Q. 7.48 - 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)$2\mathrm{Li}\left(\mathrm{s}\right) + {\mathrm{F}}_2\left(\mathrm{g}\right) \to 2\mathrm{LiF}\left(\mathrm{s}\right)$ 

b)${\mathrm{Cl}}_2\left(\mathrm{g}\right) + 2\mathrm{KI}\left(\mathrm{aq}\right) \to \mathrm{I}\left(\mathrm{s}\right) + 2\mathrm{KCl}\left(\mathrm{aq}\right)$ 

c)$2\mathrm{Al}\left(\mathrm{s}\right) + 3{\mathrm{Sn}}^{2+}\left(\mathrm{aq}\right) \to 3\mathrm{Sn}\left(\mathrm{s}\right) + 2{\mathrm{Al}}^{3+}\left(\mathrm{aq}\right)$ 

d)$\mathrm{Fe}\left(\mathrm{s}\right) + {\mathrm{CuSO}}_4\left(\mathrm{aq}\right) \to \mathrm{Cu}\left(\mathrm{s}\right) + {\mathrm{FeSO}}_4\left(\mathrm{aq}\right)$ 

Step-by-Step Solution

Answer

(a) Li is oxidized and F is reduced.

(b) I is oxidized and Cl is reduced.

(c) Sn is reduced and Al is oxidized.

(d) Fe is oxidized and Sn is reduced.

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 $2\mathrm{Li}\left(\mathrm{s}\right) + {\mathrm{F}}_2\left(\mathrm{g}\right) \to 2\mathrm{LiF}\left(\mathrm{s}\right)$

$2$electron is getting added from Li on the product side and $2$electron is reduced from F . The reactant in which an atom loses its electron is reducing, and the reactant in which an atom gains its electron is oxidizing.

Hence, Li is oxidizing and F 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 ${\mathrm{Cl}}_2\left(\mathrm{g}\right) + 2\mathrm{KI}\left(\mathrm{aq}\right) \to \mathrm{I}\left(\mathrm{s}\right)+2\mathrm{KCl}\left(\mathrm{aq}\right)$

$1$electron is being added from I 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, I 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 $2\mathrm{Al}\left(\mathrm{s}\right) + 3{\mathrm{Sn}}^{+2}\left(\mathrm{aq}\right) \to 3\mathrm{Sn}\left(\mathrm{s}\right) + 2{\mathrm{Al}}^{+3}\left(\mathrm{aq}\right)$

$3$electron is being added from Al on the product side and $2$electron is removed from Sn. 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, Al is oxidizing and Sn 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 $\mathrm{Fe}\left(\mathrm{s}\right) + {\mathrm{CuSO}}_4\left(\mathrm{aq}\right) \to \mathrm{Cu}\left(\mathrm{s}\right) + {\mathrm{FeSO}}_4\left(\mathrm{aq}\right)$

$2$electron is being added from Cu on the product side and $2$electron is being reduced from Fe. The reactant in which an atom loses its electron is reducing, and the reactant in which an atom gains its electron is oxidizing.

Hence, Cu is oxidizing and Fe is reduced.