Question

The $p{K_{sp}}$ of $AgI$ is $16.07$. If the ${E^{\circ} }$ value for $A{g^ + }/Ag$ is $0.7991V$. Find the ${E^{\circ} }$ for the half cell reaction $AgI/\left( s \right) + {e^ - } \to Ag + I$.

Answer 1

Hint: Half- cell potential is defined as the potential which developed at the electrode of each half cell in an electrochemical cell and total potential is calculated from the potentials of two half cells and a half cell reaction is a reaction which either includes oxidation reaction or reduction reaction.

Complete step by step answer:
The standard reduction potential can be calculated by subtracting the standard reduction potential of any reaction held at the anode from the standard reduction potential of any reaction held at the cathode. The minus sign is used to indicate that oxidation is the reverse of reduction.
The reaction involves:
$AgI/\left( s \right) + {e^ - } \to Ag + I$
By using the formula of solubility equilibrium
$\left( {{K_{sp}}} \right)$= solubility product constant
$\left[ {A{g^ + }} \right]$= cation in an aqueous solution
$\left[ {{I^ - }} \right]$= anion in an aqueous solution
The rate of reaction $\left( {{K_{sp}}} \right)$= $\left[ {A{g^ + }} \right]\left[ {{I^ - }} \right]$
$A{g^ + }/Ag$ means silver cation reduced to silver atom by gaining one electron to fill its valency.
At cathode: $AgI/\left( s \right) + {e^ - } \to Ag + I$
$E_{Red}^{\circ} $=?
At anode : $Ag \to A{g^ + } + {e^ - }$
Silver atoms oxidized to silver cation by releasing one electron.
$E_{Oxide}^{\circ} $= $0.7991V$
To calculate ${E^{\circ} }$ (standard electrode potential), we need to calculate ${E_{cell}}$ (electrode potential of a cell).
By using the formula, ${E_{cell}}$= $E_{cell}^{\circ} - \dfrac{{0.0591}}{n}\log X$
$ \Rightarrow $$E_{cell}^{\circ} = \dfrac{{ + 0.0591}}{n}\log {K_{sp}}$$\left[ {{E_{cell}} = 0} \right]$
$\left[ {\log {K_{sp}} = P{K_{sp}}} \right]$
Therefore, $ \Rightarrow $ $E_{cell}^{\circ} = \dfrac{{ + 0.0591}}{n}P{K_{sp}}$
$p{K_{sp}}$= $16.07$ ( given)
$ \Rightarrow $$E_{cell}^{\circ} = \dfrac{{ + 0.0591}}{1} \times 16.07$ (n=1)
= $ - 0.949V$.
${E_{cell}}$= $E_{Red}^{\circ} $ + $E_{Oxide}^{\circ} $
$E_{Red}^{\circ} $ = $ - 0.949V$ $ - $ $0.7991V$
= $ - 0.2159V$
Hence, ${E^{\circ} }$at cathode is equal to $ - 0.2159V$.

Note:
The standard reduction potential is the reduction potential of a molecule under specific and standard conditions. It can be useful in determining the directionality of a reaction and can be considered to be the negative of the oxidation potential.
Fluorine gas is the best oxidizing agents with the highest positive standard potential \[\left( { + 2.87{\text{ }}V} \right).\] The difference between oxidation potential and reduction potential is that oxidation potential determines the ability of a chemical element to be oxidized, whereas the reduction potential determines the capacity of a chemical element to be reduced.