which is the correct balanced equation for a given titration process, (a) or (b)?
a) 6H+ + 2MnO4- + 3H2O2 -> 4O2 + 2Mn2+ + 6H2O
b) 6H+ + 2MnO4- + 5H2O2 -> 5O2 + 2Mn2+ + 8H2O
Process:
Make a solution in a 100mL volumetric flask using 1g of NaBO3*4H2O, 20mL of 1m H2SO4, and distilled water. Using this solution take 10mL of it and place in a beaker and add 20mL of 1M H2SO4. Now take the 30mL solution beaker and titrate with 0.0103M of KMnO4 to the endpoint. (endpoint found to be ~29mL)The H2O2 come from NaBO3*4H2O in acidic solution.
**I think the answer is (b) because this is an acidic solution, but i'm not sure. Can you please tell me the error of my ways if I am incorrect? Also, is that a way to find this out using calculations? (stoichiometry perhaps or another way like?) I think knowing the theory behind this would be helpful in the future. please help me, i'd love your feedback.
You can't tell from the H^+ since both a and b show H^+, an acidic solution.
b is correct. The long way is to balance the two half equations,
MnO4^- ==> Mn^+2
H2O2 ==>O2
Doing so will give you (b).
A quickie, but a little confusing with this problem, is to make sure the loss of electrons equals the gain of electrons.
For (b), Mn goes from +7 on the left to +2 on the right and that for 2 atoms Mn = 10 electrons gained. For H2O2, O goes from -2 (for both O atoms) to zero, a loss of 2 electrons and for 5 molecules = 10 electrons. gain = loss. ok. and the atoms balance.
For (a), the atoms balance but electron gain is not equal to electron loss. Mn goes from +7 on the left to +2 on the right and for 2 atoms that is 10 electrons gained. For 3H2O2 you have -6 charge total (for 6 O atoms) going to zero on the right for 6 electrons lost (not 10) and in fact those O atoms don't balance (6 on left going to 8 on right) although the total on both sides is the same.
thank you so much. you're my hero =)
Hello sarah!
I had the same dilemma in my chemistry class today and I asked my chemistry teacher about it; he could not give a clear answer to this question.
DrBob222's analyse is quite correct but for one thing: there are two O-atoms from MnO_4^- that could theoretically give away together 4 electrons, balancing for the 10 electrons gained by the two Mn^+7, which then goes to Mn^+5, when you take into counting the other 6 electrons that are at disposal from the O-atoms in hydrogenperoxide. This is the only solution to the problem as far as I can see.
Simply put:
4 electrons from 2 O-atoms in MnO_4^- + 6 electrons from 6 O-atoms in H_2O_2 = 10 electrons
Now you will have as well in total eight O-atoms all giving away their extra electrons and forming new covalent bondings with each other, i.e. 4 O_2. Now the two supposedly missing O-atoms are accounted for.
Since my chemistry teacher could not give a clear answer to this problem, I went on for a search on the Internet and found a PDF-file where the balancing of this equation came into questioning. This equation was labelled as an "undetermined system" because it could be described as following:
2MnO_4^- + 6H^+ + (5-2n)H_2O_2 --> 2Mn^+2 + (5-n)O_2 + (8-2n)H_2O
Any given value of n (with the exception of any value that makes the coefficients negative) will give a valid equation that in no way violates the laws of conservation of mass or charge, and the transaction of electrons will be balanced.
However, theory is theory. Whether equation (a) is valid or not in praxis with given standard conditions remains a question. With this particular equation I would use the half-reaction method instead of the oxidation number method.
I hope I may have shed some light on the subject!
To determine the correct balanced equation for the titration process, we need to analyze the reaction stoichiometry and consider the oxidation states of the elements involved.
In this titration process, we have 10mL of a solution containing NaBO3·4H2O, H2SO4, and distilled water. The NaBO3·4H2O is the source of H2O2 in the acidic solution. This solution is then titrated with KMnO4.
The balanced equation for the reaction between KMnO4 and H2O2 in an acidic solution is given as follows:
5H2O2 + 2MnO4- + 6H+ -> 5O2 + 2Mn2+ + 8H2O
Now, let's analyze the options given:
a) 6H+ + 2MnO4- + 3H2O2 -> 4O2 + 2Mn2+ + 6H2O
b) 6H+ + 2MnO4- + 5H2O2 -> 5O2 + 2Mn2+ + 8H2O
Looking at the balanced equation for the reaction between KMnO4 and H2O2, we can see that for every 2 moles of MnO4-, we need 5 moles of H2O2. Therefore, option (b) is the correct balanced equation for this titration process.
To determine the correct balanced equation, it is important to understand the stoichiometry of the reaction and the oxidation states of the elements involved. In this case, we know that Mn is present in its +7 oxidation state in MnO4-, and it gets reduced to Mn2+. H2O2 is the oxidizing agent and gets reduced to O2.
Using stoichiometry and oxidation states, we can deduce and balance the equation. It is a good approach to familiarize yourself with the oxidation states and stoichiometry of common chemical reactions to accurately determine balanced equations.
Hope this explanation helps!