<p>How do I balance these? How much do I need to know about these for Saturday's test? Help!!</p>
<p>Not a giant Chemistry freak – but, uh, you need to generally have the same amount of elements on the reactant and the product side.</p>
<p>what is an ionic equation? I’m currently taking college ochem and I’ve never heard of them :P</p>
<p>If you mean reactions that involve ionic compounds, my advice is to know where the electrons are going.</p>
<p>Net ionic/complete ionic equations? Just like normal equations…</p>
<p>It’s a topic that’s in Kaplan…but I haven’t found it in PR so I guess it’s superfluous?</p>
<p>Dont you just cross out everything that is soluble to get the ionic equation?</p>
<p>Correct me if I’m wrong, but here is what an ionic equation should look like.</p>
<p>EX:</p>
<p>HCl + NaOH –> H2O + NaCl</p>
<p>H+ + Cl- + Na+ + OH- –> H2O + Na+ + Cl-</p>
<p>H+ + OH - –> H2O
(because both sides have a Na+ and Cl-, they cancel each other out and are not included in the final simplified ionic equation.)</p>
<p>^ That’s right. It’s just that the complicatd ones, like the ones involving potassium permanganate, are really annoying to balance</p>
<p>Okay tetrisfan since you mentioned permanganate which is a very strong oxidizing agent I’m going to assume that you are asking about redox reactions.</p>
<p>A redox reaction is a reaction where there is a transfer of electrons. (Some elements are reduced by gaining electrons and others are oxidized by losing electrons.)</p>
<p>The easiest way to balance a redox reaction is to first split the reaction into two half reactions</p>
<p>ex.</p>
<p>CuS(s) + NO3-(aq) —> Cu2+(aq) + SO42-(aq) + NO(g)</p>
<p>The first thing you would do is eliminate the Cu2+ because it is not reduced or oxidized.</p>
<p>S2-(aq) + N03-(aq) —> SO42-(aq) + NO(g)</p>
<p>This is called an unbalanced net ionic equation. Next split it up into half reactions.</p>
<p>S2- —> SO42-
NO3- — NO</p>
<p>These are unbalanced half reactions, as you can see oxygen is unbalanced as is charge.
Balance the oxygen first by adding H2O.</p>
<p>S2- + 4H2O —> SO42-
NO3- —> NO + 2H2O</p>
<p>Now oxygen is balanced but hydrogen is unbalanced as is charge. Now if the reaction occurs in an acidic solution add H+ to the side deficient in hydrogen, if the reaction occurs in a basic solution add H20 to the side deficient in hydrogen and the same number of OH- to the other side.</p>
<p>In acidic solution:</p>
<p>S2- + 4H2O —> SO42- + 8H+
NO3- + 4H+ —> NO + 2H2O</p>
<p>In basic solution:</p>
<p>S2- + 4H2O + 8OH- —> SO42- + 8H20
NO3- + 4H2O —> NO + 2H2O + 4OH-</p>
<p>Next balance the charge by adding electrons which have a charge of -1</p>
<p>S2- + 4H2O —> SO42- + 8H+ + 8e-
NO3- + 4H+ +3e- —> NO + 2H2O</p>
<p>Now multiply all of the coefficients in each equation so that the number of electrons transferred in each is equal.</p>
<p>(S2- + 4H2O —> SO42- + 8H+ + 8e-)<em>3
(NO3- + 4H+ +3e- —> NO + 2H2O)</em>8</p>
<p>By multiplying you get:</p>
<p>3S2- + 12H2O —> 3SO42- + 24H+ + 24e-
8NO3- + 32H+ +24e- —> 8NO + 16H2O</p>
<p>Now add the equations together</p>
<p>3S2- + 12H2O + 8NO3- + 32H+ + 24e- —> 3SO42- + 24H+ + 24e- + 8NO + 16H2O</p>
<p>Now cancel what appears on both sides</p>
<p>3S2- + 8NO3- + 8H+ —> 3SO42- + 8NO + 4H2O</p>
<p>And now you’re done. This will work no matter what the equation is as long as it is a redox reaction. There are multiple ways of balancing these equations and this is probably the easiest. I highly doubt you’ll have to do any really difficult redox balancing on the chemistry SAT, but its always good to know.</p>
<p>^ Redox and ionic equations aren’t the same.</p>
<p>I don’t think there will be very much stuff about ionic equations. It’ll all be pretty simple, if there is anything. Just don’t worry.</p>
<p>Obviously redox equations are different than ionic equations, but I assumed the OP used the wrong term considering that a 5 year old could balance ionic equations whereas redox equations are much more difficult.</p>