Question

Balance the following oxidation-reduction reactions on paper. Show your work. Assume all reactions occur in an aqueous solution.

a) H2O2+Ce(SO4)2>>>Ce2(SO4)3+H2SO4+O2

b) MnCl2+NaOH+Br2>>>MnO2+NaCl+NaBr

c) CrO7^-2+I^-1>>>Cr^+3+I2 (acidic)

d)TeO2+BrO3^-1>>>H6TeO6+Br2 (acidic)

e) N2O4+Be^-1>>>NO2^-1+BrO3^-1 (basic)

f) Sb^+1+HClO>>>Sb2O5+Cl^-1

g) I2+KIO3+HCl>>>KCl+ICI

h) Cl2>>>ClO3^-1+Cl^-1

Answer 1

H₂O₂ + 2 Ce(SO₄)₂ = H₂SO₄ + O₂ + Ce₂(SO₄)₃

H2O2 + Ce(SO4)2 --> H2SO4 + O2 + Ce2(SO4)3
O2^2- --> O2 losing 2 electrons
2 Ce(+4) --> Ce2(+3) gaining 2 electrons

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MnCl2 + 4NaOH + Br2 = MnO2 + 2NaCl + 2NaBr + 2H2O

Mn2++ 2H2O >> MnO2 + 4H+ + 2e-
Br2 + 2e- >> 2Br-
--------------------------------------...
Mn2+ + Br2 + 2H2O >> MnO2 + 2Br- + 4H+
we add on the left and on the right side 4 OH- ( 4OH-+ 4H+ >> 4 H2O)

Mn2+ + Br2 + 4 OH- >> MnO2 + 2 Br- + 2H2O
This is the ionic reaction.
MnCl2 + 4NaOH + Br2 >> MnO2 + 2NaBr + 2NaCl + 2H2O

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