Given the following standard electrode potential, ${\mathrm{PbBr}}_2\left(\mathrm{s}\right)+2{\mathrm{e}}^{-}\to$$\mathrm{Pb}\left(\mathrm{s}\right)+2{\mathrm{Br}}^{-}\left(\mathrm{aq}\right);{\mathrm{E}}^0=-0.248{\mathrm{VPb}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}\to \mathrm{Pb}\left(\mathrm{s}\right);{\mathrm{E}}^0=-0.126\mathrm{V}\text{.If the }{\mathrm{K}}_{\mathrm{sp}}$ for PbBr₂ is 7.4 x 10^-x; x is

(1) ${\mathrm{PbBr}}_2+2{\mathrm{e}}^{-}\to \mathrm{Pb}+2{\mathrm{Br}}^{(-)};{\mathrm{\Delta G}}_1^0=-2\mathrm{F}(-0.248)$

(2) ${\mathrm{Pb}}^{+2}+2{\mathrm{e}}^{-}\to \mathrm{Pb};{\mathrm{\Delta G}}_2^0=-2\mathrm{F}(-0.126)$

(3) ${\mathrm{PbBr}}_2\to {\mathrm{Pb}}^{+2}+2{\mathrm{Br}}^{(-)};{\mathrm{\Delta G}}_3^0=-\mathrm{RTln}{\mathrm{K}}_{\mathrm{sp}}$

(3) =  (1) +  (2)

${\mathrm{\Delta G}}_3^0=\left({\mathrm{\Delta G}}_1^0-{\mathrm{\Delta G}}_2^0\right)$

$\mathrm{R}=8.314\mathrm{J}/\text{mole}/\mathrm{K}, \mathrm{T}=298\mathrm{K}$

$-\mathrm{RT}\text{ in }{\mathrm{K}}_{\mathrm{sp}}=-2\mathrm{F}\left[\right(-0.248)+0.126]$

${\mathrm{K}}_{\mathrm{sp}}=7.4\times {10}^{-5}$