Question

1. Balance the following redox equation in acid. Put the correct stochiometric coefficient (0, 1, 2, etc.). (Note: 0 if the species does not appear on that side of the equation.)

HSO₃⁻(aq) + MnO₄⁻(aq) + H⁺(aq)  → SO₄²⁻(aq) + Mn²⁺(aq) + H₂O(l)

2. Balance the following redox equation in base. Put the correct stochiometric coefficient (0, 1, 2, etc.). (Note: 0 if the species does not appear on that side of the equation.)

Fe(OH)₂ (s) + CrO₄^2-(aq) + H₂O(l)   → Fe(OH)₃(s) +  [Cr(OH)₄] ⁻(aq) + OH⁻(aq)

Answer 1

Balancing the chemical equation

1. HSO3-(aq) + MnO4-(aq) + H+(aq) ---> SO4^2-(aq) + Mn2+(aq) + H2O(l)

First half reaction,

HSO3- ---> SO4^2-

balance O and then H,

HSO3- + H2O ---> SO4^2- + 3H+

add e-'s,

HSO3- + H2O ---> SO4^2- + 3H+ + 2e-

second half-reaction,

MnO4- ---> Mn2+

balance O and H,

MnO4- + 8H+ ---> Mn2+ + 4H2O

add e-'s,

MnO4- + 8H+ + 5e- ---> Mn2+ + 4H2O

multiply first equation by 5 and second by 2 and add both equations,

5HSO3- + 5H2O ---> 5SO4^2- + 15H+ + 10e-

2MnO4- + 16H+ + 10e- ---> 2Mn2+ + 8H2O

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5HSO3-(aq) + 2MnO4-(aq) + H+(aq) ---> 5SO4^2-(aq) + 2Mn2+(aq) + 3H2O(l)

Is balanced equation

2. Fe(OH)2(s) + CrO4^2-(aq) + H2O(l) ---> Fe(OH)3(s) + [Cr(OH)4]-(aq) + OH-(aq)

Ist half-reaction,

Fe(OH)2(s) ---> Fe(OH)3(s)

balance O and H,

Fe(OH)2(s) + H2O ---> Fe(OH)3(s) + H+

add OH- and e-'s,

Fe(OH)2(s) + OH- ---> Fe(OH)3(s) + e-

IInd-half reaction,

CrO4^2- --> [Cr(OH)4]-

balance H and add OH-,

CrO4^2- + 4H2O --> [Cr(OH)4]- + 4OH-

add e-'s,

CrO4^2- + 4H2O + 3e- --> [Cr(OH)4]- + 4OH-

multiply first reaction with 3 and add both equations,

3Fe(OH)2(s) + 3OH- ---> 3Fe(OH)3(s) + 3e-

CrO4^2- + 4H2O + 3e- --> [Cr(OH)4]- + 4OH-

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3Fe(OH)2(s) + CrO4^2-(aq) + 4H2O(l) ---> 3Fe(OH)3(s) + [Cr(OH)4]-(aq) + OH-(aq)

Is balanced equation