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I don't understand the first equation. The chromium atoms in the dichromate ion receive 6 electrons, and their oxidation state declines from 6+ to 3+. But where do the 14 hydrogen cations get their 14 electrons in order to form water with the seven oxygens?


1 Answer 1


Look closely at the oxidation state of all atoms in the equation. $$\ce{Cr2O7^2- (aq) + 14 H+ (aq) + 6e- <=> 2Cr^3+ (aq) + 7 H2O (l)}$$

The oxidation state in chromium changes from +6 to +3. This is what the electrons in the equation are necessary for. Oxygen retains its oxidation state of -2 in both sides of the equation, as well as hydrogen which remains its +1 oxidation state.

In the other equation $$\ce{O2 (g) + 4H+ (aq) +4e- <=> 2H2O (l)}$$

the oxygen gets reduced by changing its oxidation state form 0 to -2. Again for hydrogen, the oxidation state of +1 is retained.

The total equation for this process would therefore be $$\ce{Cr2O7^2- (aq) + 14 H+ (aq) <=> 4Cr^3+ (aq) + 8H2O (l) + 3O2}.$$

From the standard potentials you can see, that this reaction may happen, it does not tell you anything about the conditions under which this reaction will happen. There are a lot more factors to consider.


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