PLEASE HELP In the redox conversion of Cr(OH)− 4 to CrO2− 4 , the oxidation number of Cr goes from (−1, 0, +3) to (−2, +6, +8).?
In the redox conversion of Cr(OH)−
4
to CrO2−
4
,
the oxidation number of Cr goes from (−1,
0, +3) to (−2, +6, +8). Recall that the
oxidation number of oxygen is typically −2.
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[Cr(OH)4]^1- =====>> [CrO4[^2-
Charge of ion = -1
4(O) + 4(H) = 4(-2) + 4(+1) = -4 ;
Cr + -4 = -1 ; Cr = +4 -1 = +3
One could also use fact that the hydroxide ion( OH)^1- is minus 1. Four of them = -4.
[CrO4]^2- = Cr + 4O = -2 ; Cr + 4(-2) = -2 ;
Cr - 8 = -2 ; Cr = -2 + 8 , Cr = +6
Answer is Cr goes from a +3 to a +6.