Question

The standard reduction potentials for two reactions are given below:
$
  AgCl(s) + {e^ - } \to Ag(s) + C{l^ - }(aq) \\
  {E^o} = 0.22V \\
  A{g^ + }(aq) + {e^ - } \to Ag(s) \\
  {E^o} = 0.80V \\
 $
The solubility product of AgCI under standard conditions of temperature (298 K) is given by:
$
  A.{\text{ 1}}{\text{.6}} \times {\text{1}}{{\text{0}}^{ - 5}} \\
  B.{\text{ 1}}{\text{.5}} \times {\text{1}}{{\text{0}}^{ - 8}} \\
  C.{\text{ 3}}{\text{.2}} \times {\text{1}}{{\text{0}}^{ - 10}} \\
  D.{\text{ 1}}{\text{.5}} \times {\text{1}}{{\text{0}}^{ - 10}} \\
 $

Answer 1

Hint: The solubility product is K and can be found out from the Nernst equation. After reversing the second equation, you can find the standard reduction potential of the equation.

Complete step by step answer:
First let us understand the related terms:
The standard reduction potential is called the reduction potential of a molecule under particular, standard conditions. Standard reduction potentials are used in determining the direction of a reaction. The reduction potential of a given species is assumed to be the negative of the oxidation potential.
Solubility is the property of a substance which is solute to get dissolved in a solvent in order to form a solution.
The solubility product is one of the equilibrium constants where its value depends on temperature. It usually increases as the temperature increases due to increase in solubility.
The given equation are:
$
  AgCl(s) + {e^\_} \to Ag(s) + C{l^ - }(aq) \\
  {E^o} = 0.22 \\
  A{g^ + }(aq) + {e^ - } \to Ag(s) \\
  {E^o} = 0.80 \\
  Ag(s) \to A{g^ + }(aq) + {e^ - } \\
  {E^o} = - 0.80 \\
 $
As to get the required reduction potential, we add the equations.
Adding equation (1) & (2), we get
$
  AgCl(s) \to A{g^ + }(aq) + C{l^ - }(aq) \\
  {E^o} = 0.58V \\
    \\
 $
In electrochemistry, the Nernst equation is an equation which gives relation between reduction potential of an electrochemical reaction (half-cell or full cell reaction) and the standard electrode potential, temperature, and activities (often approximated by concentrations) of the chemical species which undergoes reduction and oxidation.
From Nernst equation
$
  {E_{cell}} = {E^o}_{cell} - \dfrac{{RT}}{{nF}}\ln K \\
  0 = - 0.58 - \dfrac{{8.314 \times 298 \times 2.303}}{{1 \times 96500}}\log K \\
  \log K = \dfrac{{0.58}}{{0.0591}} = - 9.81 \\
  K = 1.5 \times {10^{ - 10}} \\
    \\
 $
Hence the solubility product of AgCl is $K = 1.5 \times {10^{ - 10}}$
The correct option is D.

Note:
Reduction potential (Eo) is defined as a tendency of a chemical species to be reduced by gaining an electron and is defined with electrochemical reference of hydrogen, which is globally given the reduction potential of zero.