Question

The standard reduction potential of a silver|silver chloride|chloride ion electrode is 0.2 V and that of a silver electrode is 0.79V. The maximum amount of AgCl that can dissolve in 106 L of a 0.1 M AgNO3 solution is

Answer 1

Answer: $143.32\times 10^{-3}g$ of AgCl can be dissolved in the given $AgNO_3$ solution.

Explanation: For a given cell $Ag\mid AgCl\mid Cl^-$, following half cell reactions are:

Reduction:      $AgCl(s)+e^-\rightarrow Ag(s)+Cl^-(aq.)$   $E^o=0.2V$

Oxidation:        $Ag(s)\rightarrow Ag^+(aq.)+e^-$                  $E^o=-0.79V$

Net Reaction:  $AgCl(s)\rightarrow Ag^+(aq.)+Cl^-(aq.)$

$E^o_{cell}=E^o_{reduction}+E^o_{oxidation}$

$E^o_{cell}=0.2V+(-0.79V)=-0.59V$

Complete ionization of $AgNO_3$ is:

$AgNO_3\rightarrow Ag^++NO_3^-$

As 1 mole of $AgNO_3$ produces 1 mole of $Ag^+$ ion, so for 0.1M $AgNO_3$ concentration of silver ion will be 0.1M

Equation for the cell potential will be

$E_{cell}=E^o_{cell}-\frac{0.059}{n}log\frac{[Ag^+][Cl^-]}{[AgCl]}$

As the cell is in equilibrium, $\Delta G=0\text{ hence, }E_{cell}=0$

Solubility product of AgCl is given by:

$K_{sp}=[Ag^+][Cl^-]$

Above equation becomes,

$E^o_{cell}=\frac{0.059}{n}logK_{sp}$  (As AgCl is pure solid, so its concentration will be 1)

$-0.59=\frac{0.059}{1}logK_{sp}$

$K_{sp}=10^{-10}$

$AgCl(s)\rightleftharpoons Ag^+(aq.)+Cl^-(aq.)$

As 1 mole of AgCl produces 1 mole of $Ag^+$ and 1 mole of $Cl^-$, so S moles of AgCl will produces S moles of $Ag^+$ and S moles of $Cl^-$ ions

Total concentration of $Ag^+=(S+0.1)M$

Total concentration of $Cl^-=S$

$10^{-10}=(S+0.1)(S)$

Neglecting small value of S as compared to 0.1, we get

$S=10^{-9}mol/L$

If 1 liter of $AgNO_3$ dissolves $10^{-9}mol/L$ of AgCl

So, $10^6\text{ liter of }AgNO_3\text{ will dissolve }=\frac{10^{-9}mol/L\times 10^6L}{1}$

                                                   = $10^{-3}moles$

Amount of AgCl that can be dissolved in $AgNO_3$ will be = Moles × Molar mass of AgCl

Molar mass of AgCl = 143.32 g/mol

Amount of AgCl = $10^{-3}mol\times 143.32g/mol$

                           = $143.32\times 10^{-3}grams$

Answer 2

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