I think you meant for the F to be F^-. The complete equation is
4XeF6 + 36OH- → 3XeO64- + Xe + 24F- + 18H2O
Half equations follow: Ignore the F for the moment.
Xe^6+ + 6e ==> Xe
Xe^6+ + 12OH^- --> [XeO6]^4- + 2e + 6H2O
To make the electrons the same multiply eqn 1 by 1 and eqn 2 by 3 for
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Xe^6+ 6e ==> Xe
3Xe^6+ 36OH^- ==> 3[XeO6]^4- + 6e + 18H2O . Now add
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4Xe^6+ +6e + 36OH^- ==> Xe + 3[XeO6]^4- + 6e + 18H2O. Now add F^-
4XeF6 +6e + 36OH^- ==> Xe + 3[XeO6]^4- + 6e + 18H2O + 24F^-
Done
Check it to make sure it balances as follows:
The elements are the same on both sides.
The charge is the same on both sides.
The electron change for Xe is correct at 6 electrons.
If those three don't balance I have made an error.
Balance the following electrochemistry reaction using the half-reaction method in basic conditions.
XeF6 -> Xe + XeO6^4- + F
1 answer
1 answer