Redox Reactions on Chem Sat?

Test Preparation SAT Subject Test Preparation
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<p>How hard are the redox reactions on the chem sat? I was taking a practice test from barron's and their redox reactions are extremely hard to balance/solve. ex: KMnO4 +H2So3 -> K2So4 +MnSo4 +H2So4 +h2o. Apparently I was supposed to balance it with the method of electron exchange, a method my teacher never taught us to do. Do we need to know how to balance redox reactions for these hard barron-like equations on the chem sat?</p>

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<p>This is an important inquiry.</p>

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<p>lmao bro isn’t that from the first practice test? or something…
anyways electron exchanging is basically like doing reduction oxidation i think?(lmao just guessing. why don’t u skim through barrons for it?)</p>

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<p>I know that electron exchanging is explained in the Barron’s book. My question is whether or not the technique is actually applied on the actual Chemistry SATII test, whether such a question will appear at all. I’ve taken AP chem and the USNCO and never heard of the electron exchanging technique until now. There’s only two days left so I’d rather review stuff I aleady know rather than learn a new concept.</p>

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<p>It’s the same concept. IDK what electron exchange is but I just knew how to balance oxidation reduction equations and I got that question…</p>

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<p>You need to be able to identify redox reactions (bare minimum). It certainly wouldn’t hurt to know how to balance them as well, but the balancing eq. questions I got were not redox eq.</p>

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<p>If I remember correctly, there were a couple REDOX balancing on the May 2011 test. I use the oxidation/reduction half-reaction method, which may or may not be “electron exchange” as mentioned above.</p>

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<p>Do we need to know any electrochem for the sat 2? Im kinda confused about it.</p>

<p>Ex) Electrolysis of Nacl (aq) </p>

<p>What appears at the anode? Chlorine, Oxygen, or both and why??</p>

<p>And also, how can you use the E values to predict what appears?</p>

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<p>@shining knight: Electrolysis just means breaking the thing apart using water (I think) , so it will be NaCl –> Na+ + Cl-. And the way you can predict which is the anode is that anode = oxidation, Na is oxidized (Na –> Na+) so it is at the anode. I might be wrong, if so, then I’m sorry for confusing you further. Barrons’ is really stupid about Oxidation-Reduction.
If the reaction occurs, that means E is positive, so the more positive REDUCTION value will be the cathode.</p>

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<p>So the redox problem I had balancing was KMnO4 +H2So3 -> K2So4 +MnSo4 +H2So4 +h2o
Barron’s says the oxidation half reaction is:
5So3^-2 +5H2O -> 5SO4^-2 +10H+ +10 e-</p>

<p>Does that imply that H2So3 is soluble in water and completely ionizes, or is it a mistake.</p>

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<p>^Yes, sulfuric acid is a strong acid.</p>

<p>Shingingknight: [electrolysis</a> of nacl - Google Search](<a href=“electrolysis of nacl]electrolysis - Google Search”>electrolysis of nacl - Google Search)</p>

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<p>^H2so3 is sulfurOUS acid (a weak acid) not sulfuric acid. So barron’s was wrong?</p>

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<p>Oops, sorry. Missed the subscript.</p>

<p>I did it for you without dissociating H2SO3:</p>

<p>10e- + 16H+ + 2MnO4- > 2Mn2+ + 8H2O (half-reaction)
5H2O + 5H2SO3 > 5SO42- + 20H+ + 10e- (half-reaction)
2MnO4- + 5H2SO3 > 2Mn2+ + 3H2O + 5SO42- + 4H+</p>

<p>That’s in net ionic form; the products are really K2SO4 + 2MnSO4 + 2H2SO4 + 3H2O. Is this the answer from the book?</p>

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<p>You did it completely right and got the right products. Good job!</p>