Complete and balance the following half-reactions. In each case indicate whether the half-reaction is an oxidation or a reduction.
A. H2SO3(aq)--> SO4 (2-)(aq) (acidic solution)
B. Cr(OH)3(s) --> CrO4 (2-)(aq) (basic solution)
Any help would be appreciated. I am desperate at this point.
1 Answer
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A. H2SO3 = 2H+ and (SO3)2-
lookin at just (SO3)2-, the oxidation number of S is +4
in (SO4)2-, the oxidation number of S is +6
since the oxidation number of S increases from +4 to +6, then u can conclude that S undergoes oxidation (or SO3).
since SO3 undergoes oxidation, u can assume that it loses 2 electrons.
H2SO3 + ______ -------> SO4(2-) + 2e- +_____
balance out the right hand side of the equation (which has 4 negative charges in total) by adding 4H+
H2SO3 + ______ --------> SO4(2-) + 2e- + 4H+
now, when u look at the equation, u'll notice that on the right side there is an additional of O in SO4, and 2 extra H. so u realise, hey! that's a water molecule, which makes sense since u never added anything to H2SO3 in the first place, but its an aq solution, so logically water is part of the reactants!
thus, H2SO3(aq) + H2O(l) --------> SO4(2-)(aq) + 2e- + 4H+(aq)
B. in Cr(OH)3, since its a neutral compound, and OH carries a charge of -1 each, then the oxidation number of Cr is +3.
for chromate ion, Cr has an oxidation number of +6. so, since the oxidation number increase, Cr, like the previous question, undergoes oxidation. similarly, we work out the balanced equation in the same manner.
since oxidation increases by 3, u can assume that Cr(OH)3 loses 3 electrons
Cr(OH)3 + ______ ------> CrO4(2-) + 3e- + _____
now, the right side has 5 additional negative charges. but since tis is a basic solution, instead of using H+ to balance the charges, OH- is used instead
Cr(OH)3 + 5OH- ------> CrO4(2-) + 3e- + ______
so, now u can see that u haf an additional 8H and 4O at the left side. tat's 4 water molecules that should be added to the right side!
Cr(OH)3(aq) + 5OH-(aq) ------> CrO4(2-)(aq) + 3e- + 4H2O(l)
tadaa!