How to separate this redox reaction into two half reactions?

[Dhamnekar Winod]

How to separate this redox reaction into two half reactions?

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

[ChenBeier]

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

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

[Dhamnekar Winod]

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?

[ChenBeier]

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.

[Dhamnekar Winod]

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?

[ChenBeier]

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.