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In each of the following elements:-Assign oxidation numbers.-Identify the oxidizing and reducing agents.-Identify the change in oxidation number. Cr2O7^2- + 2OH^- —> 2CrO4^2- + H2O

User Notepad
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To dicover which compound is being oxidized and which one is being reduced, we need to know the oxidation number of each specie.

The oxidation number (NOX) of an element is the electrical charge it acquires when it forms an ionic bond or the partial character (δ) it acquires when it forms a predominantly covalent bond.

There are some rules for NOX, like:

The NOX of simple substances is always zero.

The NOX of ions is equal to their charge.

The sum of the NOX elements of a compound always equals zero.

The sum of the NOXs of the elements in a composite ion is always equal to the charge on the ion.

The NOX of Oxygen in an compound is -2 almost in all cases.

The NOX of hydrogen is +1

With this in mind, let's calculate the NOX of each specie of the following equation:

Cr2O7^2- + 2OH^- —> 2CrO4^2- + H2O

Reactants:

Cr2O7^2-

NOX of oxygen: -2

NOX of Cr: +6

2x + (-2*7) = -2

2x = 14 - 2

x = 6

OH-:

O: -2

H: +1

Products

CrO4^2-

Nox of oxygen: -2

Nox of Cr: +8

x + (-2*4) = -2

x - 8 = -2

x = +6

H2O

H: +1

O: -2

Answer:

-Assign oxidation numbers:

Products: Cr: +6; O: -2; H: +1

Reactants: Cr: +6; O: -2; H: +1

-Identify the oxidizing and reducing agents:

No compounds oxidize or reduce. This reaction is not a redox reaction.

-Identify the change in oxidation number: there is no change in oxidation number, so this is not a redox reaction.

User Dusean Singh
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