The trouble is together follows: "Calculate Ecell values for the following cells. I beg your pardon reactions room spontaneous as created (under traditional conditions)? Balance the reaction that are not already balanced."a.) MnO4– (aq) + I–(aq) ——> I2(aq) + Mn2+(aq)So I set up my 2 half-reactions: 5 < 2I– ——> I2 + 2e– Ecell = –0.54v 2 < 5e– + 8H+ + MnO4– ——> Mn2+ + 4H2O > Ecell=1.51vand as soon as you combine them friend get: 16H+ + 10I– + 2MnO4– ——> 5I2 + 2Mn2+ + 8H2O Ecell = 0.97vSo I'm relatively positive that component of the question is right. However, our teacher hasn't yet defined how to find out if the reaction is spontaneous. I would certainly guess the it could have something to execute with the reactivity series, but I wasn't sure and didn't want to opportunity it.Thanks in breakthrough for your *delete me*Lucas

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mikeRetired StaffSr. Member
Posts: 1246Mole Snacks: +121/-35Gender:

If the reduction potential is positive it is voluntarily in the direction written. If the reduction potential was an unfavorable the turning back reaction would certainly be spontaneous.
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lucasloredoNew MemberPosts: 8Mole Snacks: +0/-0
Thanks because that the quick reply, yet I'm quiet a bit confused. So, for instance, the palliation potential because that the first half reaction, together I created it, is –0.54v, so does that make the totality equation spontaneous? i am really confused
might you please sophisticated a small further? We simply learned this principle a couple days back and I'm still no comfortable with the material.Thanks again,Lucas

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Remember that reduction is a gain in electron (OILRIG - oxidation is loss/reduction is gain).So palliation potentials are for the reaction in the reduction direction, if you turning back the reaction (like friend did with iodine) it i do not care an oxidation reaction.The reaction v the greater more optimistic reduction potential will continue spontaneously in the reduction direction as soon as reacted with another compound v a smaller reduction potential (in which instance it will proceed in the oxidation direction).