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\[{E_{el}} = E_{el}^0 - \dfrac{{0.0592}}{n}{\log _{10}}{\left[ {{\rm{C}}{{\rm{l}}^ - }} \right]^2}\]

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:

$\left[ {{\rm{C}}{{\rm{l}}^ - }} \right] = \sqrt {0.00939}$

$\left[ {{\rm{C}}{{\rm{l}}^ - }} \right] = 0.1$