Calculate and find the reduced form of the oxidizing agent CrO4^2-
To find the reduced form of the oxidizing agent CrO4^2-, we need to determine the oxidation state of chromium (Cr) in this species.
The oxidation state of oxygen in a compound is usually -2, so the total negative charge from the four oxygen atoms (-2 x 4 = -8) should be balanced by the overall charge of the CrO4^2- ion, which is -2.
Let the oxidation state of chromium be x. Since the overall charge on the ion is -2, we can set up an equation:
x + (-8) = -2
x - 8 = -2
x = -2 + 8
x = +6
Therefore, the oxidation state of chromium in CrO4^2- is +6.
To find the reduced form of this oxidizing agent, we need to decrease the oxidation state of chromium from +6 to a lower state. One common reduced form of this oxidizing agent is Cr^3+.
So, the reduced form of the oxidizing agent CrO4^2- is Cr^3+.