Oxidation & Reduction (chem)
-
Hi all. I have a question regarding this topic which i am unsure of. please try your best to help.i would appreciate it real much!
To test for a reducing agent
Acidified potassium dichromate(VI) can be used to test for the presence of a reducing agent. Acidified potassium dichromate(VI) is made by adding H2SO4 to aqueous potassium dichromate(VI). colour of acidified potassium dichromate(VI) solution changes from orange to green in presence of reducing agent.
Cr2O7 2- + 14H+ + 6e- -> 2 Cr 3+ + 7 H2O
then it made this funny statement which i completely couldn't comprehend. (in this reaction, the dichromate (VI) ion Cr2O7 2- is reduced to the chromium (III) ion Cr 3+. Cr2O7 2- loses oxygen and oxidation state of chromium decreases from +6 to +3.
My own explanation/observation: from the equation, you can see that the Cr2O7 2- loses oxygen to become the Cr3 atoms. (oxidation). however, what if i were to compare using increase/decrease in oxidation? i thought you would see the -2 charge of Cr2O7 and the charge of Cr, which is 3+ ? then, isn't this an increase in oxidation state, instead of reduction as said in the example? I also do not know how you see that oxidation state of chromuium decreases from +6 to -3.
eh, a bit too lengthy. sorry! and thanks for reading! and yea, do assist me!
-
Decrease in oxidation state is reduction.
The charge of Cr atoms in (Cr2O7)2- is +6.
(+6)×2+(-2)×7=12-14=-2
After the reaction, Cr6+ is being reduced to Cr3+. -
To truly understand RedOx and Oxidation States (O.S.), you need to understand about Electronegativity, and be able to draw out the Kekule structures of the compounds (to see for yourself how electrons in such-and-such a number of bond pairs, are polarized and drawn away from the less electronegative species toward the more electronegative species; to make sense of the MEANING of Oxidation State).
However, because at 'O' levels you're not taught about these, it then becomes an issue of oversimplified principles, blind faith, and "just memorize, don't ask so much" pedagogical limitations.
Assuming you're 'O' levels then, I'll just give the simplified explanation and you'll just have to do your best to work with it.
Reduction and Oxidation is primarily a matter of electrons, gaining and losing electrons. Oxygen is often used to define redox, simply because oxygen is the 2nd most electronegative element (after F), and the Redox definition of losing and gaining oxygen works in virtually all contexts.
The mistake you made, in thinking that the oxidation state had actually increased from -2 to +3, is due to the fact that you can only specify the oxidation state of (atom(s) of) an ELEMENT in a particular compound, and not the compound itself.
That's right. You CAN say the dichromate(VI) ion has a CHARGE of 2-, but you CANNOT say the dichromate(VI) ion has an oxidation state of -2.
You can ONLY say, the oxidation state of CHROMIUM (the element) has decreased, from +6 (in the dichromate(VI) ion) to +3 (in the chromium(III) ion).
Hence, in terms of electrons, the (atom(s) of the) ELEMENT in the compound has GAINED electrons when going from the LHS (ie. the dichromate(VI) ion), to the RHS (ie. the chromium(III) ion).
In terms of oxygen, the COMPOUND on the LHS (ie. the dichromate(VI) ion) has LOST oxygen to become the compound on the RHS (ie. the chromium(III) ion).
>>> I also do not know how you see that oxidation state of chromuium decreases from +6 to -3. <<<
For 'A' levels :
Draw the Kekule structure of the dichromate(VI) ion, note the number of bond pairs each chromium atom shares with oxygen (ie. 6), realize that in each bond pair, the electrons are polarized toward the more electronegative oxygen and hence each chromium atom has 'macham' lost 6 electrons; then compare it to the (obvious) oxidation state of +3 in the chromium(III) ion.
(Image owned by and courtesy of http://pubchem.ncbi.nlm.nih.gov )
For 'O' levels :
1. The oxidation state of oxygen is -2 (it's always -2 except in the case of either bonded with lagi electronegative fluorine (whereupon F gets priority of -1 to achieve stable octet, and O.S. of oxygen is variable), or in the case of peroxide ion -O-O-; since oxygen kakilang, oxygen equally electronegative with oxygen, so they won't bully themselves, so each share the electrons in the bond between the 2 Os equally, so O.S. of oxygen is -1 in peroxide ion)
2. Let the oxidation state of Cr be X (algebraic variable). Hence (2)(X)+(7)(-2)=-2; X = +6.
3. Hence oxidation state of Cr on LHS is +6, on RHS is +3.
We Love Chemistry! -
Thanks guys for helping!
UltimaOnline, may I clarify with you some of my doubts?
