91Ó°ÊÓ

Balance the following reaction: \(\operatorname{Cr}(s)+\mathrm{CrO}_{4}^{2-}(a q) \rightarrow \operatorname{Cr}(\mathrm{OH})_{3}(s)\) To balance this reaction, we need to follow these steps: Step 1: Divide the reaction into half-reactions \[\text{Oxidation:} \quad \operatorname{Cr}(s) \rightarrow \operatorname{Cr}(\mathrm{OH})_{3}(s)\] \[\text{Reduction:} \quad \mathrm{CrO}_{4}^{2-}(a q) \rightarrow \operatorname{Cr}(\mathrm{OH})_{3}(s)\] Step 2: Balance the atoms involved in the redox process (excluding O and H) Oxidation: The Cr atoms are already balanced. Reduction: The Cr atoms are already balanced. Step 3: Balance the O atoms by adding H_2O Oxidation: Not necessary as there are no O atoms. Reduction: Add 3 H_2O to the right side. \[\mathrm{CrO}_{4}^{2-}(a q) \rightarrow \operatorname{Cr}(\mathrm{OH})_{3}(s)+3\mathrm{H}_{2}\mathrm{O}(l)\] Step 4: Balance the H atoms by adding OH^- Oxidation: Add 3 OH^- to the right side. \[\operatorname{Cr}(s) \rightarrow \operatorname{Cr}(\mathrm{OH})_{3}(s)+3\mathrm{OH}^-\] Reduction: Add 6 OH^- to the left side. \[\mathrm{CrO}_{4}^{2-}(a q)+6\mathrm{OH}^- \rightarrow \operatorname{Cr}(\mathrm{OH})_{3}(s)+3\mathrm{H}_{2}\mathrm{O}(l)\] Step 5: Balance charges by adding electrons (e^-) Oxidation: Add 3 e^- to the right side. \[\operatorname{Cr}(s) \rightarrow \operatorname{Cr}(\mathrm{OH})_{3}(s)+3\mathrm{OH}^-+3\mathrm{e}^-\] Reduction: Add 3 e^- to the left side. \[\mathrm{CrO}_{4}^{2-}(a q)+6\mathrm{OH}^-+3\mathrm{e}^- \rightarrow \operatorname{Cr}(\mathrm{OH})_{3}(s)+3\mathrm{H}_{2}\mathrm{O}(l)\] Step 6: Multiply each half-reaction by the necessary factor to make the number of electrons equal Oxidation: No need to multiply, as the number of electrons is already the same. Reduction: No need to multiply, as the number of electrons is already the same. Step 7: Add both half-reactions and cancel electrons and common species \[\operatorname{Cr}(s)+\mathrm{CrO}_{4}^{2-}(a q)+6\mathrm{OH}^- \rightarrow \operatorname{Cr}(\mathrm{OH})_{3}(s)+3\mathrm{H}_{2}\mathrm{O}(l)+3\mathrm{OH}^-\] Simplify the reaction: \[\operatorname{Cr}(s)+\mathrm{CrO}_{4}^{2-}(a q)+3\mathrm{OH}^- \rightarrow 2\operatorname{Cr}(\mathrm{OH})_{3}(s)\] Now, the reaction is balanced.

