I need chemistry help. I think 1. c, 2.b, & 3.a (please tell me why or why not and help me set up the problems if possible)
1.) The following half reaction has been balanced except for the electrons. How many and where should the electrons be included?
14 H+(aq) + Cr2O72-(aq) ---> 2 Cr3+(aq) +7 H2O(l)
a.) 6 electrons product side
b.) 6 electrons reactant side
c.) 3 electrons product side
d.) 9 electrons product side
e.) 9 electrons reactant side
2.) Which one of the following statements is true?
a.) Electrons are considered as reactants in oxidation half reactions.
b.) For some balanced REDOX reactions it is possible that
electronslost ≠ electronsgained .
c.) When Ecell<0, the reaction is non-spontaneous.
3.) Balance the following redox reaction if it occurs in acidic solution. What are the coefficients in front of H2C2O4 and H2O in the balanced reaction? MnO4– (aq) + H2C2O4(aq) → Mn2+(aq) + CO2(g)
a.) H2C2O4 = 5 H2O = 8
b.) H2C2O4 = 5 H2O = 1
c.) H2C2O4 = 1 H2O = 4
d.) H2C2O4 = 3 H2O = 2
Thank you DrBob222
1.) To determine the number and placement of electrons in the balanced half-reaction, we need to examine the changes in oxidation numbers. In this half-reaction:
14 H+(aq) + Cr2O72-(aq) ---> 2 Cr3+(aq) + 7 H2O(l)
The oxidation number of Cr changes from +6 to +3, indicating a reduction. The oxidation number of H changes from +1 to 0, indicating a reduction as well.
Since the reduction involves the gain of electrons, we need to include electrons on the reactant side so that the charges balance. The coefficient in front of Cr2O72- is 1, and the total charge of Cr2O72- is -2. The coefficient in front of Cr3+ is 2, and the total charge of Cr3+ is +6. Therefore, to balance the charges, 6 electrons are needed on the product side.
So the correct answer is c.) 3 electrons product side.
2.) The correct answer is b.) For some balanced REDOX reactions, it is possible that electrons lost do not exactly equal electrons gained. This is because in some reactions, there may be multiple steps that involve the transfer of electrons. As long as the overall reaction is balanced and the total electrons lost equal the total electrons gained, the reaction is considered balanced.
3.) To balance this redox reaction in an acidic solution, we first need to assign oxidation numbers to each element. In this reaction:
MnO4– (aq) + H2C2O4(aq) → Mn2+(aq) + CO2(g)
The oxidation number of Mn changes from +7 to +2, indicating a reduction. The oxidation number of C changes from +3 to +4, indicating an oxidation.
To balance the reaction, we can follow these steps:
1. Balance the atoms other than hydrogen and oxygen. In this case, the only non-H and non-O atoms are Mn and C.
MnO4– (aq) + H2C2O4(aq) → Mn2+(aq) + CO2(g)
2. Balance the oxygen atoms by adding water (H2O) molecules to the side that needs more oxygen. Since we are in an acidic solution, we should add H2O on the side that needs more oxygen and balance the hydrogen ions (H+) accordingly.
MnO4– (aq) + H2C2O4(aq) → Mn2+(aq) + CO2(g) + H2O
3. Balance the hydrogen atoms by adding hydrogen ions (H+) to the side that needs more hydrogen.
MnO4– (aq) + H2C2O4(aq) + 5 H+ → Mn2+(aq) + CO2(g) + H2O
4. Balance the charges using electrons. In this case, the charge on the left side is -1 (MnO4-) and the charge on the right side is +2 (Mn2+). Therefore, 3 electrons should be added to the left side to balance the charges.
MnO4– (aq) + H2C2O4(aq) + 5 H+ + 3 e- → Mn2+(aq) + CO2(g) + H2O
Therefore, the correct coefficients for H2C2O4 and H2O are:
H2C2O4 = 1
H2O = 4
So the correct answer is c.) H2C2O4 = 1, H2O = 4.
1. c won't do it.
Count the charge. On the left there is 14+ (from 14H^+) + 2- (from Cr2O7^2-). On the right there is 2*3 = 6+. So you
Add 6e to the left side.
2. b won't get it. In redox reactions electron gain is ALWAYS ALWAYS ALWAYS = electrons lost (if the equation is balanced. The flip side of that is that if the electrons lost do not equal electrons gained then the equation is NOT balanced no matter how you slice it.
Hint: The reaction is spontaneous when Ecell = +
3. You should be able to balance these. What's the problem. What do you not understand. I can give you a web site to help if you need it.