Chem p4 help Please!

[Mcdoodoo]

s10 p42 q4-c
can anyone explain to me how to do this?

 

[MagicmanFarhanul]

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

 

[fluffycloud]

How did you calculate the e not value for Fe(OH)3

 

[beacon_of_light]

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

You wont do it like that... the questions says specifically about "acidic" and alkaline conditions... so you need to compare these two E*s like this
O2 + 4H+/2H2O = +1.23V Fe3+/Fe2+ = +0.77 V "acidic" medium since H+ is involved" E* of reaction = +0.46V

O2+ 2H2O/4OH– = +0.40V Fe(OH)3/Fe(OH)2 = –0.56V "alkaline medium since OH- is involved" E* of reaction = +0.96V
Since E* for the second reaction is more positive (higher) than the E* for the first reaction, the second reaction is more likely to take place...