\]
Here, \( n = 2 \) (number of electrons transferred).

### Step 1: Calculate the Equivalent of \( I^- \) Added
Since **normality (N) = equivalent/L**, we find the equivalents of \( I^- \):

\[
\text{Equivalents of } I^- = \text{Volume (L)} \times \text{Normality}
\]

\[
= \left(\frac{18}{1000}\right) \times 0.1
\]

\[
= 1.8 \times 10^{-3} \text{ equivalents}
\]

Since \( I^- \) gets oxidized to \( I_2 \), it acts as a reducing agent, meaning it reduces arsenate \( AsO_4^{3-} \) to arsenite \( AsO_2^- \).

### Step 2: Find the Ratio \(\frac{[AsO_4^{3-}]}{[AsO_2^-]}\)

- Initial moles of \( AsO_4^{3-} \) (assuming a **1N** arsenate solution means 1 eq/L):

Since no concentration for \( Na_3AsO_4 \) is given, assume the equivalents of \( AsO_4^{3-} \) initially present in 20mL are **x**.

After the reaction, a portion of \( AsO_4^{3-} \) is converted into \( AsO_2^- \) due to the reduction by iodide.

The moles of \( AsO_4^{3-} \) reduced are equal to the equivalents of \( I^- \) added, since **1 mole of \( I^- \) reduces 1 mole of \( AsO_4^{3-} \)**.

\[
[ AsO_4^{3-} ]_{\text{final}} = [ AsO_4^{3-} ]_{\text{initial}} - 1.8 \times 10^{-3}
\]

\[
[ AsO_2^- ] = 1.8 \times 10^{-3}
\]

Thus, the ratio:

\[
\frac{[ AsO_4^{3-} ]}{[ AsO_2^- ]} = \frac{[ AsO_4^{3-} ]_{\text{initial}} - 1.8 \times 10^{-3}}{1.8 \times 10^{-3}}
\]

If we assume that the initial concentration of \( AsO_4^{3-} \) was much higher than the amount reduced (say **y** equivalents), we approximate:

\[
\frac{[ AsO_4^{3-} ]}{[ AsO_2^- ]} \approx \frac{y}{1.8 \times 10^{-3}}
\]

### Step 3: Apply Nernst Equation

\[
E = -0.70 - \frac{0.0591}{2} \log \frac{[ AsO_4^{3-} ]}{[ AsO_2^- ]}
\]

Substituting values:

\[
E = -0.70 - 0.02955 \log \left(\frac{y}{1.8 \times 10^{-3}}\right)
\]

If the initial arsenate concentration was significantly large compared to \( 1.8 \times 10^{-3} \), we can approximate the logarithm term accordingly. However, without explicit data on \( y \), we can only express the reduction potential in terms of the unknown **initial concentration of \( AsO_4^{3-} \)**.

Would you like to specify an assumed concentration for \( AsO_4^{3-} \) to get a numerical answer?