When the following reaction is balanced in acid solution, the coefficient of H+(aq) will be:
Cr2O72-(aq) + Sn2+(aq) --> Cr3+(aq) + Sn4+(aq)
5 answers
You need to be able to do these yourself. What do you not understand?
I understand you are supposed to separate into half reactions and balance using H20 H+ and e- however, when i start doing that I get confused halfway through.
this is as far as I have gotten:
4e- + 7H+ + Cr2O72- = 2Cr3+ +7H20
2Sn2+ = 2Sn4+ +4e-
then cancel electrons...is this right?
this is as far as I have gotten:
4e- + 7H+ + Cr2O72- = 2Cr3+ +7H20
2Sn2+ = 2Sn4+ +4e-
then cancel electrons...is this right?
4e- + 7H+ + Cr2O72- = 2Cr3+ +7H20
Since you didn't write the entire equation I can't tell if everything is separated or not; however, the Cr half reaction is separated properly. The trouble you have in balancing is because you have assigned the wrong number of electrons. Cr in Cr2O7^2- is +12 for the two and +6 on the right for the two. That makes all of the other numbers wrong.
6e + Cr2O^2- ==> 2Cr^+3
Now you count charges on the left (-8) and on the right(+6) and add 14 H^+ on the left.
14H^+ + 6e + Cr2O7^2- ==> 2Cr^3+
Now add 7H2O on the right to balancae.
14H^+ + 6e + Cr2O7^2- ==> 2Cr^3+ + 7H2O.
I don't know if you've separated the Sn half cell properly or not.
Since you didn't write the entire equation I can't tell if everything is separated or not; however, the Cr half reaction is separated properly. The trouble you have in balancing is because you have assigned the wrong number of electrons. Cr in Cr2O7^2- is +12 for the two and +6 on the right for the two. That makes all of the other numbers wrong.
6e + Cr2O^2- ==> 2Cr^+3
Now you count charges on the left (-8) and on the right(+6) and add 14 H^+ on the left.
14H^+ + 6e + Cr2O7^2- ==> 2Cr^3+
Now add 7H2O on the right to balancae.
14H^+ + 6e + Cr2O7^2- ==> 2Cr^3+ + 7H2O.
I don't know if you've separated the Sn half cell properly or not.
28H+ 3Sn2+ + 2(Cr2O7)2- ------> 3Sn4+ + 4Cr3+ + 14H2O
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