Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers and students. I am trying to balance the REDOX equation shown below, but I got stuck after balancing the major atoms. Whenever you balance a redox reaction in acidic or basic solution, you have to account for interactions with the solution itself. This is a key detail that I find a lot of students miss - you cannot balance a reaction like this without considering the solution, because the solution itself is participating.
At any rate, once you know that you need to use the half-reaction method, the next step is to use it! This tutorial has a good description of the method, your book probably does as well, and I'll give a brief description here for good measure.
Multiply the half-reactions by the appropriate coefficients so that when they are added together, the charges (electrons) cancel - this is necessary because electrons are not created or destroyed in chemical reactions. Add the equations together, and if the solution is basic, add $\ce{OH^-}$ to both sides to remove all protons.


While this is easy to understand, I have found that in practice, a majority of mistakes happen here. This part is also prone to error - be careful and remember to multiply every species by the coefficent.
If the electrons don't cancel out here, then you made a mistake - go back and check each step. If the solution is basic, at this point you would add $\ce{OH-}$ to both sides of the equation until the 16 protons were eliminated. Not the answer you're looking for?Browse other questions tagged homework redox reaction or ask your own question. What is the balanced equation for the reaction of potassium permanganate, iron sulfate, and sulfuric acid?
Your clue that this is happening is the phrase "in acidic (or basic) solution" - any time you see that in a redox problem, you should be thinking "half-reaction method." Another clue is that H and O appear on only one side of the equation. The quick way is to make sure that the elements that are not O and H are in their own equations.


Be careful when counting charges - remember to multiply by the coefficients, and remember that the charges don't have to add up to zero yet - they just have to have the same sum on both sides. We can add anything we want as long as it's to both sides (since this is, after all, just algebra). If you are familiar with the algebraic method of balancing equations, another clue is that the system of equations would be unsolvable. The proper way is to figure out what is being reduced and what is being oxidized, then split them so that the oxidation happens in one reaction and the reduction happens in the other. Each $\ce{OH^-}$ will react with a proton to form one water molecule, and so the waters on the right would cancel out some of the waters on the left.



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