91Ó°ÊÓ

Chemical synthesis or, alternatively, chemical breakdown into two or more separate chemicals occurs when one component interacts with another to generate new material. These processes are known as chemical reactions, and they are generally irreversible until followed by other chemical reactions.

$$\begin{gathered} {\text{Oxidisation: 2Hg}}_{\text{(l)}}\text{+ 2C}{{\text{l}}^{\text{-}}}_{\text{(aq)}}→{\text{2e}}^{\text{-}}{\text{+ 2Hg}}_{\text{2}}{\text{Cl}}_{\text{2(s)}} \\ \text{Reduction: A}{{\text{g}}^{+}}_{\text{(aq)}}{\text{+e}}^{\text{-}}→{\text{Ag}}_{\text{(s)}} \end{gathered}$$

$\begin{array}{rcl}& & {\text{E}}_{\text{cell}}^{\text{o}}{\text{= E}}_{\text{reduced}}^{\text{o}}{\text{- E}}_{\text{oxidised}}^{\text{o}}\\ & & {\text{E}}_{\text{cell}}^{\text{o}}{\text{= E}}_{{\text{Ag}}^{\text{+}}}^{\text{o}}{\text{- E}}_{{\text{Hg}}_{\text{2}}{\text{Cl}}_{\text{2}}}^{\text{o}}\\ & & \text{= 0.80 V - 0.24 V}\\ & & \text{= 0.56 V}\\ & & \end{array}$

It is given the we have the value as:

${\text{E}}_{\text{cell}}\text{= 0.60 V}$

The Q may be extracted from the Nernst equation.

${\text{E}}_{\text{cell}}{\text{= E}}_{\text{cell}}^{\text{o}}\text{-}\frac{\text{RT}}{\text{nF}}\text{×±õ²Ô²Ï}$

The process involves two electrons, and the standard temperature is 298 K.

$$\begin{gathered} {\text{E}}_{\text{cell}}{\text{= E}}_{\text{cell}}^{\text{o}}\text{-}\frac{{\displaystyle \text{8.314}\frac{\text{J}}{\text{mol K}}\text{×298 K}}}{n×{\displaystyle \text{96,500 C}}}\text{×±ô²Ô²Ï} \\ {\text{E}}_{\text{cell}}\text{= 0.56 -}\frac{{\displaystyle \text{8.314}\frac{\text{J}}{\text{mol K}}\text{×298 K}}}{2×{\displaystyle \text{96,500 C}}}\text{×±ô²Ô²Ï} \\ \text{0.60 = 0.56 -}\frac{{\displaystyle \text{8.314}\frac{\text{J}}{\text{mol K}}\text{×298 K}}}{2×{\displaystyle \text{96,500 C}}}\text{×±ô²Ô²Ï} \\ \text{lnQ =}\frac{{\displaystyle \text{- (0.60 - 0.56) ³Õ×2×96,500 C}}}{\text{8.314}\frac{\text{J}}{\text{mol K}}\text{×298 K}} \\ \frac{\text{1}}{{\text{[}\stackrel{¯}{\text{Cl}}\text{- ]}}^{\text{2}}{\left[{\text{Ag}}^{\text{+}}\right]}^{\text{2}}}\text{=e}\frac{{\displaystyle \text{- (0.60 - 0.56) ³Õ×2×96,500 C}}}{\text{8.314}\frac{\text{J}}{\text{mol K}}\text{×298 K}} \end{gathered}$$

Assume [ ${\text{Cl}}^{\text{-}}$] is constant and [ ${\text{Cl}}^{\text{-}}$]= 1

$$\begin{gathered} \frac{\text{1}}{{\text{[1]}}^{\text{2}}{\left[{\text{Ag}}^{\text{+}}\right]}^{\text{2}}}\text{=}\frac{{\displaystyle \text{- (0.60 - 0.56) ³Õ×2×96,500 C}}}{{\displaystyle \text{8.314}\frac{\text{J}}{\text{mol K}}\text{×298°­}}} \\ {\left[{\text{Ag}}^{\text{+}}\right]}^{\text{2}}\text{=}\frac{1}{{\displaystyle \text{e}\frac{-\; (0.60\; -\; 0.56)\mathrm{³Õ×2×96,500}C}{\text{8.314}\frac{\text{J}}{\text{mol K}}\text{×298°­}}}} \\ \left[{\text{Ag}}^{\text{+}}\right]\text{=}\sqrt{\frac{1}{{\displaystyle \text{e}\frac{-\; (0.60\; -\; 0.56)V×2×96,500C}{\text{8.314}\frac{\text{J}}{\text{mol K}}\text{×298°­}}}}} \\ \left[{\text{Ag}}^{\text{+}}\right]\text{= 4.75M} \end{gathered}$$

Therefore, the value is: $\left[{\text{Ag}}^{\text{+}}\right]\text{= 4.75 M}$

$$\begin{gathered} \left[{\text{Ag}}^{\text{+}}\right]\text{=}\sqrt{\frac{1}{{\displaystyle \text{e}\frac{{\displaystyle \text{- (0.53 - 0.56) ³Õ×2×96,500 C}}}{{\displaystyle \text{8.314}\frac{\text{J}}{\text{mol K}}\text{×298°­}}}}}} \\ \left[{\text{Ag}}^{\text{+}}\right]\text{= 0.311 M} \end{gathered}$$

Therefore, the value is: $\left[{\text{Ag}}^{\text{+}}\right]\text{= 0.311 M}$