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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: ChemGuy on January 23, 2005, 03:10:53 PM

Title: Balancing Redox Equations
Post by: ChemGuy on January 23, 2005, 03:10:53 PM

Aside from the fact some think Redox reactions are difficult, I think they are rather easy...for the most part...just one question:

Fe^2+(aq) + Cr2O7^2-(aq) --> Cr^3+(aq) + Fe^3+(aq)

After all the addiding of the water and protons, I ended up with thisbalanced redox:

14H+(aq) + 3Fe^2+(aq) + Cr2O7^2-(aq) --> 3Fe^3+(aq) + 2Cr^3+(aq) + 7H2O(l)

(Iron is being oxidized, and chromium is being reduced)

Title: Re:Balancing Redox Equations
Post by: Donaldson Tan on January 23, 2005, 06:49:04 PM

Fe²⁺ <-> Fe³⁺ + e

Cr₂O₇^2- + 6e + 14H⁺ <-> 7H₂O + 2Cr³⁺

combining these half equations, u need to add 6sets of the first equation into the 2nd one to balance the electrons.

Cr₂O₇^2- + 6Fe²⁺ 14H⁺ <-> 6Fe³⁺ + 7H₂O + 2Cr³⁺