Struggling with Redox Reaction Equations?

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The discussion centers on the challenges of balancing redox reaction equations, particularly the calculation of electron transfer in half-equations. The original poster understands oxidation states but struggles with accurately balancing electrons. A response provides a method for balancing electrons by analyzing the charges on both sides of the half-equation and adjusting the electron count accordingly. It emphasizes the importance of considering coefficients in front of ions when calculating total charges. This guidance aims to clarify the process of balancing redox reactions effectively.
koosney@stcc
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hey I'm in yr 12 chemistry in Australia and I am having problems understanding redox reactions equations, i understand oxidation states and a little bit but i can't figure our about the electrons calculus, i can do the rest of the formula or steps but i cannot seem to be able to calculate the electron transfer accurately i was hoping some one would be able to help with methods or something.
thank you
 
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thanks but that doesn't really help me calculate how to balance electrons in half equations (thats the only problem i am having)
 
Hi,
I think I understand what you are having trouble with. When you are finished balancing a half-equation and let's say it looks like this:
Au3+(aq) + 2I−(aq) -> Au(s) + I2(s)
(3+ and - on left side are charges), so you have Au (3+) and I (2-) on left, and on right you don't have any charges as you can see. Notice that when you have a number in front like 2I, make sure you don't forget that number and see only one (-) at the top, you actually have (2-). So, now 2 out of 3 protons of Au on left will balance 2 electrons on left and you will be left with one proton (1+) on left. Then, if you want to balance electrons, you see that you have zero on right and you want same on left so you add one e- to the left to cancell the proton (+) you had left over after you balanced your equation. Same thing if you had in the end 4+ on left and 2- on right of equation you would add 6- to left to cancell 4+ and have 2- left over to balance the right side.

Hope this helps
Jaska
 
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