Consider the change in the oxidation state of bromine corresponding to different emf values as shown in the diagram below:
${{\mathrm{BrO}}^{-}}_4 \stackrel{1.82\mathrm{V}}{\to } {{\mathrm{BrO}}^{-}}_3 \stackrel{1.5\mathrm{V}}{\to }\mathrm{HBrO} \stackrel{1.595\mathrm{V}}{\to }{\mathrm{Br}}_2 \stackrel{1.0652\mathrm{V}}{\to }{\mathrm{Br}}^{-}$
Then the species undergoing disproportionation is

Consider the following ${\mathrm{E}}_{\mathrm{o}}$ values:

${\mathrm{E}}_{{\mathrm{Fe}}^{3+}/{\mathrm{Fe}}^{2+}}^0=+0.77\mathrm{V};{\mathrm{E}}_{{\mathrm{Sn}}^{2+}/\mathrm{Sn}}^0=-0.14\mathrm{V}$

Under standard conditions, the cell potential for the reaction given below is:

${\mathrm{Sn}}_{\left(9\right)}+2{\mathrm{Fe}}_{\left(\mathrm{aq}\right)}^{3+}\to 2{\mathrm{Fe}}_{\left(\mathrm{aq}\right)}^{2+}+{\mathrm{Sn}}_{\left(\mathrm{aq}\right)}^{2+}$

Aluminium oxide may be electrolyzed at $1000°\mathrm{C}$ to furnish aluminium metal.

(Atomic mass $=27 \mathrm{amu}$. $1$ Faraday $=96500 \mathrm{C}$) The cathode reaction is: ${\mathrm{Al}}^{3+}+3{\mathrm{e}}^{-}⟶\mathrm{Al}.$

To prepare $5.12 \mathrm{kg}$ of aluminium metal by this method would require:

Given data is at $25°\mathrm{C}$:

${\mathrm{Ag}}^0+{\mathrm{I}}^{-}\to \mathrm{AgI}+{\mathrm{e}}^{-}; \mathrm{E}°=0.152 \mathrm{V}$

$\mathrm{Ag}\to {\mathrm{Ag}}^{+}+{\mathrm{e}}^{-}; \mathrm{E}°=-0.800 \mathrm{V}$

What is the value of ${\mathrm{logK}}_{\mathrm{sp}}$ for $\mathrm{AgI}$ (Take $\frac{0.474}{0.059}=8.065$)