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I have got the homework to balance the following redox reaction.

$\ce{Cr(OH)3- + HO2- -> Cr2O7^{-2} + OH- }$(Basic medium)

To balance this reaction first thing we need to do is to find the oxidation number of each element.

My efforts:

I have found out oxidation number of each element on product side.. But it is bit confusing in reactant side.

Can any help me in finding the oxidation number of elements on reactant side ?

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    $\begingroup$ Hydroxide is a - 1ion (OH)- so you can figure out the first reactant. For HO2- hydrogen rule overrides the oxygen rule so H+ and Oxygen is -1. Certainly makes more sense than Hydrogen being +3 right? $\endgroup$ – rch Jul 18 '14 at 7:32
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About oxidation states, I find it easier to follow the extended definition, as it is explained in this question: Electronegativity Considerations in Assigning Oxidation States

Look at the first compound $\ce{Cr(OH)3- }$. What is the most electronegative element? Assume it gets all the electrons to fill the octet.

Oxygen, it shall have the oxidation state $-2$

What about the other atoms?

Hydrogen can only receive or lose an electron, in this case it is bonded to the much more electronegative element oxygen, it shall have the oxidation state of $+1$
Chromium gets the oxidation state of $+2$, since $\ce{O}(3\cdot -2) + \ce{H}(3\cdot +1) -\ce{e-}(1\cdot-1)=-Ox(\ce{Cr}) = +2$

Look at the hydrogen peroxide ion, apply what you know about Hydrogen to calculate the oxidation state of oxygen.

Hydrogen gets $+1$ (see above)
Charge has to remain, i.e. $-1$ Oxygen gets $-1$, since $\ce{H}(1\cdot +1) -\ce{e-}(1\cdot-1)=-\frac12Ox(\ce{O})=-1$

Now the hydroxyl ion is straightforward.

Hydrogen gets $+1$ (see above)
Oxygen gets $-2$

Look at dichromate and apply the above.

Oxygen gets $-2$, most electronegative
Chromium gets $+6$, since $\ce{O}(7\cdot -2) -\ce{e-}(2\cdot-1)=-\frac12Ox(\ce{Cr}) = +6$

In total that will give you:

$\ce{\overset{{+II}}{Cr}~(~\overset{{-II}}{O}~\overset{{+I}}{H}~)3{}^{-} + \overset{{+I}}{H}~\overset{{-I}}{O}_2{}^{-} -> \overset{{+IV}}{Cr}_2~\overset{{-II}}{O}_7{}^{2-} + {}^{-}\overset{{-II}}{O}~\overset{{+I}}{H} }$

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