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I read that silver in the presence of hydrogen sulfide corrodes to form silver sulfide and hydrogen.

$$\ce{Ag + H2S -> AgS + H2}$$

But in the reactivity series silver is placed much below hydrogen.

So, how is this displacement reaction taking place at all?

Reference: J. Chem. Educ. 2000, 77, 3, 328A; doi 10.1021/ed077p328A

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    $\begingroup$ Are you aware the reactivity serie applies to hydrated ions ? $\endgroup$ – Poutnik Oct 11 '20 at 15:45
  • $\begingroup$ @Poutnik Franky, no. I only know the reactivity series of metals and anions. $\endgroup$ – Shub Oct 11 '20 at 15:47
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    $\begingroup$ I don't quite believe this happens. Oxygen is a part of the process too. $\endgroup$ – Ivan Neretin Oct 11 '20 at 15:51
  • $\begingroup$ @Shub Please note ChemSE uses mhchem to ease writing chemical equations / formulae. It functions well in questions, answers, comments; but because it is something special, you should not use it to format the title of a question. Its scope is described on mhchem.github.io/MathJax-mhchem. $\endgroup$ – Buttonwood Oct 11 '20 at 16:29
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    $\begingroup$ And AgS is not formed. $\endgroup$ – Ed V Oct 11 '20 at 19:16
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This reaction happens in nature so I can't deny it of course.

This is what I thought about the Half cell reactions

Oxidizing half cell reaction

The oxidizing half cell potential value was taken by this document link below pg 3.

https://hal.sorbonne-universite.fr/hal-01022872/file/Post-print_P1425.pdf

Ag2S + 2e- → 2Ag + S2- E = -0.691 V SHE

Reducing half cell reaction

2H2O + 2 e- -> H2 + 2OH- E = -0.828 V SHE

Ecell = Ecathode(re) - Eanode(ox)

Thus

Ecell = -0.828 V + 0.691 V = -0.137 V

We get a negative value for Ecell thus non spontaneous of course.

According to the equation of thermodynamics

ΔGo=−nFEocell

But aren't we talking about Silver Tarnish so its a poluted media with acidic H+ from H2S

Reducing reaction depends on pH acidity off the polluted media

Reducing reaction pH dependence equation

Reffer here link

https://web.viu.ca/krogh/chem301/water%20redox%20boundaries.pdf

E = Ecell - (RT/2F)ln([OH-]2) (H2 is in 1 atm pressure standard state)

See at acidic medium the E value gets higher thus

Ecell = (-0.828 V <-- ++++) + 0.691 V = (-0.137 V <--- ++++);

When the cell voltage is positive thermodynamic-ally feasible thus reaction happens

(ΔG < 0 spontaneous reaction)

Final reaction,

2Ag + S2- + 2H2O -> H2 + 2OH- + Ag2S

But where dose S2- come from its H2S

So

2Ag + 2H+ + S2- + 2H2O -> H2 + 2H+ + 2OH- + Ag2S

Simplifying water goes away!

There we have it final answer

2Ag + H2S⟶Ag2S + H2

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