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Mcdoodoo said:s10 p42 q4-c
can anyone explain to me how to do this?
MagicmanFarhanul said:Mcdoodoo said:s10 p42 q4-c
can anyone explain to me how to do this?
hey dude ,heres the solution
the half equations
Fe(OH)3+e-------Fe(OH)2+oh ,the e not value is -.56
for Fe 2+ +2e---------Fe,the enot value is.-.44
so more postive the enot value means ,its a more stronger reducing agent,thus gets oxidized more easily,
so in this case fe 2+ is having a more positive value then fe(oh)3 so fe2+ is more stable to oxidation as its a stronger reducing agent .
hope it helps u out
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