Write balanced net ionic equations for the following reactions in basic solution: ...

We start with the base reaction:

{eq}H_2O_2(aq)+Cr_2O_7^{2-}(aq)-->O_2(g)+Cr^{3+}(aq) {/eq}

The oxygen in the peroxide has an oxidation state of -1. The oxygen in the dichromate is -2 leading to the Cr being at a state of +6. The oxygen on the product side is elemental and has an oxidation state of 0. The Cr on the product side has an oxidation state of 3+. Cr goes from 6+ to 3+ so it is reduced. Oxygen in the peroxide goes from -1 to ) so it is oxidized. We will show his in the half reactions:

{eq}Oxidation:\\ H_2O_2(aq) \rightarrow O_2 (g) +2e^-\\ Reduction:\\ Cr_2O_7^{2-}(aq)+6e^- \rightarrow 2Cr^{3+}(aq) {/eq}

Because we are in basic solution we will balances the charges, hydrogen and oxygen with hydroxide anion and water:

{eq}Oxidation:\\ H_2O_2(aq) +2OH^-(aq) \rightarrow O_2 (g)+2H_2O(l) +2e^-\\ Reduction:\\ Cr_2O_7^{2-}(aq)+7H_2O(l)+6e^- \rightarrow 2Cr^{3+}(aq)+14OH^-(aq) {/eq}

Before the two halves can be put together they need to show the same number of electrons. We will multiply the oxidation step by 3:

{eq}Oxidation:\\ 3H_2O_2(aq) +6OH^-(aq) \rightarrow 3O_2 (g)+6H_2O(l) +6e^-\\ Reduction:\\ Cr_2O_7^{2-}(aq)+7H_2O(l)+6e^- \rightarrow 2Cr^{3+}(aq)+14OH^-(aq) {/eq}

Finally we will combine the halves and remove anything that appears on both sides of the reaction:

{eq}3H_2O_2(aq) +6OH^-(aq)+Cr_2O_7^{2-}(aq)+7H_2O(l)+6e^- \rightarrow 3O_2 (g)+6H_2O(l) +6e^-+2Cr^{3+}(aq)+14OH^-(aq)\\ Balanced\:reaction: 3H_2O_2(aq) +Cr_2O_7^{2-}(aq)+H_2O(l) \rightarrow 3O_2 (g) +2Cr^{3+}(aq)+8OH^-(aq)\\ {/eq}