question on redox

[issa]

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why is the answer 3? i thought the answer should be 1.


Shot at 2007-08-07

 

[fancymylotus]

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.

 

[issa]

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?

 

[fancymylotus]

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.

 

[doc toothache]

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.