complete and balance te following half-reaction. in each case indicate whether t

Question

complete and balance te following half-reaction. in each case indicate whether the half-reaction is an oxidation or a reduction. 1.Mo^3+(aq)-->Mo(s)(acidic or basic solution) 2.H2SO3(aq)-->SO4^2-(aq)(acidic solution) 3.NO3^-(aq)-->NO(g)(acidic solution) 4.O2(g)-->H2O(l)(acidic solution) 5.MN^2+(aq)-->MnO2(s)(basic solution) 6.Cr(OH)3(s)--CrO4^2-(aq)(basic solution) 7.O2(g)-->H2O(l)(basic solution)

Explanation / Answer

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) Balance the oxygen: O2 --> 2H2O Balance the hydrogen by adding H+: O2 + 4H+ ---> 2H2O Balance the charges by adding electrons: O2 + 4H+ + 4e- --> 2H2O (yes it does go from 0 to -2, but since the oxidation number it lowering, it is REDUCTION, not oxidation).

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