Chapter 19, Problem 75E

Interpretation Introduction

Interpretation: The cell potential at $25°C$ for the given electrochemical cell based on the following two half-reactions is to be calculated:

$Oxidation: Pb\left(s\right)\to P{b}^{2+}(aq,0.10M)+2e^{-}$

$\begin{array}{l}\mathrm{Re}duction:Mn{O}_4{}^{–}(aq,1.50M)+4H^{+}(aq,2.0M)+3e^{–}\to Mn{O}_2\left(s\right)+2H_{2}O(l)\hfill \\ \hfill \end{array}$