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

A) The half-reactions and the balanced redox reaction are shown below.

reduction half-reaction: ({eq}\rm MnO_4^-(aq)+4H_2O(l)+3e^{-} \rightarrow MnO_2(s) + 2H_2O(l)+4OH^{-}(aq) {/eq}) x 4

• The number of Mn atoms is already balanced.
• The number of O atoms is balanced by adding 2 moles of H{eq}_2 {/eq}O on the product side.

• The number of H atoms is balanced by adding 4 moles of H{eq}_2 {/eq}O on the reactant side and another 4 moles of OH{eq}^{-} {/eq} on the product side.

• The charge is balanced by adding 3 moles of electrons on the reactant side.

oxidation half-reaction: ({eq}\rm C_2H_5OH(aq) + H_2O(l)+4OH^{-}(aq) \rightarrow CH_3CO_2H(aq)+4H_2O(l)+4e^{-} {/eq}) x 3

• The number of C atoms is already balanced.
• The number of O atoms is balanced by adding 1 mole of H{eq}_2 {/eq}O on the reactant side.

• The number of H atoms is balanced by adding 4 moles of H{eq}_2 {/eq}O on the product side and another 4 moles of OH{eq}^{-} {/eq} on the reactant side.

• The charge is balanced by adding 4 moles of electrons on the product side.

balanced redox reaction: {eq}\mathbf{4MnO_4^-(aq)+3C_2H_5OH(aq)\rightarrow 4MnO_2(s) +3CH_3CO_2H(aq)+4OH^{-}(aq)+H_2O(l)} {/eq}

• The reduction half-reaction is multiplied by 4 and the oxidation half-reaction is multiplied by 3 to obtain equal transfer of electrons.
• The half-reactions are added and the spectator terms are eliminated.

B) The half-reactions and the balanced redox reaction are shown below.

oxidation half-reaction: ({eq}\rm H_2O_2(aq)+2OH^{-}(aq)\rightarrow O_2(g)+2H_2O(l)+2e^{-} {/eq}) x 3

• The number of O atoms is already balanced.
• The number of H atoms is balanced by adding 2 moles of H{eq}_2 {/eq}O on the product side and another 2 moles of OH{eq}^{-} {/eq} on the reactant side.

• The charge is balanced by adding 2 moles of electrons on the product side.

reduction half-reaction: {eq}\rm Cr_2O_{7}^{2-}(aq)+14H_2O(l)+6e^{-}\rightarrow 2Cr^{3+}(aq)+7H_2O(l)+14OH^{-}(aq) {/eq}

• The number of Cr atoms is balanced by multiplying the number of moles of Cr{eq}^{3+} {/eq} by 2.

• The number of O atoms is balanced by adding 7 moles of H{eq}_2 {/eq}O on the product side.

• The number of H atoms is balanced by adding 14 moles of H{eq}_2 {/eq}O on the reactant side and another 14 moles of OH{eq}^{-} {/eq} on the product side.

• The charge is balanced by adding 6 moles of electrons on the reactant side.

balanced redox reaction: {eq}\mathbf{Cr_2O_{7}^{2-}(aq)+3H_2O_2(aq)+H_2O(l)\rightarrow 2Cr^{3+}(aq)+3O_2(g)+8OH^{-}(aq)} {/eq}

• The oxidation half-reaction is multiplied by 3 to have equal number of electrons in both half-reactions.
• The half-reactions are added and the spectator terms are eliminated.

C) The half-reactions and the balanced redox reaction are shown below.

oxidation half-reaction: ({eq}\rm Sn^{2+}(aq) \rightarrow Sn^{4+}(aq)+2e^{-} {/eq}) x 4

• The number of Sn atoms is already balanced.
• The charge is balanced by adding 2 moles of electrons on the product side.

reduction half-reaction: {eq}\rm IO_4^-(aq)+8H_2O(l)+8e^{-} \rightarrow I^-(aq)+4H_2O(l)+8OH^{-}(aq) {/eq}

• The number of I atoms is already balanced.
• The number of O atoms is balanced by adding 4 moles of H{eq}_2 {/eq}O on the product side.

• The number of H atoms is balanced by adding 8 moles of H{eq}_2 {/eq}O on the reactant side and another 8 moles of OH{eq}^{-} {/eq} on the product side.

• The charge is balanced by adding 8 moles of electrons on the reactant side.

balanced redox reaction: {eq}\mathbf{4Sn^{2+}(aq)+IO_4^-(aq)+4H_2O(l) \rightarrow Sn^{4+}(aq) +I^-(aq)+8OH^{-}(aq)} {/eq}

• The oxidation half-reaction is multiplied by € to have equal number of electrons in both half-reactions.
• The half-reactions are added and the spectator terms are eliminated.

D) The redox reaction can be balanced by inspection.

balanced redox reaction: {eq}\mathbf{PbO_2(s)+Cl_2(g)\rightarrow PbCl_2(s)+O_2(g)} {/eq}