Earlier you mentioned "Hence, in terms of electrons, the (atom(s) of the) ELEMENT in the compound has GAINED electrons when going from the LHS (ie. the dichromate(VI) ion), to the RHS (ie. the chromium(III) ion)."
How do you know that the compound has gained electrons? through the charges of Cr2O7 and Cr, which is -2 and +3 respectively? Does this mean an increase in electrons?
Thanks again for helping. (=
-
Hi anpanman,
>>> How do you know that the compound has gained electrons? through the charges of Cr2O7 and Cr, which is -2 and +3 respectively? Does this mean an increase in electrons? <<<
It's not the compound that has gained electrons, it's the (atoms of the) element chromium that has gained electrons. We know this, not through the charges directly (because you have to take into consideration the other atoms/ions/species within the compound that contribute to the overall charge), but directly, through the oxidation state of the (atoms of the) element chromium.
Gaining electrons (which are -vely charged) means a decrease of the oxidation state, ie. REDUCTION.
So simple bottomline answer - always work out the oxidation state of the element (that you've identified to have variable oxidation states, eg. transition metals) on the LHS, and compare it with it's oxidation state on the RHS; to determine if it has been reduced or oxidized.
Furthermore, you should practice writing reduction and oxidation half-equations, ie. with electrons in the equation. Electrons appear on the LHS for reduction half-equations, and on the RHS for oxidation half-equations.
For a good practice question (for both 'O' levels and 'A' levels, even though it may be somewhat challenging for 'O' level students, but it's do-able), see the 6th post in my thread below :
http://www.sgforums.com/forums/2297/topics/320107
>>> Thanks again for helping. (= <<<
You're welcome, may you enjoy Chemistry!
-
sorry to ask, but my teacher told me that to determine the number of electrons gain or lost, we simply look at the oxidation state number. is this true? sounds so weird.
-
Originally posted by anpanman:
sorry to ask, but my teacher told me that to determine the number of electrons gain or lost, we simply look at the oxidation state number. is this true? sounds so weird.
Yes, that's right.For instance, if you determine by stoichiometry that the number of mol of electrons lost PER MOLE of a compound oxidized is x (eg. 2), and if the final oxidation state is +y (eg. +5), then the intial oxidation state (let it be z) must have been (z) + (+2) = +5; z = +3.
As another example, if you determine by stoichiometry that the number of mol of electrons gained PER MOLE of a compound reduced is x (eg. 2), and if the initial oxidation state is +y (eg. +5), then the final oxidation state (let it be z) must have been (+5) + (-2) = z; z = +3.
Fun?
-
Originally posted by UltimaOnline:
Yes, that's right.For instance, if you determine by stoichiometry that the number of mol of electrons lost PER MOLE of a compound oxidized is x (eg. 2), and if the final oxidation state is +y (eg. +5), then the intial oxidation state (let it be z) must have been (z) + (+2) = +5; z = +3.
As another example, if you determine by stoichiometry that the number of mol of electrons gained PER MOLE of a compound reduced is x (eg. 2), and if the initial oxidation state is +y (eg. +5), then the final oxidation state (let it be z) must have been (+5) + (-2) = z; z = +3.
Fun?
Yea, think I'm catching it.
I have got 2 examples which I have difficulties understanding, with regards to what I mentioned above(the oxidation state no., etc)
say an equation goes something like:
O2 + 4e- -> 2O 2-
(0) (-2)
take that the numbers in brackets are the oxidation state numbers. so, in this case, we have a decrease in oxidation state from 0 to -2. so shouldn't it be a loss of 2 electrons? i think i'm having some misconceptions about stuff, but i do not know where i went wrong.
next is 2N 3- -> N2 + 6e-
(-3) (0)
we can see that this is a GAIN OF ELECTRONS judging from the equation given. then this should be a case of reduction. however, the oxidation state increases. shouldn't it decrease instead?
yea, i think that's as far as i do not understand. I would really need you to help me clear up my "wrong" thinking if there's any.
-
1. Gain of electron(s) result in reduction or decrease in oxidation number (more negative). 2. The N atoms LOSES electrons. The "excess" electrons on the RHS of the equation are what is lost; only do electrons on LHS combine with other atoms for them to gain electrons.
-
Tamago has already addressed your queries. Is it ok now?
Remember, electrons on LHS means adding e- to reactants so reactants kena reduced. Likewise, electrons on RHS means removing e- from reactants so reactants kena oxidized.
>>> in this case, we have a decrease in oxidation state from 0 to -2. so shouldn't it be a loss of 2 electrons? <<<
The reactants have gained 4 electrons, each O atom (of the O2 molecule) has gained 2 e- to form 2 O2- ions.
Understand liao?
-
Thanks. Think i should be able to understand better now!
