Help with chemistry please?

J

Jessica

New member
How do I balance a redox equation in basic media if a solid precipitate is formed ? The equation is as follows : Br aqueus with a neg charge plus CrO4 with a minus 2 charge aqueus yields BrO3 minus 1 charge aqueus plus Cr(OH)3 solid
 
Br-(aq) +CrO4 2-(aq) --->BrO3-(aq) +Cr(OH)3(s)
find the alternation of oxidation number,
Br- = -1 -> Br=+5 that's oxidation (6) => multiply by 1(coeficient)
Cr=+6 -> Cr=+3 that's reduxtion(3) => multiply by 2(coeficient)
put the numbers into each element
Br- +2CrO4 2- --->BrO3- +2Cr(OH)3
give the H+ at left part so that it can be balanced,
Br- +3CrO4 2- +9H+ --->BrO3- +3Cr(OH)3
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