C6H5OH + 14 Mn{4+} + 11 H2O ---> 6 CO2 + 14 Mn{2+} + 28 H{+} This is the balance

Question

C6H5OH + 14 Mn{4+} + 11 H2O ---> 6 CO2 + 14 Mn{2+} + 28 H{+} This is the balanced equation for Phenol and Manganese (IV) in water, why is the number of electrons transferred to the Manganese atoms equal to the Number of Hydrogens (28), are the 28 hydrogens giving up 28 electrons? But isn't the oxidation state of Hydrogen <+1>? That would mean it wouldn't have any electrons to give. Can anyone answer this for me as to why there are 28 electrons being transferred if hydrogen has a oxidation state of +1 meaning it doesn't have any electrons to give?

Explanation / Answer

PLEASE RATE ME AND AWARD ME KARMA POINTS IF IT IS HELPFUL FOR YOU Vanadium forms a number of different ions - for example, V2+ and V3+. If you think about how these might be produced from vanadium metal, the 2+ ion will be formed by oxidising the metal by removing two electrons: The vanadium is now said to be in an oxidation state of +2. Removal of another electron gives the V3+ ion: The vanadium now has an oxidation state of +3. Removal of another electron gives a more unusual looking ion, VO2+. The vanadium is now in an oxidation state of +4. Notice that the oxidation state isn't simply counting the charge on the ion (that was true for the first two cases but not for this one). The positive oxidation state is counting the total number of electrons which have had to be removed - starting from the element. It is also possible to remove a fifth electron to give another ion (easily confused with the one before!). The oxidation state of the vanadium is now +5. Every time you oxidise the vanadium by removing another electron from it, its oxidation state increases by 1. Fairly obviously, if you start adding electrons again the oxidation state will fall. You could eventually get back to the element vanadium which would have an oxidation state of zero. What if you kept on adding electrons to the element? You can't actually do that with vanadium, but you can with an element like sulphur. The sulphur has an oxidation state of -2. Summary Oxidation state shows the total number of electrons which have been removed from an element (a positive oxidation state) or added to an element (a negative oxidation state) to get to its present state. Oxidation involves an increase in oxidation state Reduction involves a decrease in oxidation state Recognising this simple pattern is the single most important thing about the concept of oxidation states. If you know how the oxidation state of an element changes during a reaction, you can instantly tell whether it is being oxidised or reduced without having to work in terms of electron-half-equations and electron transfers.

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