Balance the following oxidation-reduction reactions that occur in basic solution. a. \(\mathrm{Cr}(s)+\mathrm{CrO}_{4}^{2-}(a q) \rightarrow \mathrm{Cr}(\mathrm{OH})_{3}(s)\) b. \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{S}^{2-}(a q) \rightarrow \mathrm{MnS}(s)+\mathrm{S}(s)\) c. \(\mathrm{CN}^{-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{CNO}^{-}(a q)+\mathrm{MnO}_{2}(s)\)

Step 1: Determine the oxidation states of elements involved: In the given equation, the oxidation states of the elements are as follows: - For \(Cr(s)\), the oxidation state is 0. - For \(CrO_{4}^{2-}\), the oxidation state of Cr is +6. - For \(Cr(OH)_{3}\), the oxidation state of Cr is +3. Step 2: Write the half-reactions for oxidation and reduction: Oxidation: \(Cr(s) \rightarrow Cr^{3+}\) (oxidation state change from 0 to +3) Reduction: \(Cr^{6+} \rightarrow Cr^{3+}\) (oxidation state change from +6 to +3) Step 3: Balance the half-reactions: Oxidation: \(Cr(s) \rightarrow Cr^{3+} + 3e^{-}\) (Balance charge by adding electrons) Reduction: \(Cr^{6+} + 3e^{-} \rightarrow Cr^{3+}\) (Balance charge by adding electrons) Step 4: Balance hydrogens and oxygens in the reduction half-reaction: Add 4 \(OH^{-}\) to balance the oxygens in the \(CrO_{4}^{2-}\). Then add 4 \(H_{2}O\) to balance the hydrogens: Reduction: \(CrO_{4}^{2-} + 3e^{-} + 8H^{+} \rightarrow Cr^{3+} + 4H_{2}O\) Step 5: Eliminate the extra species in the balanced half-reactions: Combine the two half-reactions, eliminating \(Cr^{3+}\) and \(3e^{-}\), obtaining: \(Cr(s) + CrO_{4}^{2-} + 8H^{+} \rightarrow Cr^{3+} + 4H_{2}O + Cr^{3+}\) Step 6: Balance the final equation by simplifying and combining: \(Cr(s) + CrO_{4}^{2-} + 8H^{+} \rightarrow 2Cr^{3+} + 4H_{2}O\) Add 6 \(OH^{-}\) on both sides to obtain balanced equation in basic solution: \(Cr(s) + CrO_{4}^{2-} + 2H_{2}O \rightarrow Cr(OH)_{3} (s) + 3OH^{-}\)

Step 1: Determine the oxidation states of elements involved: - For \(MnO_{4}^{-}\), the oxidation state of Mn is +7. - For \(MnS\), the oxidation state of Mn is +2. - For \(S^{2-}\), the oxidation state of S is -2. - For \(S(s)\), the oxidation state of S is 0. Step 2: Write the half-reactions for oxidation and reduction: Oxidation: \(S^{2-} \rightarrow S(s)\) (oxidation state change from -2 to 0) Reduction: \(Mn^{7+} \rightarrow Mn^{2+}\) (oxidation state change from +7 to +2) Step 3: Balance the half-reactions: Oxidation: \(S^{2-} \rightarrow S(s) + 2e^{-}\) (Balance charge by adding electrons) Reduction: \(Mn^{7+} + 5e^{-} \rightarrow Mn^{2+}\) (Balance charge by adding electrons) Step 4: Balance the hydrogens and oxygens in the reduction half-reaction: Add 4 \(OH^{-}\) to balance the oxygens in the \(MnO_{4}^{-}\). Then add 4 \(H_{2}O\) to balance the hydrogens: Reduction: \(MnO_{4}^{-} + 5e^{-} + 8H^{+} \rightarrow Mn^{2+} + 4H_{2}O\) Step 5: Multiply half-reactions to make electrons equal and add them: \[2*(MnO_{4}^{-} + 5e^{-} + 8H^{+} \rightarrow Mn^{2+} + 4H_{2}O)\] \[5*(S^{2-} \rightarrow S(s) + 2e^{-})\] Sum: \(2MnO_{4}^{-} + 10e^{-} + 16H^{+} + 5S^{2-} \rightarrow 2Mn^{2+} + 8H_{2}O + 10S(s) + 10e^{-}\) Step 6: Balance the final equation by simplifying and combining: \(2MnO_{4}^{-} + 16H^{+} + 5S^{2-} \rightarrow 2Mn^{2+} + 8H_{2}O + 10S(s)\) Add 16 \(OH^{-}\) on both sides to obtain: \(2MnO_{4}^{-} + 5S^{2-} + 8H_{2}O \rightarrow 2MnS(s) + 10S(s) + 16OH^{-}\)

Step 1: Determine the oxidation states of elements involved: - For \(CN^{-}\), the oxidation state of C is +2. - For \(CNO^{-}\), the oxidation state of C is +4. - For \(MnO_{4}^{-}\), the oxidation state of Mn is +7. - For \(MnO_{2}\), the oxidation state of Mn is +4. Step 2: Write the half-reactions for oxidation and reduction: Oxidation: \(C^{2+} \rightarrow C^{4+}\) (oxidation state change from +2 to +4) Reduction: \(Mn^{7+} \rightarrow Mn^{4+}\) (oxidation state change from +7 to +4) Step 3: Balance the half-reactions: Oxidation: \(C^{2+} \rightarrow C^{4+} + 2e^{-}\) (Balance charge by adding electrons) Reduction: \(Mn^{7+} + 3e^{-} \rightarrow Mn^{4+}\) (Balance charge by adding electrons) Step 4: Balance the hydrogens and oxygens in the reduction half-reaction: Add 4 \(OH^{-}\) to balance the oxygens in the \(MnO_{4}^{-}\). Then add 4 \(H_{2}O\) to balance the hydrogens: Reduction: \(MnO_{4}^{-} + 3e^{-} + 8H^{+} \rightarrow MnO_{2}(s) + 4H_{2}O\) Step 5: Balance the hydrogens and oxygens in the oxidation half-reaction: Add 1 \(OH^{-}\) and 1 \(H_{2}O\) to \/ the oxidation half-reaction: Oxidation: \(CN^{-} + H_{2}O \rightarrow CNO^{-} + 2e^{-} + 2H^{+}\) Step 6: Multiply half-reactions to make electrons equal and add them: \[3*(CN^{-} + H_{2}O \rightarrow CNO^{-} + 2e^{-} + 2H^{+})\] \[2*(MnO_{4}^{-} + 3e^{-} + 8H^{+} \rightarrow MnO_{2}(s) + 4H_{2}O)\] Sum: \(3CN^{-} + 3H_{2}O + 2MnO_{4}^{-} + 24H^{+} \rightarrow 3CNO^{-} + 6e^{-} + 6H^{+} + 2MnO_{2}(s) + 8H_{2}O\) Step 7: Balance the final equation by simplifying and combining: \(3CN^{-} + 2MnO_{4}^{-} + 10H^{+} \rightarrow 3CNO^{-} + 2MnO_{2}(s) + 5H_{2}O\) Add 10 \(OH^{-}\) on both sides to obtain: \(3CN^{-} + 2MnO_{4}^{-} + 5H_{2}O \rightarrow 3CNO^{-} + 2MnO_{2}(s) + 10OH^{-}\)