Question

Consider the following standard reduction potentials in acid solution:
E^o(V)

Al3+ + 3e– ? Al(s) –1.66
AgBr(s) + e– ? Ag(s) + Br– +0.07
Sn4+ + 2e– ? Sn2+ +0.14
Fe3+ + e– ? Fe2+ +0.77
The strongest reducing agent among those shown above is

Answer 1

Thehalf-reactions with themorepositive electrode potential will undergo

reduction, so they are goodoxidizing agents.

Thehalf-reactions with themorenegative electrodepotential will undergo oxidation, so they aregoodoxidizing agents.

The providedstandardreduction potentialsareasfollows:

\(\mathrm{Al}^{3+}(a q)+3 e^{-} \longrightarrow \mathrm{Al}(s) \quad E^{\circ}=-1.66 \mathrm{~V}\)

\(\operatorname{AgBr}(s)+e^{-} \longrightarrow \mathrm{Ag}(s)+\mathrm{Br}^{-} \quad E^{\circ}=+0.07 \mathrm{~V}\)

\(\mathrm{Sn}^{4+}(a q)+2 e^{-} \longrightarrow \mathrm{Sn}^{2+}(a q) \quad E^{\circ}=+0.14 \mathrm{~V}\)

\(\mathrm{Fe}^{3+}(a q)+e^{-} \longrightarrow \mathrm{Fe}^{2+}(a q) \quad E^{\circ}=+0.77 \mathrm{~V}\)

Thestrongerreducingagent will bethemorenegativevalue.

Here, \(\mathrm{Al}(s)\) hasmorenegative value. Thus, thestrongest reducing agentis:

\(\mathrm{Al}^{3+}(a q)+3 e^{-} \longrightarrow \mathrm{Al}(s) \quad E^{\circ}=-1.66 \mathrm{~V}\)