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if the solution in calomel electrode was dried is that mean that is useless, if it is not how to do and how we test it for that works

if the solution in calomel electrode was dried is that mean that is useless, if it is not how to do and how we test it for that works

Grade:12th Pass

1 Answers

Aman Bansal
592 Points
8 years ago

Dear Fairouz,

Theory of operation

The electrode is based on the redox reaction

\text{Hg}_2^{2+} + 2\text{e}^- \rightleftarrows 2\text{Hg(l)}

The Nernst equation for this reaction is

 E = E^0_{\text{Hg}_2^{2+}/\text{Hg}} - \frac{RT}{2F} \ln\frac{1}{a_{\text{Hg}_2^{2+}}}

where E0 is the standard electrode potential for the reaction and aHg is the activity for the mercury cation (the activity for a liquid of 1 Molar is 1). This activity can be found from the solubility product of the reaction

 \text{Hg}_2^{2+} + 2\text{Cl}^- \rightleftarrows \text{Hg}_2\text{Cl}_2\text{(s)},\qquad  K_{\text{sp}} = a_{\text{Hg}_2^{2+}} a_{\text{Cl}^-}^2

By replacing the activity in the Nernst equation with the value in the solubility equation, we get

 E = E^0_{\text{Hg}_2^{2+}/\text{Hg}} + \frac{RT}{2F} \ln K_{\text{sp}} - \frac{RT}{2F} \ln a^2_{\text{Cl}^-}

The only variable in this equation is the activity (or concentration) of the chloride anion. But since the inner solution is saturated with potassium chloride, this activity is fixed by the solubility of potassium chloride. When saturated the redox potential of the calomel electrode is +0.2444 V vs. SHE at 25 °C, but slightly higher when the chloride solution is less than saturated. For example, a 3.5M KCl electrolyte solution increases the reference potential to +0.250 V vs. SHE at 25 °C, and a 0.1 M solution to +0.3356 V at the same temperature.

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Aman Bansal

Askiitian Expert

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