Chemistry Forum

How to separate this redox reaction into two half reactions?

\( HCHO(l) + 2[Ag(NH_3)_2]^+(aq) + 3OH^-(aq) \rightarrow 2Ag(s) + HCOO^-(aq) + 4NH_3(aq) + 2H_2O(l)\)

HCHO + 3 OH- => HCOO- + 2 H₂O + 2 e- Oxidation

[Ag(NH₃)₂]+ + e- => Ag + 2 NH₃ Reduction

You have given oxidation and reduction reactions in your reply. But in other chemistry notes, these reactions are given as below:

\(H_2O(l) + HCHO(l) \rightarrow HCOO^-(aq) +3H^+ +2e^-\Rightarrow\) OXIDATION


\( [Ag(NH_3)_2]^+ (aq) +e^- \rightarrow Ag(s) + 2NH_3(aq) + 2H_2O(l) \Rightarrow\) REDUCTION

What is your explanation for these half reactions?

It depends which pH you have. If its acidic environment then H+ is used, in alcaline condition OH-.

But the right one is the alcaline one in this case. The Diamminsilver is not stable in acidic condition.

We get the following redox reactions by adding the two half reactions \(H_2O (l) + HCHO(l) \rightarrow HCOO^-(aq) + 3H^+ + 2e^- \tag{Oxidation} \) and \([Ag(NH_3)_2]^+(aq) + e^- \rightarrow Ag(s) + 2NH_3(aq) + 2H_2O(l) \times 2 \tag{Reduction}\)

\(= HCHO(l) + 2[Ag(NH_3)_2]^+(aq) \rightarrow HCOO^-(aq) +3H^+ + 2Ag(s) +4NH_3(aq) + 3H_2O(l)\)

which is different from the original redox equation namely \( HCHO(l) + 2[Ag(NH_3)_2]^+ (aq) + 3OH^- (aq) \rightarrow 2Ag(s) + HCOO^-(aq) +4NH_3(aq) + 2H_2O(l) \)

Where is the mistake? where are we wrong? Why did such discrepancy occur?

Would any chemistry expert answer these questions?

Did you read my answer above. You are wrong because [Ag(NH₃)₂] + is not existing in acidic conditions.

Mathematically you can write both equations, but only one will happen in practise.