11.44 Use half-reaction method to balance the following Red-ox reaction. Pay attention to the media. You need supply two balanced half reactions and one balanced overall reaction. (a) AsH_(3)(g)+ClO_(3)^-(aq)arrow H_(3)AsO_(4)(aq)+Cl^-(aq) acidic media (b) S_(2)O_(3)^2-(aq)+Cl_(2)(g)arrow HSO_(4)^-(aq)+Cl^-(aq) acidic media )[Pb(OH)_(4)]^2-(aq)+ClO^-(aq)arrow PbO_(2)(s)+Cl^-(aq) basic media (d) MnO_(2)(s)+Cr^3+(aq)arrow Mn^2+(aq)+CrO_(4)^2-(aq) basic media g)MnO_(4)^-(aq)+SO_(3)^2-(aq)arrow MnO_(2)(s)+SO_(4)^2-(aq) neutral media

1. Balance the half-reactions for (a) in acidic media<br /> Oxidation half-reaction: $AsH_3 \rightarrow H_3AsO_4$. Add water, $H^+$, and electrons:<br />$AsH_3 + 4H_2O \rightarrow H_3AsO_4 + 8H^+ + 8e^-$.<br /><br /> Reduction half-reaction: $ClO_3^- \rightarrow Cl^-$. Add water, $H^+$, and electrons:<br />$ClO_3^- + 6H^+ + 6e^- \rightarrow Cl^- + 3H_2O$.<br /><br /> Multiply oxidation by 3 and reduction by 4 to equalize electrons:<br />$3(AsH_3 + 4H_2O \rightarrow H_3AsO_4 + 8H^+ + 8e^-)$ <br />$4(ClO_3^- + 6H^+ + 6e^- \rightarrow Cl^- + 3H_2O)$.<br /><br />2. Combine and simplify for overall reaction<br /> Adding both reactions:<br />$3AsH_3 + 12H_2O + 4ClO_3^- + 24H^+ \rightarrow 3H_3AsO_4 + 32H^+ + 4Cl^- + 12H_2O$.<br /><br /> Simplify:<br />$3AsH_3 + 4ClO_3^- + 12H^+ \rightarrow 3H_3AsO_4 + 4Cl^-$. <br /><br />---<br /><br />3. Balance the half-reactions for (b) in acidic media<br /> Oxidation half-reaction: $S_2O_3^{2-} \rightarrow HSO_4^-$. Add water, $H^+$, and electrons:<br />$S_2O_3^{2-} + 5H_2O \rightarrow 2HSO_4^- + 8H^+ + 8e^-$.<br /><br /> Reduction half-reaction: $Cl_2 \rightarrow 2Cl^-$. Add electrons:<br />$Cl_2 + 2e^- \rightarrow 2Cl^-$.<br /><br /> Multiply oxidation by 1 and reduction by 4 to equalize electrons:<br />$1(S_2O_3^{2-} + 5H_2O \rightarrow 2HSO_4^- + 8H^+ + 8e^-)$ <br />$4(Cl_2 + 2e^- \rightarrow 2Cl^-)$.<br /><br />4. Combine and simplify for overall reaction<br /> Adding both reactions:<br />$S_2O_3^{2-} + 5H_2O + 4Cl_2 \rightarrow 2HSO_4^- + 8H^+ + 8Cl^-$.<br /><br /> Simplify:<br />$S_2O_3^{2-} + 4Cl_2 + 5H_2O \rightarrow 2HSO_4^- + 8Cl^- + 8H^+$. <br /><br />---<br /><br />5. Balance the half-reactions for (c) in basic media<br /> Oxidation half-reaction: $[Pb(OH)_4]^{2-} \rightarrow PbO_2$. Add $OH^-$ and electrons:<br />$[Pb(OH)_4]^{2-} \rightarrow PbO_2 + 2H_2O + 2e^-$.<br /><br /> Reduction half-reaction: $ClO^- \rightarrow Cl^-$. Add $OH^-$ and electrons:<br />$ClO^- + H_2O + 2e^- \rightarrow Cl^- + 2OH^-$.<br /><br /> Multiply both reactions by 1 to equalize electrons:<br />$1([Pb(OH)_4]^{2-} \rightarrow PbO_2 + 2H_2O + 2e^-)$ <br />$1(ClO^- + H_2O + 2e^- \rightarrow Cl^- + 2OH^-)$.<br /><br />6. Combine and simplify for overall reaction<br /> Adding both reactions:<br />$[Pb(OH)_4]^{2-} + ClO^- + H_2O \rightarrow PbO_2 + 2H_2O + Cl^- + 2OH^-$.<br /><br /> Simplify:<br />$[Pb(OH)_4]^{2-} + ClO^- \rightarrow PbO_2 + Cl^- + 2OH^-$. <br /><br />---<br /><br />7. Balance the half-reactions for (d) in basic media<br /> Oxidation half-reaction: $MnO_2 \rightarrow Mn^{2+}$. Add $OH^-$ and electrons:<br />$MnO_2 + 4H^+ + 2e^- \rightarrow Mn^{2+} + 2H_2O$.<br /><br /> Reduction half-reaction: $Cr^{3+} \rightarrow CrO_4^{2-}$. Add $OH^-$ and electrons:<br />$Cr^{3+} + 4H_2O \rightarrow CrO_4^{2-} + 8H^+ + 3e^-$.<br /><br /> Multiply oxidation by 3 and reduction by 2 to equalize electrons:<br />$3(MnO_2 + 4H^+ + 2e^- \rightarrow Mn^{2+} + 2H_2O)$ <br />$2(Cr^{3+} + 4H_2O \rightarrow CrO_4^{2-} + 8H^+ + 3e^-$).<br /><br />8. Combine and simplify for overall reaction<br /> Adding both reactions:<br />$3MnO_2 + 12H^+ + 2Cr^{3+} + 8H_2O \rightarrow 3Mn^{2+} + 6H_2O + 2CrO_4^{2-} + 16H^+$.<br /><br /> Simplify:<br />$3MnO_2 + 2Cr^{3+} + 4H_2O \rightarrow 3Mn^{2+} + 2CrO_4^{2-} + 8H^+$. <br /><br />---<br /><br />9. Balance the half-reactions for (e) in neutral media<br /> Oxidation half-reaction: $SO_3^{2-} \rightarrow SO_4^{2-}$. Add water and electrons:<br />$SO_3^{2-} + H_2O \rightarrow SO_4^{2-} + 2H^+ + 2e^-$.<br /><br /> Reduction half-reaction: $MnO_4^- \rightarrow MnO_2$. Add water and electrons:<br />$MnO_4^- + 4H^+ + 3e^- \rightarrow MnO_2 + 2H_2O$.<br /><br /> Multiply oxidation by 3 and reduction by 2 to equalize electrons:<br />$3(SO_3^{2-} + H_2O \rightarrow SO_4^{2-} + 2H^+ + 2e^-)$ <br />$2(MnO_4^- + 4H^+ + 3e^- \rightarrow MnO_2 + 2H_2O)$.<br /><br />10. Combine and simplify for overall reaction<br /> Adding both reactions:<br />$3SO_3^{2-} + 3H_2O + 2MnO_4^- + 8H^+ \rightarrow 3SO_4^{2-} + 6H^+ + 2MnO_2 + 4H_2O$.<br /><br /> Simplify:<br />$3SO_3^{2-} + 2MnO_4^- + 2H_2O \rightarrow 3SO_4^{2-} + 2MnO_2 + 4H^+$. <br /><br />---