Chapter 21, Problem 21.36P

Interpretation Introduction

Interpretation:

The negative ${\text{E}}^{\text{o}}{}_{cell}$ indicate about what of a redox reaction and about of reverse reaction has to be indicated.

Concept Introduction:

When a metal M is dipped in its electrolyte MX, and coupled against an SHE, the following equilibrium exists at electrode M as follows:

  $\text{M}\rightleftharpoons {\text{M}}^{\text{+}}+{\text{e}}^{-}$

At the SHE:

  $\frac{\text{1}}{\text{2}}{\text{H}}_{\text{2}}\rightleftharpoons {\text{H}}^{\text{+}}\text{+}{\text{e}}^{\text{-}}$

Therefore, M oxidizes and ${\text{H}}^{\text{+}}$ is reduced in this cell.

The reaction could be written as follows:

  $\text{M}\text{+}{\text{H}}^{\text{+}}\to {\text{M}}^{\text{+}}\text{+}\frac{\text{1}}{\text{2}}{\text{H}}_{\text{2}}$

The cell could described using a ‘cell notation’ as follows:

  ${\text{M}}_{\text{(s)}}{\text{|M}}^{\text{+}}{}_{\text{(aq)}}{\text{||H}}^{\text{+}}{}_{\text{(aq)}}\text{|}\frac{\text{1}}{\text{2}}{\text{H}}_{\text{2}}{}_{\text{(g)}}{\text{;Pt}}_{\text{(s)}}$

The direction of spontaneity is from left to right across the cell notation. Here, M oxidizes to ${\text{M}}^{\text{+}}$ spontaneously, while ${\text{H}}^{\text{+}}$ is reduced to hydrogen gas.

The ‘||’ divides the cell into two half cells. On the left of ‘||’, the oxidation half-cell; on the right the reduction half-cell.

The electrode potential of the oxidation half-cell is written as ${\text{E}}_{{\text{M/M}}^{\text{+}}}$.

The electrode potential of the reduction half-cell is written as ${\text{E}}_{{\text{H}}^{\text{+}}\text{/}\frac{\text{1}}{\text{2}}{\text{H}}_{\text{2}}}$.

The cell potential is calculated as $\text{E}{°}_{\text{cell}}\text{=}\text{E}{°}_{{\text{M/M}}^{\text{+}}}\text{+}\text{E}{°}_{{\text{H}}^{\text{+}}\text{/}\frac{\text{1}}{\text{2}}{\text{H}}_{\text{2}}}$ .