When non metal react they form molecule, and whether the bond is polar or not, the elements in the molecular compound will not have charges what so ever, because they are not ionic. But can non metals be the only that is involved Ina redox reaction ?
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asked Jun 23, 2014 at 22:29
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3 Answers
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A redox reaction is one in which oxidation state of some atom changes, and such reactions need not involve metals, nor any ionic species. Oxidation state is the idealized charge that a given atom would have if all its bonds are treated as completely ionic. Thus, strictly for the purpose of calculating oxidation state, the electrons of a covalent bond are typically assigned to the more electronegative atom. (I'd note, however, that IUPAC provides a different set of rules for calculating oxidation number.) Hence, in order for a redox reaction to occur, there does not need to be a complete transfer of electrons (in the sense of ionization). Examples of redox reactions involving molecular compounds are numerous.
For example, consider the reaction $\ce{N2 + 3H2 <=> 2NH3}$. This is the Haber process for the production of ammonia. Here nitrogen is reduced from an oxidation state of $0$ to $-3$, while hydrogen is oxidized from $0$ to $+1$. Again, nitrogen being the more electronegative atom, is assigned the electrons of the bonds. This idealization is useful in that it provides some rough idea of electron distribution, and hence has some utility in predicting reactivity.
Edit: Also, the statement "[...] the elements in the molecular compound will not have charges what so ever [...]" is inaccurate. There is a concept of formal charge that applies to neutral molecules as well as ions.
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1$\begingroup$ Related: chemistry.stackexchange.com/q/10944/4945 $\endgroup$– Martin - マーチン ♦Jun 24, 2014 at 2:55
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$\begingroup$ @Martin, thanks, and +1 for your answer to that question. The IUPAC choice of seemingly arbitrary rules is indeed bizarre. $\endgroup$– Greg E.Jun 24, 2014 at 2:58
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$\begingroup$ Thanks. I find it a kind of reassuring, that the IUPAC sometimes makes mistakes too/ does not choose the most convenient pattern. $\endgroup$– Martin - マーチン ♦Jun 24, 2014 at 3:03
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1$\begingroup$ @Martin, yeah, I can understand that. I do feel that this is one of their more egregiously bad decisions, however. Maybe their objection is to the fact that electronegativity has many definitions. That said, I'm fairly sure the various different major scales preserve the same relative ordering the overwhelming majority of the time. $\endgroup$– Greg E.Jun 24, 2014 at 3:06
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A redox reaction is any reaction in which the oxidation states of the reactants change. There are certainly reactions involving metals which fit that classification. A relatable example would be the rusting of iron (one of the possible reactions involved in rusting):
$$\ce{2Fe + O_2 + 2H_2O -> 2Fe(OH)_2}$$
Here iron is oxidized and its oxidation state changes from 0 to +2.
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$\begingroup$ All imma asking is could there only exist $non metals$? $\endgroup$ Jun 23, 2014 at 23:51
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$\begingroup$ If that doesn't answer your question, I don't know what you're asking... $\endgroup$ Jun 23, 2014 at 23:53
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$\begingroup$ If elements in a molecule do not have charges, how will the redox reaction go on? Since there is no way to know the increase or decrease in charges. $\endgroup$ Jun 23, 2014 at 23:56
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Yes. Apart from Greg's answer there are many more examples. But since you asked specifically about a redox reaction that may produce ions, I am glad to provide an example. This reaction is very common chemistry, as it is used in public pools. It is the redox dissolution of chlorine in water (coupled with an acid-base reaction). It is also a disproportionation reaction. \begin{array}{lcll} \ce{1/2 Cl2 + e- &-> & Cl-} &\text{reduction}\\ \ce{1/2 Cl2 + H2O &-> & HClO + H+ + e-} &\text{oxidation}\\\hline \ce{Cl2 + H2O &-> & HClO + HCl} &\text{redox}\\\hline \ce{HClO + H2O &<=>& H3+O + ClO-} &\text{acid-base}\\ \ce{HCl + H2O &<=>& H3+O + Cl-} &\text{acid-base}\\\hline \ce{\overset{\color{\green}{0}}{Cl2} + 3H2O &<=>& 2H3+O + \overset{\color{\green}{+I}}{Cl}O- + \overset{\color{\green}{-I}}{Cl- } } &\text{total}\\\hline \end{array}
Depending on the concentration and the $\ce{pH}$ of the solution, the extent of ionisation will vary.
answered Jun 24, 2014 at 3:34