Balance the following equations by the half-reaction method. (Use the lowest possible coefficients. These may be 0 (zero).)
(a) Fe(s) + HCl(aq) HFeCl4(aq) + H2(g)
(b) IO3-(aq) + I -(aq) + H+(aq) + H2O(l) I3-(aq) + H+(aq) + H2O(l)
(c) Cr(NCS)64-(aq) + Ce4+(aq) + H+(aq) + H2O(l) Cr3+(aq) + Ce3+(aq) + NO3-(aq) + CO2(g) + SO42-(aq) + H+(aq) + H2O(l)
(d) CrI3(s) + Cl2(g) + OH -(aq) + H2O(l) CrO42-(aq) + IO4-(aq) + Cl-(aq) + OH -(aq) + H2O(l)
(e) Fe(CN)64-(aq) + Ce4+(aq) + OH -(aq) + H2O(l) Ce(OH)3(s) + Fe(OH)3(s) + CO32-(aq) + NO3-(aq) + OH -(aq) + H2O(l)
(f) Fe(OH)2(s) + H2O2(aq) Fe(OH)3(s)
I can't understand how to do these so if you want to choose any of the six and explain how one would go about doing it. Thanks
(a) Fe(s) + HCl(aq) HFeCl4(aq) + H2(g)
(b) IO3-(aq) + I -(aq) + H+(aq) + H2O(l) I3-(aq) + H+(aq) + H2O(l)
(c) Cr(NCS)64-(aq) + Ce4+(aq) + H+(aq) + H2O(l) Cr3+(aq) + Ce3+(aq) + NO3-(aq) + CO2(g) + SO42-(aq) + H+(aq) + H2O(l)
(d) CrI3(s) + Cl2(g) + OH -(aq) + H2O(l) CrO42-(aq) + IO4-(aq) + Cl-(aq) + OH -(aq) + H2O(l)
(e) Fe(CN)64-(aq) + Ce4+(aq) + OH -(aq) + H2O(l) Ce(OH)3(s) + Fe(OH)3(s) + CO32-(aq) + NO3-(aq) + OH -(aq) + H2O(l)
(f) Fe(OH)2(s) + H2O2(aq) Fe(OH)3(s)
I can't understand how to do these so if you want to choose any of the six and explain how one would go about doing it. Thanks