Balance the following equations by using OH- but not H^+
A) PbO2 + Cl^- --> CLO^- + Pb(OH)3^-
B) HNO2 + SbO^+ --> NO + Sb2O5
C) Ag2S + CN^- + O2 --> S + Ag(CN)2^- + OH^-
D) HO2^- + Cr(OH)3^- --> CrO4^2- + OH-
E) ClO2 + OH^- --> ClO2^- + ClO3^-
F) WO3^- + O2 --> HW6O21^-5 + OH^-
G) Mn2O3 + CN^- --> Mn(CN)6^4- + (CN)2
H) Cu^2+ + H2 --> Cu + H2O
A)
Now, we can divide the given reaction into two following parts,
Reduction:
Oxidation:
Therefore, adding up these two we get the following balanced equation,
B)
Now, we can divide the given reaction into two following parts,
Reduction: . . . (1)
Oxidation: . . . (2)
Therefore, by doing we get the following balanced equation,
C)
Now, we can divide the given reaction into two following parts,
Reduction: . . . (1)
Oxidation: . . . (2)
Therefore, by doing we get the following balanced equation,
D)
In this problem, I am assuming that the left side it is Cr(OH)3 i.e., chromium(III) hyroxide as it a stable form (in the question the formula of the corresponding species is little confusing).
Now, we can divide the given reaction into two following parts,
Reduction: . . . (1)
Oxidation: . . . (2)
Therefore, by doing we get the following balanced equation,
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