Chapter 18 - Electrochemistry - Questions and Exercises - Exercises - Page 819: 18.31

a) $Fe(s)+2Ag^{+}(aq)\rightarrow Fe^{2+}(aq)+2Ag(s)$ b) $2S_{2}O_{3}^{2-}(aq)+I_{2}(s)\rightarrow S_{4}O_{6}^{2-}(aq)+2I^-(aq)$ c) $MnO_{4}^{-}(aq)+8H^{+}(aq)+5Fe^{2+}(aq)\rightarrow 5Fe^{3+}(aq)+Mn^{2+}(aq)+4H_{2}O(l)$

a) Separating the equation to half reactions, we get Oxidation half: $Fe\rightarrow Fe^{2+}$ Reduction half: $Ag^{+}\rightarrow Ag$ The atoms are already balanced in each half reaction. Now, adding electrons to balance the charges, we have Oxidation half: $Fe\rightarrow Fe^{2+}+2e^{-}$ Reduction half: $Ag^{+}+e^{-}\rightarrow Ag$ To equalize the number of electrons in both the half reactions, we multiply the reduction half reaction by 2 and write as $2Ag^{+}+2e^{-}\rightarrow 2Ag$ Adding the two half reactions and cancelling the electrons on each side, we get the balanced equation as below: $Fe+2Ag^{+}\rightarrow Fe^{2+}+2Ag$ b) Separating the equation to half reactions, we get Oxidation half: $S_{2}O_{3}^{2-}\rightarrow S_{4}O_{6}^{2-}$ Reduction half: $I_{2}\rightarrow I^-$ Now, we balance the atoms other than O and H to obtain: Oxidation half: $2S_{2}O_{3}^{2-}\rightarrow S_{4}O_{6}^{2-}$ Reduction half: $I_{2}\rightarrow 2I^-$ Add $H_{2}O$ to balance O atoms and $H^+$ to balance H atoms. But it is already balanced. Now, adding electrons to balance the charges, we have Oxidation half: $2S_{2}O_{3}^{2-}\rightarrow S_{4}O_{6}^{2-}+2e^-$ Reduction half: $I_{2}+2e^-\rightarrow 2I^-$ The number of electrons is equal in both the half reactions. By adding the two half reactions and cancelling the electrons on each side, we obtain the balanced equation as below: $2S_{2}O_{3}^{2-}+I_{2}\rightarrow S_{4}O_{6}^{2-}+2I^-$ c) Separating the equation to half reactions, we get Oxidation half: $Fe^{2+}\rightarrow Fe^{3+}$ Reduction half: $MnO_{4}^{-}\rightarrow Mn^{2+}$ The atoms other than O and H are already balanced in each half reaction. Add $H_{2}O$ to balance O atoms and $H^+$ to balance H atoms. Thus, we get $MnO_{4}^{-}+8H^+\rightarrow Mn^{2+}+4H_{2}O$ Now, adding electrons to balance the charges, we have Oxidation half: $Fe^{2+}\rightarrow Fe^{3+}+e^-$ Reduction half: $MnO_{4}^{-}+8H^{+}+5e^-\rightarrow Mn^{2+}+4H_{2}O$ To equalize the number of electrons in both the half reactions, we multiply the oxidation half reaction by 5 and write as $5Fe^{2+}\rightarrow 5Fe^{3+}+5e^-$ Adding the two half reactions and cancelling the electrons on each side, we get the balanced equation as below: $MnO_{4}^{-}+8H^{+}+5Fe^{2+}\rightarrow 5Fe^{3+}+Mn^{2+}+4H_{2}O$