May Z.

asked • 02/14/21

2MnO4- + 5H2S+ 6H+2Mn2+ + 5S+ 8H2O. The oxidation state of manganese changes from what to what? How many electrons are transferred in the reaction?

Stanton D.

Hi May, Well, first of all, you didn't write a reaction there. No = sign! But if you had done so, you would have been able to trace oxidation states. It;s done by difference. Watch: For MnO4-, you add up the oxygens, each at formal charge -2. That's because O is very electronegative, so it gets assigned those electrons it craves! At 4 O's per MnO4- ion, that would be -8. So Mn(+?) - 8 = -1 (the net charge on the permanganate ion). Solve for ?, find that it's +7. Now how about Mn(2+)? Obviously +2, there it is, so to speak. So the change per Mn is -5, and per 2Mn, -10. The Mn's gain 10 electrons in all, and the S's lose 10 electrons in all, converting from S(-2) to S(0), each. Those are the two atoms which undergo transfer of charge; the H's just hop from one electrophile (S) to another (O); they stay at +1 formal charge, which is what they are in most compounds. --Cheers -- Mr. d.
Report

02/14/21

1 Expert Answer

By:

Still looking for help? Get the right answer, fast.

Ask a question for free

Get a free answer to a quick problem.
Most questions answered within 4 hours.

OR

Find an Online Tutor Now

Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.