Question

Standard Reduction Potentials

Answer 1

For a reaction \[Fe(OH)_{2} + O_{2} \rightarrow Fe(OH)_{3}(s) , unbalanced \]

the potential reductions are \[4(Fe^{2+} \rightarrow Fe^{3+} + e^{-})\]\[O_{2} + H_{2}O + 4e^{-} \rightarrow 4(OH)^{-}\]

somebody explain how this reaction balanced becomes \[4Fe(OH)_{2} + 2H_{2}O + O_{2} \rightarrow 4(Fe[OH]_{3})\]

Answer 2

my bad the second part is actually \[O_{2} + 2H_{2}O + 4e^{-} \rightarrow 4(OH)^{-}\]

Answer 3

what exactly u didnt get may i know?
Fe+2 conveerted to Fe+3 by losing electrons
those electron are used in the reduction of O2 and H2O into OH- ions

Answer 4

how does the final result in that form. There was no Hydroxide on the reactants side but the addition of the half reactions show 4Fe(OH)_2

@chmvijay

Answer 5

i didn't get u :(

Answer 6

alright. There is no FeOH on the reactants side for the first half reaction of
Fe2+ -> fe3+ + e so at the end why is there an 4FeOH_2

Answer 7

@chmvijay Why couldn't it be like

2OH + 2H20 + 4e ---.. 6(OH)

Answer 8

THe reaction should be \[4Fe^{+2} + 2H_{2}O + O_{2} \rightarrow 4Fe(OH)_{3}\]

Answer 9

what they say is first it forms Fe+2 hydroxide and then it turns to the FE+3 hydroxide

Answer 10

alright but, why doesn't it show that intermediate step?

Answer 11

Its like they come up by adding the OH_{2} to the Fe+2 out of thin air. when in the half cell reaction there was no OH_{2} on that side. (reactants (left))

Answer 12

yup they have different reduction poetical in first case its reduction taking place mean OH are formed on next reaction its oxidation taking place

Answer 13

So I just stick to the original equation and write it out like a basic reaction with the OH involved. for any case?

Answer 14

for any case?

Answer 15

like If I had to do that reaction again and calculate the kb or something.

Answer 16

i don't know what ur realting this to KB

Answer 17

thanks

Answer 18

yw:)