GRE Subject Test: Chemistry - Oxidation-Reduction Analysis

What is the oxidation state of manganese in $K_{3}MnO_{4}$ ?

Explanation

In this compound oxygen has an oxidation number of . There are 4 oxygens, therefore the total oxidation number for the oxygens in $K_{3}MnO_{4}$ is: $Total\ O\ oxidation\ number=(-2) \times 4=-8$ . Potassium, , is a group 1 metal and has an oxidation number of . There are 3 potassium ions present in $K_{3}MnO_{4}$ , therefore the total oxidation number for the potassium ions is:

$Total\ K\ oxidation\ number=(+1) \times 3=+3$ .The sum of all oxidation number in a neutral compound such as $K_{3}MnO_{4}$ is zero. We will give the oxidation number of equal to :

Simplifying the above equation:

Rearranging to solve for :

Which of the following would be considered a Lewis base?

$Br^{-}$

$H^{+}$

$Mn^{2+}$

$CH_{2}Cl_{2}$

Explanation

A lewis base is an electron pair donor. $Br^{-}$ would be considered a lewis base based on the definition. Because $Br^{-}$ is electron rich based on its oxidation number, it is able to donate an electron pair to an electron deficient ion.

Which of the following types of reactions best describes the following reaction:

$Ce^{4+}+Fe^{2+}\rightarrow\ Ce^{3+}+Fe^{3+}$

Redox reaction

Double replacement

Combustion

Catalytic

Combination

Explanation

A redox reaction is also known as an oxidation/reduction reaction. This type of reaction involves the transfer of electrons from on reactant to another. In this reaction, an electron is transferred from $Fe^{2+}$ to $Ce^{4+}$ forming the products $Ce^{3+}$ and $Fe^{3+}$ . The substance that gains an electron is referred to as being reduced. The substance that loses an electron is referred to as being oxidized.

What is the oxidation state of manganese in $KMnO_{4}$ ?

Explanation

In this compound oxygen has an oxidation number of . There are 4 oxygens, therefore the total oxidation number for the oxygens in $KMnO_{4}$ is: $Total\ oxygen\ oxidation\ number=(-2) \times 4=-8$ . Potassium,, is a group 1 metal and has an oxidation number of . The sum of all oxidation number in a neutral compound such as $KMnO_{4}$ is zero. We will give the oxidation number of equal to :

$((-2) \times 4)+(+1)+x=0$

Simplifying the above equation:

Rearranging to solve for :

Which molecule would be considered a Lewis acid?

$Zn^{2+}$

$NH_{3}$

$OH^{-}$

$O_{2}$

Explanation

A lewis acid is an electron pair acceptor. $Zn^{2+}$ would be considered a lewis acid based on the definition. Because $Zn^{2+}$ is electron-deficient based on its oxidation number, it is able to accept an electron pair.

Oxidation-Reduction Analysis

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