question on redox

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.

issa

Senior Member
10+ Year Member
5+ Year Member
15+ Year Member
Joined
Aug 24, 2005
Messages
561
Reaction score
0
Points
0
Advertisement - Members don't see this ad
why is the answer 3? i thought the answer should be 1.


Shot at 2007-08-07
 
In order to get the electrons to cancel out, you have to multiply the second equation(the Sn one) by 3, cause you have to add 6 electrons to the reactant side of the Cr2O7 equation.

i am still confused🙁
can you explain how you did it in more detail?
 
i am still confused🙁
can you explain how you did it in more detail?

If you break apart the original redox eq into its two respective parts(the Cr2O7-->Cr+3 and then the Sn one) and then balance them both individually, you will see that you add 6 electrons to the reactants side of the Cr2O7 equation, and 2 electrons to the product side of the Sn equation. In order to cancel electrons you have to make them equal, so you multiply the second(Sn) equation by 3, cause 2X3=6, so then you can cancel out those electrons. Thats why you have the 3 in front of the Sn.
 
why is the answer 3? i thought the answer should be 1.


Shot at 2007-08-07

Sn +2...............Sn+4 (-2 e)
Cr2O7 -2.......... Cr+3
(+6................+3) (+3e)
Need to balance loss and gain of electrons. Therefore multiply (-2e)x 3 and (+3e) x 2.
 
Top Bottom