Question #a49bd

Question

Question #a49bd

1 Answer

Impact of this question

2058 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-05-21T01:26:57

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

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

$F e_{(a q)}^{3 +} + A g_{(a q)}^{+} ⇌ F e_{(a q)}^{3 +} + A g_{(s)}$ $Fe_((aq))^(3+) + Ag_((aq))^(+) rightleftharpoons Fe_((aq))^(3+) + Ag_((s))$

The second one is a little tricky because you have two species that do not ionize in aqueous solution, phosphoric acid, $H_{3} P O_{4}$ $H_3PO_4$ , and hydrogen sulfide, $H_{2} S$ $H_2S$ .

The complete ionic equation looks like this

$2 H_{3} P O_{4 (a q)} + 6 N a_{(a q)}^{+} + 3 S_{(a q)}^{2 -} \to 3 H_{2} S_{(g)} + 6 N a_{(a q)}^{+} + 2 P O_{4 (a q)}^{3 -}$ $2H_3PO_(4(aq)) + 6Na_((aq))^(+) + 3S_((aq))^(2-) -> 3H_2S_((g)) + 6Na_((aq))^(+) + 2PO_(4(aq))^(3-)$

If you eliminate the spectator ions, you'll get

$2 H_{3} P O_{4 (a q)} + 3 S_{(a q)}^{2 -} \to 3 H_{2} S_{(a q)} + 2 P O_{4 (a q)}^{3 -}$ $2H_3PO_(4(aq)) + 3S_((aq))^(2-) -> 3H_2S_((aq)) + 2PO_(4(aq))^(3-)$

Science

Math

Humanities

... and beyond

Question #a49bd

1 Answer

Related questions

Impact of this question