Refer to the galvanic cell below (the contents of each half-cell are written beneath each compartment). The standard reduction potentials are as follows: What is the oxidation state of Cr in Cr₂O₇²^-?

Question

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In order to determine the oxidation state of Cr in the given compound, we need to consider the overall charge of the compound and the known oxidation states of the other elements present. Oxygen is typically assigned an oxidation state of -2, and hydrogen is usually +1. The sum of the oxidation states in a neutral compound is zero. ₂O₇²^- has a net charge of -1, which means the sum of the oxidation states in the compound must be -1. The only element with a variable oxidation state in the compound is Cr. From the given standard reduction potentials, we can see that Cr can have an oxidation state of +3 or +6. However, neither of these values would result in a sum of oxidation states that equals -1. To balance the oxidation states in the compound, we must assume that sulfur has an oxidation state of -2 (its most common oxidation state) and phosphorus has an oxidation state of +5 or +1 (the possibilities that give a sum of oxidation states of -1). With these assumptions, we can solve for the oxidation state of Cr. (-2 x 14) + (-1 x 3) + (+1) + (-2) + (+5) + (-1) + (+1) + (+7) + (-2 x 2) = -1 -28 - 3 + 1 - 2 + 5 - 1 + 1 + 7 - 4 = -1 -24 = -1 + 3x x = +7 Therefore, the oxidation state of Cr in the compound is +7.

Refer to the Galvanic Cell Below (The Contents of Each