Balancing redox reactions

[pauly11235]

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MnO4- + I- --> I2 + Mn2+

I dont understand why we have to place a 2 in front of MnO4- and a 10 in front of I-. It's number 69 for Gen chem in TopScore test 3.

 

[yakuza]

MnO4- + I- --> I2 + Mn2+

I dont understand why we have to place a 2 in front of MnO4- and a 10 in front of I-. It's number 69 for Gen chem in TopScore test 3.

MnO4- + I- <----> Mn2+ + I2

Redox: ( 5e- + 8H+ + MnO4- ---> Mn2+ + 4H2O) multiply by 2

Oxidize: ( 2I- ----> I2 + 2e-) multiply by 5

you multiply to balance the e-, which gives you.......


10 I- + 10e- +16H+ + 2MnO4- ---> 2Mn2+ + 8H2O + 5 I2 + 10e-

BUT! this is for the acidic balanced reaction..its a whole extra step for the basic balanced reaction

Also when balancing half reactions, balance the O's and H's last, then add electrons to the more positive side

 

[Albuterol]

why are we adding 5e- to this step?
( 5e- + 8H+ + MnO4- ---> Mn2+ + 4H2O)

 

[yakuza]

why are we adding 5e- to this step?
( 5e- + 8H+ + MnO4- ---> Mn2+ + 4H2O)

The product side (Mn2+ + 4H2O) = 2+ charge.........+2 from Mn
The reactant side (8H+ + MnO4-) = +7 charge.......+8 from H, -1 from MnO4-

so you just take the difference of 7 and 2

 

[Albuterol]

got it..thanks for the explanation!