Asked by bear
balance equation
2H2O + 2e >H2 +2OH^-
Y^3 + 3e > Y
2H2O + 2e >H2 +2OH^-
Y^3 + 3e > Y
Answers
Answered by
DrBob222
2H2O + 2e >H2 +2OH^-
Y^3+ + 3e > Y
----------------------------------
Can't do it. You have written TWO half reactions and BOTH are reductions. In redox reactions ONE must be an oxidation and ONE must be a reduction. I correct the second equation by adding a + sign to the 3 you have for Y^3 to make that Y^3+.
Y^3+ + 3e > Y
----------------------------------
Can't do it. You have written TWO half reactions and BOTH are reductions. In redox reactions ONE must be an oxidation and ONE must be a reduction. I correct the second equation by adding a + sign to the 3 you have for Y^3 to make that Y^3+.
Answered by
bear
that is the problem it is for a battery.
Answered by
DrBob222
You can make a reaction out of it by reversing the Y half cell but I don't know i that will fit your criteria.
2H2O + 2e >H2 +2OH^-
Y > Y^3+ + 3e
-------------------------------
Multiply equation 1 by 3 and add to equation 2 multiplied by 2.
2H2O + 2e >H2 +2OH^-
Y > Y^3+ + 3e
-------------------------------
Multiply equation 1 by 3 and add to equation 2 multiplied by 2.
Answered by
bear
12H2O+2y+12e>6H2+12OH+2y^(3+)+6e
Answered by
DrBob222
No. You know something is wrong if the electrons don't cancel.
3*(2H2O + 2e >H2 +2OH^-)
2*(Y > Y^3+ + 3e)
----------------------------------------
6H2O + 6e ==> 3H2 + 6OH^-
2Y ==> 2Y^3+ + 6e
-----------------------------------
6H2O + 6e + 2Y ==> 2Y^3+ + 6e + 3H2 + 6OH^-
Cancel the 6e and check it.
There are 12 H on both sides.
There are 6 O on both sides.
The charge on the left is 0. the charge on the right is 0.
The electrons lost = electrons gained.
Done up brown. However, this cell may not be spontaneous; therefore, it may not generate a voltage. You won't know that until you look up the potential of each half cell as I've written it and add them together. IF you get a + voltage is will serve as a battery. If you get a - voltage it will not generate a voltage.
3*(2H2O + 2e >H2 +2OH^-)
2*(Y > Y^3+ + 3e)
----------------------------------------
6H2O + 6e ==> 3H2 + 6OH^-
2Y ==> 2Y^3+ + 6e
-----------------------------------
6H2O + 6e + 2Y ==> 2Y^3+ + 6e + 3H2 + 6OH^-
Cancel the 6e and check it.
There are 12 H on both sides.
There are 6 O on both sides.
The charge on the left is 0. the charge on the right is 0.
The electrons lost = electrons gained.
Done up brown. However, this cell may not be spontaneous; therefore, it may not generate a voltage. You won't know that until you look up the potential of each half cell as I've written it and add them together. IF you get a + voltage is will serve as a battery. If you get a - voltage it will not generate a voltage.
Answered by
bear
I am trying to solve this problem,, DeltaG=-nF(delta E_cell) I want to known what is the value of n is. n =electrons per mole. I tried 1,2,3, and four.
no luck. that is why I wanted to balance the equation. what is the value of N?
no luck. that is why I wanted to balance the equation. what is the value of N?
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