BALANCING REDOX REACTIONS BY THE ION

     

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Topic: FeI2 + HNO3 ---> Fe(NO3)3 + HIO3+ NO +H2O is it right? (Read 586 times)




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xshadowFull MemberPosts: 427Mole Snacks: +1/-0
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Have to lớn balance this redox with ionic half reaction :FeI2+ HNO3 ---> Fe(NO3)3 + HIO3 + NO + H2Omethod a) FeI2 insolubleOX: FeI2 --> Fe3+ + 2 IO- + 12 H+ + 13e- X 3RID: NO3- + 4 H+ + 3e- --> NO + 2H2O X 13I'll get:3 FeI2 + 13 NO3- + 16 H+ --->3 sắt 3+ + 6 IO3- + 13 NO + 8 H2ONOW if I considered FeI2 as a ""soluble species "" (even if it's less soluble) I have to consider two oxditaion half reaction:OX1 sắt 2+ ---> Fe3+ + 1e-OX2 2I- + 6H2O--> 2 IO3- + 12 H+ + 12 e- OX TOT = OX1 + OX2Or OX2 is it :I- + 3H2O ---> IO3- + 6 H+ + 6e-In other word when I consider the iodine half reaction I have khổng lồ consider I- ---> IO3- or 2I- ---> IO3-Because I think only in the second way it worksThanks
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mjc123ChemistSr. Member
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Posts: 2020Mole Snacks: +292/-12
You have lớn get the correct stoichiometry between OX2 and OX1. If 2 I- react for every Fe2+ that reacts, then OX1 và OX2 are as you first wrote them, so that OX = OX1 + OX2 works out correctly (i.e. 13 electrons are produced for every Fe2+ that reacts).Also cảnh báo as a general point, here and in your other post, that half reactions must balance, just like ordinary reactions. You cannot write e.g.
BorekMr. PHAdministratorDeity Member
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Posts: 27252Mole Snacks: +1771/-408Gender:
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I am known to lớn be occasionally wrong.
There two separate oxidation processes here: Fe2+ oxidation, & I- oxidation. That usually means you can't write a single redox reaction equation, unless there is some additional information - in this case, you know you have two I- for each Fe2+, so the total reaction is a sum of one Fe2+ oxidation plus two I- oxidation reactions.


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