Question

Calculate E° for the following half-reaction:
$AgI\left(s\right)+e^{-}\to Ag\left(s\right)+I^{-}$
(Hint: Reference the K_sp value for Agl and the standard
reduction potential for Ag⁺.)

Short answer

The half reaction’s cell potential is -0.14V.

Step-by-step solution

Equation of Standard reduction potential and solubility product constant

The standard reduction potential equation of Ag⁺ can be denoted as:

$A{g}^{+}+e^{-}\to Ag\left(s\right) E^o=0.8 V--\left(1\right)$

The solubility product constant can be denoted as:

$AgI\left(s\right)\to A{g}^{+}\left(aq\right)+I^{-}\left(aq\right) K_{sp}=1.5\times {10}^{16}---\left(2\right)$

Calculation of standard reduction potential

It can be calculated as:

$$\begin{gathered} E^o=\frac{0.0591}{n}\log K_{sp} \\ =\frac{0.0591}{1}\log (1.5\times {10}^{-16}) \\ =-0.94 V \end{gathered}$$

Calculation of cell potential

The cell potential can be calculated as:

$$\begin{gathered} A{g}^{+}+e^{-}\to Ag\left(s\right) E^o=0.8 V \\ AgI\left(s\right)\to A{g}^{+}\left(aq\right)+I^{-}\left(aq\right) E^o=-0.94 V \\ AgI\left(s\right)+e^{-}\to Ag\left(s\right)+I^{-}\left(aq\right) E^o=-0.14 V \end{gathered}$$