Identify the oxidation and reduction in Tollen’s test

• An aldehyde group, $-CHO$, is oxidized to a carboxylate/carboxylic acid.
• Tollen’s reagent, $[Ag(NH_3)_2]^+$, is reduced to metallic silver, $Ag$.

Write the oxidation half-reaction for aldehyde

In basic medium, the aldehyde is oxidized as: $RCHO + 3OH^- \rightarrow RCOO^- + 2H_2O + 2e^-$

This shows that one aldehyde group loses 2 electrons.

Write the reduction half-reaction for Tollen’s reagent

Each silver ion in Tollen’s reagent gains one electron: $[Ag(NH_3)_2]^+ + e^- \rightarrow Ag + 2NH_3$

So, one mole of $[Ag(NH_3)_2]^+$ accepts 1 electron.

Compare electrons transferred

Since one aldehyde group produces $2e^-$, it can reduce two silver complexes: $RCHO + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow RCOO^- + 2Ag + 4NH_3 + 2H_2O$

Conclusion

Therefore, the overall number of electrons transferred to Tollen’s reagent per aldehyde group is: $\boxed{2}$

Rounded to the nearest integer, it remains $2$.

Comparison with stored correct answer

Stored correct answer = $2$

Our derived answer = $2$

Hence, they agree.