<p>Can't find it in the review book,does that mean its nonexistant/negligible in the actual exam?</p>
<p>im pretty sure they are…at least that’s what my ap chem teacher has told us/made us practice doing</p>
<p>Check <a href=“Supporting Students from Day One to Exam Day – AP Central | College Board”>Supporting Students from Day One to Exam Day – AP Central | College Board; for what’s covered, it starts on page 6. I haven’t taken the course or the exam yet, so I wouldn’t know otherwise, but the collegeboard outlines are pretty resourceful.</p>
<p>Yes, occasionally they do show up on the AP exam.</p>
<p>Yes.
It’s mandatory.</p>
<p>Yes, they were last year.</p>
<p>mrfairladyz, thanks for the link, but Ive already seen it (tried to edit my OP to say its in the Topic Outline but somehow failed)</p>
<p>And no, they don’t talk about it in the Princeton. But the Intro says that if its in the book, if its the test. And if you know everything here, you get a 5. So I thought the redox was negligible</p>
<p>Literally, just look at the rules. They’re pretty simple (Oxidation = H^+, Reduction = OH^-). Practice 2 problems of each and you’re golden.</p>
<p>This is how I’ve been taught to do it (the half-reaction method): [How</a> to Balance Redox Reactions - Balancing Redox Reactions](<a href=“http://chemistry.about.com/od/generalchemistry/ss/redoxbal.htm]How”>How to Balance Redox Reactions)</p>
<p>Ahh ok thanks guys, and thanks for the link.</p>
<p>There’s an easier way to balance redox reactions - you can do it without knowing oxidation states, or even any chemistry for that matter.</p>
<p>1) break the reaction into two half-reactions
2) balance the main atom (whichever one is not H or O)
3) Balance O by adding H2O to the other side
4) balance H by adding H+ to the other side
5) balance the charge by adding e- to the more positive side.</p>
<p>Then multiple the half-reactions as necessary to get the same number of electrons in each. Add the reactions and cancel spectators. All the electrons should cancel out.</p>
<p>All you’re doing is counting and adding that many of something to the other side. If any step doesn’t exist, just skip it (for example, reduction of H2O2 won’t have a step 2, so go to 3.) If the reaction takes place in basic solution, add one OH- for every H+ in the final equation to form H2O, then cancel out pectator H2O molecules.</p>
<p>Once you’ve done a handful of these, they become like a game. You can even balance half-reactions in your head.</p>
<p>Here’s a link to Mark Bishop’s Chemistry Site, with further explanations and examples. <a href=“http://www.mpcfaculty.net/mark_bishop/redox_balancing.htm[/url]”>http://www.mpcfaculty.net/mark_bishop/redox_balancing.htm</a></p>