Chapter 4, Problem 4.29P

4-29 Balance these net ionic equations.

(a) Ag⁺(aq) + Br^-(aqAgBr(s)

(b) Cd²(aq) + S^2-(aq) CdS(s)

(c) Sc³(aq) + SO₄²-aq) Sc₂(SO₄)₃(s)

(d) Sn²(aq) + Fe²(aqSn(s) + Fe³(aq)

(e) K(s) + H₂OK⁺(aq) + OH^-(aq) + H₂(g)

Interpretation Introduction

(a)

Interpretation:

The balanced net ionic equation for the given reaction should be identified.

Concept Introduction:

For a chemical reaction, when the number of atom of each element in the reactant side are equal to the number of atoms of each elements in the product side then such reactions are said to be a balanced chemical equation.

Answer to Problem 4.29P

The balanced chemical equation is.

${\text{Ag}}^{\text{+}}\left(\mathrm{aq}\right){\text{ + Br}}^{\text{-}}\left(\mathrm{aq}\right)\to \text{ AgBr }\left(\text{s}\right)$.

Explanation of Solution

The given reaction is:

${\text{Ag}}^{\text{+}}\left(\mathrm{aq}\right){\text{ + Br}}^{\text{-}}\left(\mathrm{aq}\right)\to \text{ AgBr }\left(\text{s}\right)$

1 mol of Silver ion on reacting with 1 mol of bromide produces 1 mol silver bromide.

${\text{Ag}}^{\text{+}}\left(\mathrm{aq}\right){\text{ + Br}}^{\text{-}}\left(\mathrm{aq}\right)\to \text{ AgBr }\left(\text{s}\right)$

Now, the reaction is balanced with 1 number of Ag on reactant side and on product side.

The reaction is balanced with 1 number of Br on reactant side and on product side.

Interpretation Introduction

(b)

Interpretation:

The balanced net ionic equation for the given reaction should be identified.

Concept Introduction:

For a chemical reaction, when the number of atom of each element in the reactant side are equal to the number of atoms of each elements in the product side then such reactions are said to be a balanced chemical equation.

Answer to Problem 4.29P

The balanced chemical equation is.

${\text{Cd}}^{\text{2+}}\left(\mathrm{aq}\right){\text{ + S}}^{\text{2-}}\left(\mathrm{aq}\right)\to \text{ CdS }\left(\text{s}\right)$.

Explanation of Solution

The given reaction is:

${\text{Cd}}^{\text{2+}}\left(\mathrm{aq}\right){\text{ + S}}^{\text{2-}}\left(\mathrm{aq}\right)\to \text{ CdS }\left(\text{s}\right)$

1 mol cadmium ion on reacting with 1` mol sulfide produces 1 mol cadmium sulfide.

${\text{Cd}}^{\text{2+}}\left(\mathrm{aq}\right){\text{ + S}}^{\text{2-}}\left(\mathrm{aq}\right)\to \text{ CdS }\left(\text{s}\right)$

Now, the reaction is balanced with 1 number of Cd on reactant side and on product side.

The reaction is balanced with 1 number of s on reactant side and on product side.

Interpretation Introduction

(c)

Interpretation:

The balanced net ionic equation for the given reaction should be identified.

Concept Introduction:

For a chemical reaction, when the number of atom of each element in the reactant side are equal to the number of atoms of each elements in the product side then such reactions are said to be a balanced chemical equation.

Answer to Problem 4.29P

The balanced chemical equation is.

${\text{2 Sc}}^{\text{3+}}\left(\mathrm{aq}\right){\text{ + 3 SO}}_{\text{4}}{}^{\text{2-}}\left(\mathrm{aq}\right)\to {\text{Sc}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}\left(\text{s}\right)$.

Explanation of Solution

The given reaction is:

${\text{Sc}}^{\text{3+}}\left(\mathrm{aq}\right){\text{ + SO}}_{\text{4}}{}^{\text{2-}}\left(\mathrm{aq}\right)\to {\text{Sc}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}\left(\text{s}\right)$

As, only 4 mol O are present in reactant side while 12 mol in product side and thus 3 to be multiplied with SO₄^2-.

Now, 2 mol Sc present in product side while 1 in reactant side and thus 2 to be multiplied with Sc³ +.

2 mol scandium ion on reacting with 3` mol sulfate produces 1 mol scandium sulfate.

${\text{2 Sc}}^{\text{3+}}\left(\mathrm{aq}\right){\text{ + 3 SO}}_{\text{4}}{}^{\text{2-}}\left(\mathrm{aq}\right)\to {\text{Sc}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}\left(\text{s}\right)$

Now, the reaction is balanced with 2 number of Sc on reactant side and on product side.

The reaction is balanced with 3 number of s on reactant side and on product side.

The reaction is balanced with 12 number of oxygen on reactant side and on product side.

