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Assign oxidation numbers to each of the elements in dichromate ion (Cr2O72-).

Question:

Assign oxidation numbers to each of the elements in dichromate ion {eq}(Cr_2O_7^{2-}) {/eq}.

Oxidation Numbers:

Oxidation numbers are used as an accounting system to help determine where electrons go in chemical reactions. By comparing the oxidation numbers of elements in the reactants before the reaction with their oxidation numbers after the reaction as products, we can determine what was oxidized and what was reduced. This can also help us identify oxidizing and reducing agents.

Answer and Explanation:

When assigning oxidation numbers, there is an order to follow that determines which oxidation numbers you assign first. In general, you assign oxidation numbers until you have one left to solve for. In our example, oxygen is assigned an oxidation number of -2 (which means it gains two electrons) before a transition metal like chromium (Cr) is assigned an oxidation number. As there are only two elements in our polyatomic ion, this means we will solve for chromium's oxidation number.

In polyatomic ions, the sum of the oxidation states in the compound must add up to the charge of the ion (in our case, the charge is given as 2-). When we do this, we also have to take into account how many of each atom there are by multiplying the oxidation numbers by their respective subscripts from the chemical formula. We will use the variable x to represent chromium's oxidation number. This results in the following expression:

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

Using algebra to solve for x, the oxidation number of each chromium becomes +6.

Therefore the oxidation number of oxygen is -2 and chromium is +6.


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