<p>I got a bunch of problems like this one and have absolutely no idea how to work them out. I tried using the nernst equation and plugging the molarity in in all different sort of ways but kept on getting it wrong. if someone can show me how to do it, it would be greatly appreciated</p>
<p>Given the following half-reactions:
Ce^(4+) + e^− → Ce^3+ E° = 1.72 V<br>
Fe^(3+) + e− → Fe^2+ E° = 0.771 V </p>
<p>What volume of 0.76 M Ce4+ solution must be mixed with 20. mL of 0.54 M Fe2+ solution to produce a solution which produces a platinum electrode potential (relative to SHE) of 1.2455 V. </p>
<p>Thanks</p>
<p>how the hell do you do this? I've never seen a problem like this in my life.</p>
<p>^ Tell me about it, it looks a bit too complicated to show up on the AP. ElectroChem problems tend to show up easy on the AP. Whered you get this problem from EverLane?</p>
<p>Yeah dude, I am pretty sure nothing like that will be on the AP.. I hope so at least because I am taking it next tuesday and I'm used to seeing the easy electrochem problems.</p>
<p>lol that problem is ridiculous</p>
<p>i think someting is flawed with that problem. The Voltage (SHE) is independent of the moles of reaction. The only way (that i see) to get the 1.2455V is to add up 1.72 and 0.771 and then divide by two. But unlike Enthalpy or Entropy, the E° stays the same no matter how many times you multiply or divide the equation. Plus, it's already balanced, you just need to flip the Fe one, which doesn't get you the answer anyways.</p>
<p>*** im in ap chem and i have NEVER seen this type of problem
i feel screwed lol</p>
<p>Haha.
You have to use the nerdst equation somehow.
1.2455V = .949 - .0592<em>log (Fe+3</em>Ce+3/Fe+2*Ce+4)</p>
<p>But since you know the reactants molarities, I don't know the point of solving for the product (Fe+3 and Ce+3) concentrations....</p>
<p>i got a bunch of these probs on capa, and i figured out the ones that gave the concentrations of both reactants and products</p>
<p>okay, use the nerd equation, find Q, it cant be that hard. Some of them gotta be solid. Did you consider using M1V1 = M2V2? I mean its worth a shot. I always check with that when i see 2 molarities/volumes where one is missing.</p>
<p>If they give you both the reactant and product concentrations then what is the unknown that you are trying to solve for? </p>
<p>Can someone just answer how to do this problem? lol</p>
<p>Q = .00001 = [P]/[R]</p>
<p>I think the problem is written incorrectly because there should be some solids. The reactants and products both can't be a product of two molarities..</p>