Answer:
a) [tex]Fe^{2+}[/tex]⇒[tex]Fe^{3+} + e^{-}[/tex]
[tex]MnO_{4} ^{-}[/tex] + 2[tex]H_{2}O[/tex] + 3[tex]e^{-}[/tex]⇒ [tex]MnO_{2} + 4OH^{-}[/tex]
b) [tex]Sn^{2+}[/tex] ⇒ [tex]Sn^{4+} + 2e^{-}[/tex]
[tex]IO_{3} ^{-} + 6H^{+} + 5e^{-}[/tex]⇒ [tex]I_{2} + 3H_{2}O[/tex]
[tex]I_{2} + 2e^{-}[/tex]⇒[tex]2I^{-}[/tex]
c)[tex]S^{2-}[/tex]⇒[tex]S+ 2e^{-}[/tex]
[tex]NO_{3} ^{-} + 4H^{+} + 3e^{-}[/tex]⇒[tex]NO + 2H_{2}O[/tex]
d) [tex]2NH_{3} + 6OH^{-}[/tex]⇒[tex]N_{2} + 6H_{2}O +6e^{-}[/tex]
[tex]2NO_{2} + 8H^{+} + 8e^{-}[/tex]⇒[tex]N_{2} + 4H_{2}O[/tex]
Explanation:
a) [tex]Fe^{2+}[/tex]⇒[tex]Fe^{3+} + e^{-}[/tex]
[tex]MnO_{4} ^{-}[/tex] + 2[tex]H_{2}O[/tex] + 3[tex]e^{-}[/tex]⇒ [tex]MnO_{2} + 4OH^{-}[/tex] half reaction suggest that you need water to reaction occurs.
b) [tex]Sn^{2+}[/tex] ⇒ [tex]Sn^{4+} + 2e^{-}[/tex]
[tex]IO_{3} ^{-} + 6H^{+} + 5e^{-}[/tex]⇒ [tex]I_{2} + 3H_{2}O[/tex]
[tex]I_{2} + 2e^{-}[/tex]⇒[tex]2I^{-}[/tex] half reaction suggest that you need acid to reaction acurrs.
c)[tex]S^{2-}[/tex]⇒[tex]S+ 2e^{-}[/tex]
[tex]NO_{3} ^{-} + 4H^{+} + 3e^{-}[/tex]⇒[tex]NO + 2H_{2}O[/tex]
d) [tex]2NH_{3} + 6OH^{-}[/tex]⇒[tex]N_{2} + 6H_{2}O +6e^{-}[/tex]
[tex]2NO_{2} + 8H^{+} + 8e^{-}[/tex]⇒[tex]N_{2} + 4H_{2}O[/tex] note that you can obtain the total reaction by miltiplying first half reaction by 4 and second half reaction by 3.
