Using oxidation numbers we write the reduction equation:
stackrel(IV)MnO_2(s)+4H^+ + 2e^(-)rarrMn^(2+)+2H_2OIVMnO2(s)+4H++2e−→Mn2++2H2O (i)(i)
Both mass and charge are balanced as required.........
And then we need a reduction half equation.........
Cl^(-) rarr 1/2stackrel(0)Cl_2+e^-Cl−→120Cl2+e− (ii)(ii)
Any species whose "oxidation number"oxidation number "INCREASES"INCREASES is said to be "oxidized"oxidized, and any species whose "oxidation number"oxidation number "decreases"decreases is said to be "reduced"reduced,
We add (i)(i) and (ii)(ii) in such a way as to eliminate the electrons, and we get an equation that represents the observed stoichiometric change:
(i) + 2xx(ii)(i)+2×(ii) gives...............................
MnO_2(s)+4H^(+)+ 2Cl^(-)rarrMn^(2+)+Cl_2(g) + 2H_2O(l)MnO2(s)+4H++2Cl−→Mn2++Cl2(g)+2H2O(l)
Given the half equation method, for every electron loss, "oxidation"oxidation, there is thus a FORMAL electron gain, "reduction".reduction.
The (s)(s), (g)(g), (l)(l) represents the physical state of the reactant/product, i.e. "solid/gas/liquid/"solid/gas/liquid/.........The ions, i.e. Mn^(2+)Mn2+ etc. are assumed to be present in solution, and thus do not need a descriptor of state. Happy?
