Chapter 20, Problem 51A

(a)

Interpretation Introduction

Interpretation: The oxidation number of $Cr$ in $K_{2}Cr{O}_4$ is to be determined.

Concept Introduction: Zero, positive, and negative numbers that are assigned to reactants species to track the electron flow is oxidation number. When electrons are lost (oxidation) from a species, positive numbers are given while if electrons are gained(reduction), negative numbers are given. And the reactants that don't involve any change in electron count are assigned with the number zero. Certain rules must also be followed before assigning these numbers.

• The oxidation number of oxygen is -2.
• The oxidation number of H is +1.
• For a neutral compound, the total sum of the oxidation number is zero.
• For ionic compounds with a charge, the total sum of oxidation number is equal to its charge.

Answer to Problem 51A

The oxidation number of $Cr$ in $K_{2}Cr{O}_4$ is $+6$ .

Explanation of Solution

The given compound $K_{2}Cr{O}_4$ is neutral. The oxidation state of $Cr$ can be calculated as,

Let x be the oxidation number of $Cr$ .

  $\begin{array}{c}x+\left(4\times -2\right)+2=0\\ x+2-8=0\\ x=+6\end{array}$

So, the oxidation number of $Cr$ is $+6$ .

(b)

Interpretation Introduction

Interpretation: The oxidation number of $Cr$ in $C{r}_2{O}_3$ is to be determined.

Concept Introduction: Zero, positive, and negative numbers that are assigned to reactants species to track the electron flow is oxidation number. When electrons are lost (oxidation) from a species, positive numbers are given while if electrons are gained (reduction), negative numbers are given. And the reactants that don't involve any change in electron count are assigned with the number zero. Certain rules must also be followed before assigning these numbers.

• The oxidation number of oxygen is -2.
• For a neutral compound, the total sum of the oxidation number is zero.
• For ionic compounds with a charge, the total sum of oxidation number is equal to its charge.

Answer to Problem 51A

The oxidation number of $Cr$ in $C{r}_2{O}_3$ is $+3$ .

Explanation of Solution

The given compound $C{r}_2{O}_3$ is neutral. So, the oxidation number can be calculated as,

Let x be the oxidation number of $Cr$ .

  $\begin{array}{c}2x+\left(3\times -2\right)=0\\ 2x-6=0\\ x=+3\end{array}$

So, the oxidation number of $Cr$ is $+3$ .