
The standard reduction electrode potential for two reactions are given below:
$AgCl(s) + {e^ - } \to Ag(s) + C{l^ - };{E^0} = 0.22V$
\[A{g^ + }(aq) + {e^ - } \to Ag(s);{E^0} = 0.80V\]
The solubility product of AgCl under standard conditions of temperature (298 K) is given by
1) \[1.5 \times {10^{ - 10}}\]
2) \[1.5 \times {10^{ - 8}}\]
3) $3.2 \times {10^{ - 10}}$
4) \[3.2 \times {10^{ - 8}}\]
Answer
567k+ views
Hint: To calculate the cell potential of an electrochemical cell when the temperature, pressure and reactant concentration is provided, the equation given by German chemist Walther Hermann Nernst known as Nernst equation is used.
Complete step by step answer:
The given reactions are shown below.
\[AgCl(s) + {e^ - } \to Ag(s) + C{l^ - }(aq){E^0} = 0.22V\]……(i)
$A{g^ + }(aq) + {e^ - } \to Ag(s){E^0} = 0.80V$..……(ii)
Reverse the equation (ii), we get
$Ag(s) \to A{g^ + }(aq) + {e^ - }{E^0} = - 0.80V$……(iii)
Add equation (i) and (iii), we get
$AgCl(s) \to A{g^ + }(aq) + C{l^ - }(aq){E^0} = - 0.58V$
The Nernst equation gives the relation among the cell potential of electrochemical cell, temperature, standard cell potential and reaction quotient.
The Nernst equation relates the ability of the ion or atom to accept one more electron (reduction potential) measured at any given condition compared when measured at standard condition.
The Nernst equation is shown below.
${E_{Cell}} = E_{Cell}^0 - \dfrac{{RT}}{{nF}}\ln K$
Where,
${E_{Cell}}$ is the cell potential of the cell.
${E^0}$ is the cell potential under standard condition.
R is the universal gas constant.
T is the temperature.
n is the number electrons transferred between the reactant during the redox reaction.
K is the reaction quotient.
Substitute the values in the Nernst equation
$0 = - 0.58 - \dfrac{{8.314 \times 298 \times 2.303}}{{1 \times 96500}}\log K$
$\Rightarrow \log K = \dfrac{{0.58}}{{0.0591}}$
$\Rightarrow \operatorname{logK} = - 9.81$
$\Rightarrow K = 1.5 \times {10^{ - 10}}$
Thus, the solubility product of AgCl is $1.5 \times {10^{ - 10}}$.
Therefore, the correct option is 4.
Note:
Under non-standard conditions also Nernst equation can be used to determine the cell potential of the electrochemical cell. The Nernst equation cannot be used to determine the cell potential when the current flows through the electrodes because the current affects the activity of the ion present on the surface.
Complete step by step answer:
The given reactions are shown below.
\[AgCl(s) + {e^ - } \to Ag(s) + C{l^ - }(aq){E^0} = 0.22V\]……(i)
$A{g^ + }(aq) + {e^ - } \to Ag(s){E^0} = 0.80V$..……(ii)
Reverse the equation (ii), we get
$Ag(s) \to A{g^ + }(aq) + {e^ - }{E^0} = - 0.80V$……(iii)
Add equation (i) and (iii), we get
$AgCl(s) \to A{g^ + }(aq) + C{l^ - }(aq){E^0} = - 0.58V$
The Nernst equation gives the relation among the cell potential of electrochemical cell, temperature, standard cell potential and reaction quotient.
The Nernst equation relates the ability of the ion or atom to accept one more electron (reduction potential) measured at any given condition compared when measured at standard condition.
The Nernst equation is shown below.
${E_{Cell}} = E_{Cell}^0 - \dfrac{{RT}}{{nF}}\ln K$
Where,
${E_{Cell}}$ is the cell potential of the cell.
${E^0}$ is the cell potential under standard condition.
R is the universal gas constant.
T is the temperature.
n is the number electrons transferred between the reactant during the redox reaction.
K is the reaction quotient.
Substitute the values in the Nernst equation
$0 = - 0.58 - \dfrac{{8.314 \times 298 \times 2.303}}{{1 \times 96500}}\log K$
$\Rightarrow \log K = \dfrac{{0.58}}{{0.0591}}$
$\Rightarrow \operatorname{logK} = - 9.81$
$\Rightarrow K = 1.5 \times {10^{ - 10}}$
Thus, the solubility product of AgCl is $1.5 \times {10^{ - 10}}$.
Therefore, the correct option is 4.
Note:
Under non-standard conditions also Nernst equation can be used to determine the cell potential of the electrochemical cell. The Nernst equation cannot be used to determine the cell potential when the current flows through the electrodes because the current affects the activity of the ion present on the surface.
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