Consider the following equation; Cr2O7(aq)2- + 14H+(aq) + 6e- → 2Cr3+(aq) + 7H2O(I). The oxidation number of chromium changes from

Consider the following equation; Cr₂O_7(aq)^2- + 14H⁺_(aq) + 6e^- → 2Cr³⁺_(aq) + 7H₂O_(I). The oxidation number of chromium changes from

Answer Details

In the given equation, the reactant Cr₂O₇^2- has chromium in the +6 oxidation state (since each oxygen atom has an oxidation number of -2, and there are seven of them, the total negative charge of the oxygen atoms is -14, and the total negative charge of the ion is -2, so the total positive charge of the chromium atoms must be +12 to balance it out). In the product side, the product Cr³⁺ has a +3 oxidation state. This means that the oxidation number of chromium has changed from +6 to +3, a decrease of 3. Therefore, the answer is - 2 to + 6, since chromium goes from +6 to +3.