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The standard reduction potential of normal calomel electrode and reduction potential of saturated calomel electrodes are 0.27 and 0.33 volt respectively. What is the concentration of \[{\rm{C}}{{\rm{l}}^ - }\] in a saturated solution of \[{\rm{KCl}}\]?

Answer Verified Verified
Hint: Nernst equation gives the relationship between the electrode potential, the standard electrode potential, the number of electrons participating in the half reaction, and the concentrations. The Nernst equation for the half cell reaction is given by the following formula.
\[{E_{el}} = E_{el}^0 - \dfrac{{0.0592}}{n}{\log _{10}}{\left[ {{\rm{C}}{{\rm{l}}^ - }} \right]^2}\]

Complete step by step answer:
Nernst equation gives the relationship between the electrode potential, the standard electrode potential, the number of electrons participating in the half reaction, and the concentrations. Write the reduction half reaction. A chlorine molecule gains two electrons to form two chloride ions:

\[{\rm{C}}{{\rm{l}}_2}{\rm{ + 2 }}{{\rm{e}}^ - }{\rm{ }} \to {\rm{ 2 C}}{{\rm{l}}^ - }\]

The number of electrons n is 2.
Write the Nernst equation for the half cell:
\[{E_{el}} = E_{el}^0 - \dfrac{{0.0592}}{n}{\log _{10}}{\left[ {{\rm{C}}{{\rm{l}}^ - }} \right]^2}\]… …(1)
The standard reduction potential of normal calomel electrode \[\left( {E_{el}^0} \right)\] and reduction potential of saturated calomel electrodes \[\left( {{E_{el}}} \right)\] are 0.27 and 0.33 volt respectively.
Substitute values in equation (I)
${E_{el}} = E_{el}^0 - \dfrac{{0.0592}}{n}{\log _{10}}{\left[ {{\rm{C}}{{\rm{l}}^ - }} \right]^2}$

$0.33V = 0.27V - \dfrac{{0.0592}}{2}{\log _{10}}{\left[ {{\rm{C}}{{\rm{l}}^ - }} \right]^2}$

$\dfrac{{0.0592}}{2}{\log _{10}}{\left[ {{\rm{C}}{{\rm{l}}^ - }} \right]^2} = 0.27 - 0.33$
$\dfrac{{0.0592}}{2}{\log _{10}}{\left[ {{\rm{C}}{{\rm{l}}^ - }} \right]^2} = - 0.06$

Rearrange above equation:
${\log _{10}}{\left[ {{\rm{C}}{{\rm{l}}^ - }} \right]^2} = - \dfrac{{0.06 \times 2}}{{0.0592}}$
${\log _{10}}{\left[ {{\rm{C}}{{\rm{l}}^ - }} \right]^2} = - 2.03$
Take antilog
${\left[ {{\rm{C}}{{\rm{l}}^ - }} \right]^2} = {10^{ - 2.03}}$
${\left[ {{\rm{C}}{{\rm{l}}^ - }} \right]^2} = 0.00939$

Take square root on both sides of the equation:
$\left[ {{\rm{C}}{{\rm{l}}^ - }} \right] = \sqrt {0.00939}$
$\left[ {{\rm{C}}{{\rm{l}}^ - }} \right] = 0.1$

Hence, option A ) 0.1 M is the correct answer.

Note: Nernst equation gives the relationship between the electrode potential, the standard electrode potential, number of electrons participating in the half reaction and the reaction quotient. When electrode potential, standard electrode potential and the half reaction are given, it is possible to calculate the reaction quotient.