# Question #a49bd

May 21, 2015

Your first reaction is actually a redox reaction (equilibrium, to be more precise) in which the silver cation, $A {g}^{+}$, oxidizes the iron (II) cations, $F {e}^{2 +}$, to iron (III) cations, $F {e}^{3 +}$.

In aqueous solution, the nitrate ions will act as spectator ions, so the net ionic equation will look like this

$F {e}_{\left(a q\right)}^{3 +} + A {g}_{\left(a q\right)}^{+} r i g h t \le f t h a r p \infty n s F {e}_{\left(a q\right)}^{3 +} + A {g}_{\left(s\right)}$

The second one is a little tricky because you have two species that do not ionize in aqueous solution, phosphoric acid, ${H}_{3} P {O}_{4}$, and hydrogen sulfide, ${H}_{2} S$.

The complete ionic equation looks like this

$2 {H}_{3} P {O}_{4 \left(a q\right)} + 6 N {a}_{\left(a q\right)}^{+} + 3 {S}_{\left(a q\right)}^{2 -} \to 3 {H}_{2} {S}_{\left(g\right)} + 6 N {a}_{\left(a q\right)}^{+} + 2 P {O}_{4 \left(a q\right)}^{3 -}$

If you eliminate the spectator ions, you'll get

$2 {H}_{3} P {O}_{4 \left(a q\right)} + 3 {S}_{\left(a q\right)}^{2 -} \to 3 {H}_{2} {S}_{\left(a q\right)} + 2 P {O}_{4 \left(a q\right)}^{3 -}$