I'm incredibly lost right now. I understand fully how to balance redox reactions in acidic/basic solutions, but the problem my teacher posted is not letting me use the method taught to us. I've searched online for help too, but can't find any.

The equation to balance with the solution is:

$$\ce{6 Ag + 3 NaSH + 2 K2CrO4 -> 3 Ag2S + 2 Cr(OH)3 + 4 KOH + 3 NaOH}\tag{R1}$$

The solution he provided doesn't show work, but it includes an $\ce{SH-}$ and I have no clue where that can come from.

I wrote separate half-reactions:

$$ \begin{align} \ce{6 \overset{0}{Ag} &-> 3 \overset{+1}{Ag}_2S}\tag{R2} \\ \ce{2 K2\overset{+6}{Cr}O4 &-> 2 \overset{+3}{Cr}(OH)3 + 4 KOH}\tag{R3} \\ \end{align} $$

This doesn't work unfortunately, I'm just so stuck.

  • 4
    $\begingroup$ These "half rxns" are anything but half-reactions. Where are the electrons and what happened to the mass balance? $\endgroup$
    – andselisk
    Nov 14 at 17:30
  • 1
    $\begingroup$ Please visit this page, this page and this one on how to format your posts better with MathJax and Markdown. $\endgroup$
    – andselisk
    Nov 14 at 17:32
  • $\begingroup$ The Ag is oxidized since its oxidation number changes from 0 to +1, and Cr is reduced since it changes from +6 to +3, which is why I set up my reactions as such. I really don't see why they're wrong to begin with. $\endgroup$ Nov 14 at 17:42
  • 2
    $\begingroup$ Half-reactions show electrons on one side or the other. They are also mass balanced, as @andselisk also said. And you are kidding yourself if you believe you “understand fully how to balance redox reactions in acidic/basic solutions”. $\endgroup$
    – Ed V
    Nov 14 at 17:51
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    $\begingroup$ The oxidation numbers (which I appended into formatted reactions) are fine, it's electrons that is the crucial part of the half-reactions method. And there is problematic sulfur in R2 (mass balance) and potassium has no place in R3. $\endgroup$
    – andselisk
    Nov 14 at 17:57

This equation is the superposition of two equations : 1) oxidation of $\ce{Ag}$ by $\ce{K2CrO4}$ producing the ion $\ce{Ag+}$, 2) precipitation of $\ce{Ag+}$ by $\ce{NaSH}$. These two equations are : $$\ce{3 Ag + K2CrO4 + 4 H2O -> 3 Ag+ + Cr(OH)3 + 2 KOH + 3 OH-}$$ $$\ce{2 Ag+ + Na+ + SH- + OH- -> Ag2S + H2O + Na+}$$ Multiplying the first equation by $2$ and the second by $3$, and adding the results gives, after simplification :
$$\ce{6 Ag + 2 K2CrO4 + 5 H2O + 3 NaSH -> 3 Ag2S + 2 Cr(OH)3 + 4 KOH + 3 NaOH}$$ Hopefully you have followed my development. I have forgotten intermediate calculations, which are usually well treated in classroom. In case of necessity I may develop some further steps of this reasoning.


You can do it using the standard redox half-reaction method, but because chromium does not form free ions and the products do not include free silver ions you need to incorporate additional species to match the reactants and products. This would then enable the electron-balanced sum to reflect the actual reactants and products.

Tarnishing silver

Consider first the silver oxidation reaction. With free silver ions it would simply be $\ce{Ag -> Ag^+ + e^-}$, but here the bisulfide ion must be added to provide a sulfur source for $\ce{Ag2S}$. So we render

$\ce{2Ag + SH^- -> Ag2S + 2 e^-} \text{ (unbalanced)}$

where the two electrons are determined by the change in oxidation state of two silver atoms, not by charge balance. So charges along with the hydrogen must be balanced in a second step. In basic solution, we use hydroxide ions to balance the charges and then water molecules to balance hydrogen and oxygen atoms. This gives

$\color{blue}{\ce{2Ag + SH^- +OH^- -> Ag2S +H2O +2 e^-}}$

Chromium gets down to basics

Chromium reduction is rendered as reduction of chromate to chromium(III) hydroxide. Before balancing this would be rendered as

$\ce{CrO4^{2-} + 3e^- -> Cr(OH)3}\text{ (unbalanced)}$

We need five hydroxide ions to balance charges and then four water molecules to balance hydrogen and oxygen atoms:

$\color{blue}{\ce{CrO4^{2-} + 4H2O + 3e^- -> Cr(OH)3 + 5 OH^-}}$

Adding it all up

Now add three times the silver oxidation to twice the chromium reduction to balance electrons:

$\ce{2 CrO4^{2-} + 6 Ag + 3 SH^- + \color{blue}{5 H2O} -> 2 Cr(OH)3 + 3 Ag2S + 7 OH^-}$

Note the water molecules that were missing from the given equation.

All that remains is to add the spectator ions. Since the hydrosulfide is added specifically as $\ce{(3) NaSH}$ and the chromate as $\ce{(2) K2CrO4}$, there will be three sodium ions and four potassium ions, which is reflected in the ratio of the two alkali products.

  • 1
    $\begingroup$ IMO, it would help if you made it clearer that the "reactions" you've drawn in black are deliberately unbalanced, since that's easy to miss at a glance and since the OP here specifically seems to have a poor and/or incorrect understanding of how reactions should be balanced. Maybe something like "$\ce{2Ag + SH^- -> Ag2S + 2 e^- + \dots}$" or "$\ce{2Ag + SH^- ->T[unbalanced] Ag2S + 2 e^-}$" might work? (Alas, I can't figure out how to properly typeset "$\not\to$" using mhchem.) $\endgroup$ Nov 15 at 19:08
  • $\begingroup$ Well, somebody didn't like my latest edit I guess. $\endgroup$ Nov 15 at 21:48
  • $\begingroup$ @ilmari I just wrote the word "unbalanced", but \not\to seems to work for your symbol. $\endgroup$ Nov 15 at 22:12

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