Thread: Redox in basic solution View Single Post
 Sci Advisor PF Gold P: 613 You need help, my friend First, prepare atomic-based electron balance: $$N^{2+} \longrightarrow N^{5+} + 3e^-$$ $$Mn^{7+}+5e^- \longrightarrow Mn^{2+}$$ Then balance them. $$5N^{2+}+3Mn^{7+}\longrightarrow 5N^{5+}+3Mn^{2+}$$ You'll see that both electron counts and atom counts are balanced now. When you write the "real" ions, I mean, NO and MnO4-, you'll have to add OH- and H2O to the side with less oxygen and less hydrogen, etc.