Question

Given the equation: 2cr(s) + 3pb2+(aq) → 2cr3+(aq) + 3pb(s), which is the correct reduction half reaction?a.pb2+(aq) → pb(s) + 2e-b.cr(s) → cr3+(aq) + 3e-c.cr(s) + 3e- → cr3+(aq)d.pb2+(aq) + 2e- → pb(s)

Answer 1

Answer: Reduction half reaction is $Pb^{2+}(aq) + 2e^{-} \rightarrow Pb(s)$.

Explanation:

When there is decrease in oxidation number by gain of electrons then it is known as reduction.

For example, the given reaction is as follows.

  $2Cr(s) + 3Pb^{2+}(aq) \rightarrow 2Cr^{3+}(aq) + 3Pb(s)$

Reduction half reaction: $Pb^{2+}(aq) + 2e^{-} \rightarrow Pb(s)$

Oxidation half reaction: $Cr(s) \rightarrow Cr^{3+}(aq) + 3e^{-}$

Thus, we can conclude that option (d) is the correct reduction half reaction.

Answer 2

Answer:

Pb2+(aq)+2e-=pb(s)

Explanation: