Q. 7.47 - An Introduction to General, Organic, and Biological Chemistry

Question

In each of the following, identify the reactant that is oxidized and the reactant that is reduced: 

a. $Zn\left(s\right) + C{l}_2\left(g\right)\to ZnC{l}_2\left(s\right)$

b.${\mathrm{CI}}_2\left(\mathrm{g}\right) + 2\mathrm{NaBr}\left(\mathrm{aq}\right)\to 2\mathrm{NaCl}\left(\mathrm{aq}\right) + {\mathrm{Br}}_2\left(\mathrm{l}\right)$

c.$2\mathrm{PbO}\left(\mathrm{s}\right) \to 2\mathrm{Pb}\left(\mathrm{s}\right) + {\mathrm{O}}_2\left(\mathrm{g}\right)$

d.$2\mathrm{Fe} \left(\mathrm{aq}\right) + {\mathrm{Sn}}^{2+}\left(\mathrm{aq}\right) \to 2{\mathrm{Fe}}^{2+} \left(\mathrm{aq}\right) + {\mathrm{Sn}}^{4+} \left(\mathrm{aq}\right)$

Step-by-Step Solution

Answer

(a) Zn is oxidized and $C{l}_2$is reduced.

(b) Br is oxidized and Cl is reduced.

(c) Pb is reduced and O is oxidized.

(d) Fe and Sn are oxidized.

Part (a) Step 1: Given Information

We are given the equations and we have to find out which reactant is oxidizing or reducing.

Part (a) Step 2: Explanation

Now, in $Zn\left(s\right) + C{l}_2\left(g\right)\to ZnC{l}_2\left(s\right)$

$2$electron is getting added from Zn on the product side and data-custom-editor="chemistry" $2$electron is reduced from Cl. The reactant in which an atom loses its electron is reducing, and the reactant in which an atom gains its electron is oxidizing.

Hence, Zn is oxidizing and Cl is reduced.

Part (b) Step 1: Given Information

We are given the equations and we have to find out which reactant is oxidizing or reducing.

Part (b) Step 2: Explanation

Now, in ${\mathrm{CI}}_2\left(\mathrm{g}\right) + 2\mathrm{NaBr}\left(\mathrm{aq}\right)\to 2\mathrm{NaCl}\left(\mathrm{aq}\right) + {\mathrm{Br}}_2\left(\mathrm{l}\right)$

$1$electron is being added from Br on the product side and $1$electron is being reduced from Cl. The reactant in which an atom loses its electron is reducing, and the reactant in which an atom gains its electron is oxidizing.

Hence, Br is oxidizing and Cl is reduced. 

Part (c) Step 1: Given Information

We are given the equations and we have to find out which reactant is oxidizing or reducing.

Part (c) Step 2: Explanation

Now, in $2\mathrm{PbO}\left(\mathrm{s}\right) \to 2\mathrm{Pb}\left(\mathrm{s}\right) + {\mathrm{O}}_2\left(\mathrm{g}\right)$

$1$electron is being added from O on the product side and as oxygen is removed from Pb. The reactant in which an atom loses its electron or loses oxygen is reducing, and the reactant in which an atom gains its electron is oxidizing.

Hence, O is oxidizing and Pb is reduced. 

Part (d) Step 1: Given Information

We are given the equations and we have to find out which reactant is oxidizing or reducing.

Part (d) Step 2: Explanation

Now, in $2\mathrm{Fe} \left(\mathrm{aq}\right) + {\mathrm{Sn}}^{+2}\left(\mathrm{aq}\right) \to 2{\mathrm{Fe}}^{+2}\left(\mathrm{aq}\right) + {\mathrm{Sn}}^{+4} \left(\mathrm{aq}\right)$

$2$electron is being added to Fe and Sn on the product side. The reactant in which an atom loses its electron or loses oxygen is reducing, and the reactant in which an atom gains its electron is oxidizing.

Hence, both Fe and Sn are oxidizing.