Chapter 19, Problem 19.116SP

Interpretation Introduction

Interpretation:

The value of ${\text{K}}_{\text{sp}}$ for ${\text{Hg}}_{\text{2}}{\text{Br}}_{\text{2}}$ at 25°C should be calculated if the cell potential of the following galvanic cell is 1.214 V:

${\text{Hg}}_{\text{(l)}}{\text{| Hg}}_{\text{2}}{\text{Br}}_{\text{2}}{}_{\text{(s)}}{\text{|Br}}^{\text{-}}\text{(0}{\text{.10M)||MnO}}_{\text{4}}^{\text{-}}\text{(0}{\text{.10M),Mn}}^{\text{2+}}\text{(0}{\text{.10M),H}}^{\text{+}}\text{(0}{\text{.10M)|Pt}}_{\text{(s)}}$

Concept introduction:

In the electrochemical cell, the reactions at cathode and anode occur due to the difference in their reduction electrode potential value. The EMF of the cell can be calculated with the help of electrode reduction potential values. The reaction at each electrode is called as half-reaction and the combination of both half-reactions gives the cell reaction of given electrochemical cell. The standard cell potential for an electrochemical cell can be calculated as: $\begin{array}{l}{\text{E}}_{\text{cell}}^{\text{°}}{\text{ = E}}_{\text{cathode}}^{\text{°}}\text{}{\text{ - E}}_{\text{anode}}^{\text{°}}\\ {\text{E}}_{\text{cell}}^{\text{°}}{\text{ = E}}_{\text{reduction}}^{\text{°}}\text{}{\text{ - E}}_{\text{oxidation}}^{\text{°}}\end{array}$

The potential of the cell can be calculated with the help of the Nernst equation that can be shown as:

$\begin{array}{l}{\text{E° = E°}}_{\text{cell}}\text{ - }\frac{\text{0}\text{.0592 V}}{\text{n}}\text{ log Q }\\ \text{n = number of electrons}\\ \text{Q = reaction quotient}\end{array}$