I would do this. Note N changes, Br changes BUT some of the Br^- are 1- on the left AND 1- on the right so they don't change. When this happens balance the redox part then add enough Br^- in the final equation for those Br^- that don't change.
6e + 2N^3+ ==> N2
3H2O + 3Br^- ==> 3HOBr + 6e + 3H^+
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Add the two, check to make sure everything balances, then put the 2N^3+ and the 3Br^- together to make the original 2NBr3 which means then you add another 3Br^- on the right to go with the 3H^+ already there to make 3HBr.
Then check everything.
I get
3H2O + 2NBr3 ==> 3HOBr + 3HBr + N2
:-)
I need major help with balancing redox reactions. I can't seem to get this one: (in basic conditions)
NBr_3(aq)=> N_2(g) + Br^-(aq) + HOBr(aq)
I don't know how to split this up properly, and when I do it is like impossible to balance it.
Thanks!
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1 answer