Calclate the solubility product of AgI at 25.0 °C, given the following data:

<blockquote>
<table border cellpadding=5>
<tr>
<td>Reduction half-reaction
<td align=center>E&#176; (V)

<tr>
<td>AgI(s) + e&#175; ---> Ag(s) + I&#175;
<td align=center>-0.15

<tr>
<td>I<sub>2</sub>(s) + 2e&#175; ---> 2I&#175;
<td align=center>-0.54

<tr>
<td>Ag<sup>+</sup> + e&#175; ---> Ag(s)
<td align=center>0.80
</table>
</blockquote>

<P>Solution:

<P>1) K<sub>sp</sub> equation for AgI is:

<blockquote>AgI(s) &#8660; Ag<sup>+</sup>(aq) + I&#175;(aq)</blockquote>

<P>2) The equations to use are:

<blockquote>
<table border cellpadding=5>
<tr>
<td>
<td align=center>E&#176; (V)

<tr>
<td>AgI(s) + e&#175; ---> Ag(s) + I&#175;
<td align=center>-0.15

<tr>
<td>Ag(s) ---> Ag<sup>+</sup> + e&#175;
<td align=center>-0.80
</table>

<P>Yielding E&#176; = -0.95 V
</blockquote>

<P>3) use the Nernst Equation:

<blockquote>
E<sub>cell</sub> = E&#176; - (0.0591 / n) log K

<P>0 = -0.95 - (0.0591 / 1) log K

<P>0.95 / -0.0591 = log K

<P>log K = -16.07

<P>K = 8.51 x 10&#175;<sup>17</sup>
</blockquote>

<P>Note: some instructors might insist that you round the answer off to two significant figures. On <A HREF="http://www.ktf-split.hr/periodni/en/abc/kpt.html">this site,</A> the K<sub>sp</sub> is listed as 8.52 x 10&#175;<sup>17</sup>.