Interpretation Introduction

(d)

Interpretation:

The balanced net ionic equation for the given reaction should be identified.

Concept Introduction:

For a chemical reaction, when the number of atom of each element in the reactant side are equal to the number of atoms of each elements in the product side then such reactions are said to be a balanced chemical equation.

Answer to Problem 4.29P

The balanced chemical equation is.

${\text{Sn}}^{\text{2+}}\left(\mathrm{aq}\right){\text{ + 2 Fe}}^{\text{2+}}\left(\mathrm{aq}\right)\to \text{ Sn }\left(\text{s}\right){\text{ + 2 Fe}}^{\text{3+}}\left(\mathrm{aq}\right)$.

Explanation of Solution

The given reaction is:

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

Half reaction at anode,

${\text{Fe}}^{\text{2+}}\left(\mathrm{aq}\right)\to {\text{ Fe}}^{\text{3+}}\left(\mathrm{aq}\right)$

A single electron is added to product to balance the charge.

$\text{2 *}\left[{\text{Fe}}^{\text{2+}}\left(\mathrm{aq}\right)\to {\text{ Fe}}^{\text{3+}}\left(\mathrm{aq}\right){\text{ + e}}^{\text{-}}\right]\text{ -------}\left(\text{1}\right)$

Half reaction at cathode,

Two electron is added to reactant to balance the charge.

${\text{Sn}}^{\text{2+}}\left(\mathrm{aq}\right){\text{ + 2 e}}^{\text{-}}\to \text{Sn }\left(\text{s}\right)\text{ -------}\left(\text{2}\right)$

Adding equation 1 and 2 for balanced reaction.

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

2 mol tin ion on reacting with 2` mol iron(ii) ion produces 1 mol tin with 2 mol iron (iii).

Interpretation Introduction

(e)

Interpretation:

The balanced net ionic equation for the given reaction should be identified.

Concept Introduction:

For a chemical reaction, when the number of atom of each element in the reactant side are equal to the number of atoms of each elements in the product side then such reactions are said to be a balanced chemical equation.

Answer to Problem 4.29P

The balanced chemical equation is.

$\text{2 K}\left(\text{s}\right){\text{ + H}}_{\text{2}}\text{O }\left(l\right){\text{ H}}^{\text{+ }}\to {\text{2 K}}^{\text{+}}\left(\mathrm{aq}\right){\text{ + OH}}^{\text{-}}\left(\mathrm{aq}\right){\text{ + H}}_{\text{2}}\left(g\right)$.

Explanation of Solution

The given reaction is:

$\text{K}\left(\text{s}\right){\text{ + H}}_{\text{2}}\text{O }\left(l\right)\to {\text{ K}}^{\text{+}}\left(\mathrm{aq}\right){\text{ + OH}}^{\text{-}}\left(\mathrm{aq}\right){\text{ + H}}_{\text{2}}\left(g\right)$

Half reaction at anode,

$\text{K}\left(\text{s}\right)\to {\text{ K}}^{\text{+}}\left(\mathrm{aq}\right)$

A single electron is added to product to balance the charge.

$\text{2 *}\left[\text{ K}\left(\text{s}\right)\to {\text{ K}}^{\text{+}}\left(\mathrm{aq}\right){\text{ + e}}^{\text{-}}\right]\text{ -------}\left(\text{1}\right)$

Half reaction at cathode,

${\text{H}}_{\text{2}}\text{O }\left(l\right)\to {\text{ OH}}^{\text{-}}\left(\mathrm{aq}\right){\text{ + H}}_{\text{2}}\left(g\right)$

Two electron is added to reactant along with proton to balance the charge and number of hydrogen.

${\text{H}}_{\text{2}}\text{O }\left(l\right){\text{ + H}}^{\text{+}}{\text{+ e}}^{\text{-}}\to {\text{ OH}}^{\text{-}}\left(\mathrm{aq}\right){\text{ + H}}^{+}\left(\text{a}q\right){\text{ H}}_2\text{(g) }$

${\text{2(H}}_{\text{2}}\text{O }\left(l\right){\text{ + H}}^{\text{+}}{\text{+ e}}^{\text{-}}\to {\text{ OH}}^{\text{-}}\left(\mathrm{aq}\right){\text{ + H}}^{+}\left(\text{a}q\right)\text{ +}{\text{H}}_2\text{(g)) -------}\left(\text{2}\right)$

Adding equation 1 and 2 for balanced reaction.

$\text{2 K}\left(\text{s}\right){\text{ + 2H}}_{\text{2}}\text{O }\left(l\right)\to {\text{2 K}}^{\text{+}}\left(\mathrm{aq}\right){\text{ + 2OH}}^{\text{-}}\left(\mathrm{aq}\right){\text{ + 2H}}_{\text{2}}\left(g\right)$

2 mol potassium ion on reacting with 1` mol water produces 2 mol potassium ion with 1 mol hydrogen.