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

Question:

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

H2O2(aq)+Cr2O2-7(aq)-->O2(g)+Cr3+(aq)

Balanced Redox Reactions:

In a redox reaction the oxidation states of at least two species in the reaction change. One increases and the other decreases. The species that has its oxidation state increase is being oxidized. The species whose oxidation state decreases is reduced. Oxidation is the loss of electrons, reduction the gain. In order to balance a redox reaction, the species are split into the half reactions which show the electrons. This facilitates balancing.

{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}