Write a balanced net ionic equation for the redox reaction of Ag2S and Al metal that produces Ag metal and Al(OH)3.

Question

Write a balanced net ionic equation for the redox reaction of Ag₂S and Al metal that produces Ag metal and Al(OH)₃. 

Low concentrations of hydrogen sulfide in air react with silver to form Ag₂S, more familiar to us as tarnish. Silver polish contains aluminum metal powder in an alkaline suspension.

I think the equation is 3Ag2 (s) + 2Al(s) + 6OH- --> 6 Ag(s) + 2S^2- (aq) + 2Al(OH)3 (s)

but it's wrong, and I'm trying to determine where I misplaced or misaligned the values

Answer 1

Ag₂S(s) + Al(s) ==> Ag(s) + S^2-(aq) + Al(OH)₃(s) .. unbalanced redox

Looking at the reduction half reaction:

Ag₂S(s) ==> Ag(s) + S^2-(aq) .. unbalanced

Ag₂S(s) ==> 2Ag(s) + S^2-(aq) .. balanced for Ag and S

Ag₂S(s) + 2e- ==> 2Ag(s) + S^2-(aq) .. balanced for Ag, S and charge = balanced reduction reaction

Looking at the oxidation half reaction:

Al(s) ==> Al(OH)₃(s) .. unbalanced

Al(s) + 3H₂O ==> Al(OH)₃(s) .. balanced for Al and O

Al(s) + 3H₂O + 3OH^-(aq) ==> Al(OH)₃(s) + 3H₂O .. balanced for Al, O and H using base (OH^-)

Al(s) + 3H₂O + 3OH^-(aq) ==> Al(OH)₃(s) + 3H₂O + 3e- .. balanced for Al,O,H and charge = balanced rxn

Multiply reduction reaction by 3 and oxidation reaction by 2 in order to equalize electrons. Then add the two balanced equations together and combine/cancel like terms to end up with the final balanced equation.

3Ag₂S(s) + 6e- ==> 6Ag(s) + 3S^2-(aq)

2Al(s) + 6H₂O + 6OH^-(aq) ==> 2Al(OH)₃(s) + 6H₂O + 6e-

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3Ag₂S(s) + 2Al(s) + 6OH^-(aq) ==> 6Ag(s) + 3S^2-(aq) + 2Al(OH)₃(s) .. balanced redox equation

Since there are no spectator ions, this should be the net ionic equation.