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

2 answers

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.
Thank you DrBob222