Chapter 19, Problem 19.154MP

Interpretation Introduction

(a)

Interpretation:

The equation for the reaction of ${\text{H}}_{\text{2}}{\text{MoO}}_{\text{4}}$ with elemental arsenic in acidic solution to give ${\text{Mo}}^{3+}$ and ${\text{H}}_{\text{3}}{\text{AsO}}_{\text{4}}$ should be balanced.

Concept introduction:

Oxidation; loss of one or more electrons by a substance.

Reduction; the gain of one or more electrons by a substance.

Electrons are transfer from one substance to another, in an oxidation and reduction (redox) reaction. Redox reactions are the basis for electrochemical reactions. Oxidation numbers are used to determine the atoms which get oxidized and reduced in redox reactions. Half reaction method represents the electron transfer method.

Interpretation Introduction

(b)

Interpretation:

The value of $E^0$ for the following reaction should be calculated.

$3{\text{As}}_{(aq)}+5H_{2}Mo{O}_4{}_{(aq)}+15H^{+}{}_{(aq)}\to 3{\text{H}}_{\text{3}}{\text{AsO}}_{\text{4}}{}_{(aq)}+5{\text{Mo}}^{3+}{}_{(aq)}+8{\text{H}}_{\text{2}}{\text{O}}_{(aq)}$

Concept introduction:

In an electrochemical cell; half-cell with the higher reduction potential will undergo reduction within the cell and the half-cell with the lower reduction potential will undergo oxidation within the cell. The $E^0$ value for the overall cell is the sum of $E^0$ values for the oxidation and reduction half-cell reactions.

$E^0{}_{cell}=E^0{}_{red}+E^0{}_{ox}$