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Coefficients while balancing ionic equations 
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#1
May212, 12:30 PM

P: 35

Hello. I'm having some trouble balancing ionic equations..
Are we supposed to consider the coefficients of the reactants/products? I came across a contradiction in the examples given in my book : 1) Na + H(+1) → Na(+1) + H2 So in order to figure out the oxidation/reduction part, we will find out the oxidation number. The oxidation number of Na would be 0 as it is in elemental form and since in the RHS it has an oxidation number of +1, it is being oxidized. In case of H, we will balance the element first. So, 2H(+1) → H2 My question is, in this case, what will be the oxidation number of H in the LHS? +1 or +2? According to the book it is +2 so that means we would be considering the coefficients. 2) I(1) + Br2 → I2 + Br(1) So again we would figure out what is being oxidized and what is being reduced. And here, since the oxidation number of I is being increased from 1 to 0, it is being oxidized. And in case of Br, we would balance the element first. So, Br2 → 2Br Again, what would be the oxidation number of Br? 1 or 2? Do we consider the coefficient? Please explain the answer..Thankyou 


#2
May312, 04:57 AM

Admin
P: 23,600

Oxidation number changes when the element loses or gains the electrons. So, when H^{+} (ON equal +1) gets reduced to hydrogen H_{2} (ON equal 0) half reaction equation is
2H^{+} + 2e^{} > H_{2} as you need two electrons to balance the charge. 


#3
May312, 06:28 AM

P: 35

So we do not consider the coefficients but we do add the number of electrons according to the charge that needs to be balanced. But suppose that we have to figure out whether a substance is getting oxidized or reduced, then will consider the coefficient?
For example, Cl2 + HC2O4(1) > 2CO3(2) +2Cl(1) So here the oxidation number of Cl2 on the left hand side is 0. What will it be on the right hand side? 1 or 2? Whatever it will be, it is getting reduced. Now in the case of carbon, it is 6 ( we do consider the subscript right? ) and if we balance it on the left hand side, it has a coefficient 2. So what will be the oxidation number of carbon here? 2 or 4? Whatever it will be, it would be reduction. So what is getting oxidized here? 


#4
May312, 07:35 AM

Admin
P: 23,600

Coefficients while balancing ionic equations



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