Balancing net ionic reactions?

Question

Solid calcium is reacted with 40.0mL of 0.250M potassium dichromate in acidic solution. What is the sum of the coefficients for the balanced net ionic reaction?

Ca(s) + Cr2O7^-2 (aq) ---> Ca^+2(aq) + Cr^+3 (aq)

PLEASE SHOW ME HOW YOU GOT YOUR ANSWER!!

This is what I did:::

Ca + 2Cr + 7O ---> Ca + Cr

I tried to balance here::

14H+ + Ca + 2Cr + 7O ---> Ca + 2Cr + 7H2O + 1E-

My net ionic reaction was then::

14 H+ + 7O ---> 7H20 + 1E

Which was incorrect

Answer 1

This is a redox reaction and I would use the half-reaction method to balance it.

Ca(s) + Cr2O7^-2 (aq) ---> Ca^+2(aq) + Cr^+3 (aq)

3(Ca  -->  Ca++ + 2e-)

14H+ + Cr2O7=  + 6e-  --> 2Cr3+ + 7H2O

--------------------------------------...

14H+ + 3Ca + Cr2O7=  -->  3Ca++ + 2Cr3+ + 7H2O

Answer 2

OK - You've a very strong reducer, metallic calcium wanting to donate a couple of electrons to become cations (2+).  You also have a potent oxidizer in dichromate ions, which have hexavalent chrome cations(6+) that would rather be Cr(3+).

1)  Write out the redox reactions so that the net charges are balanced:

(It looks like you got the proton count worked out for the dichromate OK.)

Ca -> Ca(2+) + 2e-

Cr2O7(2-) + 6 e-  + 14H+ -> 2 Cr(3+) + 7 H2O

2)  You need to scale the coefficients to combine them so the electrons are balanced out.



3Ca(s) -> 3Ca(2+) + 6 e-

Cr2O7(2-) + 14H+ + 6 e- -> 2Cr(3+) + 7 H2O

gives  3Ca(s) + Cr2O7 (2-) +14H+ -> 3 Ca(2+) + 2Cr(3+) +7H2O

Net charge -2 + 14 = 6 + 6  OK

3) Sum up the coefficients so the instructor can check your answer really quickly.  

BTW - This reaction is insanely spontaneous and would heat the solution up, perhaps to the boiling point.