Destroyer GC #44 2013ed

[flossit]

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Can someone explain where these 3 el are coming from, when the q. asks to balance redox rxn?

MnO4 + I + H ----- I2 + MnO2 +H2O

 

[Syxx]

Here you go. I've worked it out. The answer that they gave should be 6. 3 electrons are needed to balance the reduction half-reaction.

 

[flossit]

Here you go. I've worked it out. The answer that they gave should be 6. 3 electrons are needed to balance the reduction half-reaction.

Thank you very much!

 

[flossit]

Here you go. I've worked it out. The answer that they gave should be 6. 3 electrons are needed to balance the reduction half-reaction.

I still do not understand why MnO4 has +3? Im getting that Mn has +7. What am i doing wring?

 

[Syxx]

I still do not understand why MnO4 has +3? Im getting that Mn has +7. What am i doing wring?

I think you are confusing formal charge, overall charge of a compound, and overall charge in solution.

The formal charge of Mn in permanganate is +7. The overall charge on permanganate is -1. The +3 refers to the overall charge of the left side of that reaction, the combination of the charges of permanganate and hydrogen. When the charge of the hydrogens (+4) and the charge of permangante (-1) are added together we get an overall charge +3 for everything on the left side. Now the right side of that half-reaction reaction has an overall charge of 0 and we need the overall charges of the left side to equal the right side. That means we need to somehow add something to the left side that can account for a charge of -3 . The way we can do this is by adding 3 electrons since each electron contributes a -1 charge.

 

[flossit]

I think you are confusing formal charge, overall charge of a compound, and overall charge in solution.

The formal charge of Mn in permanganate is +7. The overall charge on permanganate is -1. The +3 refers to the overall charge of the left side of that reaction, the combination of the charges of permanganate and hydrogen. When the charge of the hydrogens (+4) and the charge of permangante (-1) are added together we get an overall charge +3 for everything on the left side. Now the right side of that half-reaction reaction has an overall charge of 0 and we need the overall charges of the left side to equal the right side. That means we need to somehow add something to the left side that can account for a charge of -3 . The way we can do this is by adding 3 electrons since each electron contributes a -1 charge.

Got it! Thank you!