Balance the following reaction: \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{S}^{2-}(a q) \rightarrow \mathrm{MnS}(s)+\mathrm{S}(s)\) To balance this reaction, we need to follow these steps: Step 1: Divide the reaction into half-reactions \[\text{Oxidation:} \quad \mathrm{S}^{2-}(a q) \rightarrow \mathrm{S}(s)\] \[\text{Reduction:} \quad \mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{MnS}(s)\] Step 2: Balance the atoms involved in the redox process (excluding O and H) Oxidation: The S atoms are already balanced. Reduction: The Mn atoms are already balanced. Step 3: Balance the O atoms by adding H_2O Oxidation: Not necessary as there are no O atoms. Reduction: Add 4 H_2O to the right side. \[\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{MnS}(s)+4\mathrm{H}_{2}\mathrm{O}(l)\] Step 4: Balance the H atoms by adding OH^- Oxidation: Not necessary as there are no H atoms. Reduction: Add 8 OH^- to the left side. \[\mathrm{MnO}_{4}^{-}(a q)+8\mathrm{OH}^- \rightarrow \mathrm{MnS}(s)+4\mathrm{H}_{2}\mathrm{O}(l)\] Step 5: Balance charges by adding electrons (e^-) Oxidation: Add 2 e^- to the right side. \[\mathrm{S}^{2-}(a q) \rightarrow \mathrm{S}(s)+2\mathrm{e}^-\] Reduction: Add 5 e^- to the left side. \[\mathrm{MnO}_{4}^{-}(a q)+8\mathrm{OH}^-+5\mathrm{e}^- \rightarrow \mathrm{MnS}(s)+4\mathrm{H}_{2}\mathrm{O}(l)\] Step 6: Multiply each half-reaction by the necessary factor to make the number of electrons equal Oxidation: Multiply by 5. \[\mathrm{5S}^{2-}(a q) \rightarrow 5\mathrm{S}(s)+10\mathrm{e}^-\] Reduction: Multiply by 2. \[\mathrm{2MnO}_{4}^{-}(a q)+16\mathrm{OH}^-+10\mathrm{e}^- \rightarrow 2\mathrm{MnS}(s)+8\mathrm{H}_{2}\mathrm{O}(l)\] Step 7: Add both half-reactions and cancel electrons and common species \[5\mathrm{S}^{2-}(a q)+2\mathrm{MnO}_{4}^{-}(a q)+16\mathrm{OH}^- \rightarrow 2\mathrm{MnS}(s)+8\mathrm{H}_{2}\mathrm{O}(l)+5\mathrm{S}(s)\] Now, the reaction is balanced.

Balance the following reaction: \(\mathrm{CN}^{-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{CNO}^{-}(a q)+\mathrm{MnO}_{2}(s)\) To balance this reaction, we need to follow these steps: Step 1: Divide the reaction into half-reactions \[\text{Oxidation:} \quad \mathrm{CN}^{-}(a q) \rightarrow \mathrm{CNO}^{-}(a q)\] \[\text{Reduction:} \quad \mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{MnO}_{2}(s)\] Step 2: Balance the atoms involved in the redox process (excluding O and H) Oxidation: The C and N atoms are already balanced. Reduction: The Mn atoms are already balanced. Step 3: Balance the O atoms by adding H_2O Oxidation: Not necessary as there are no O atoms. Reduction: Add 2 H_2O to the right side. \[\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{MnO}_{2}(s)+2\mathrm{H}_{2}\mathrm{O}(l)\] Step 4: Balance the H atoms by adding OH^- Oxidation: Not necessary as there are no H atoms. Reduction: Add 4 OH^- to the left side. \[\mathrm{MnO}_{4}^{-}(a q)+4\mathrm{OH}^- \rightarrow \mathrm{MnO}_{2}(s)+2\mathrm{H}_{2}\mathrm{O}(l)\] Step 5: Balance charges by adding electrons (e^-) Oxidation: Add 2 e^- to the right side. \[\mathrm{CN}^{-}(a q) \rightarrow \mathrm{CNO}^{-}(a q)+2\mathrm{e}^-\] Reduction: Add 3 e^- to the left side. \[\mathrm{MnO}_{4}^{-}(a q)+4\mathrm{OH}^-+3\mathrm{e}^- \rightarrow \mathrm{MnO}_{2}(s)+2\mathrm{H}_{2}\mathrm{O}(l)\] Step 6: Multiply each half-reaction by the necessary factor to make the number of electrons equal Oxidation: Multiply by 3. \[\mathrm{3CN}^{-}(a q) \rightarrow 3\mathrm{CNO}^{-}(a q)+6\mathrm{e}^-\] Reduction: Multiply by 2. \[\mathrm{2MnO}_{4}^{-}(a q)+8\mathrm{OH}^-+6\mathrm{e}^- \rightarrow 2\mathrm{MnO}_{2}(s)+4\mathrm{H}_{2}\mathrm{O}(l)\] Step 7: Add both half-reactions and cancel electrons and common species \[3\mathrm{CN}^{-}(a q)+2\mathrm{MnO}_{4}^{-}(a q)+8\mathrm{OH}^- \rightarrow 3\mathrm{CNO}^{-}(a q)+2\mathrm{MnO}_{2}(s)+4\mathrm{H}_{2}\mathrm{O}(l)\] Now, the reaction is balanced.