Q21.19P - Chemistry: Molecular Nature Of Matter And Change

Question

Balance the following skeleton reactions and identify the oxidizing and reducing agents:

$$\begin{gathered} \mathbf{\text{(a) S}}{{\mathbf{\text{O}}}_{\mathbf{\text{3}}}}^{\mathbf{\text{2 - }}}{\mathbf{\text{(aq) + Cl}}}_{\mathbf{\text{2}}}\mathbf{\text{(g) }}\mathbf{\to }\mathbf{\text{S}}{{\mathbf{\text{O}}}_{\mathbf{\text{4}}}}^{\mathbf{\text{2 - }}}{\mathbf{\text{(aq) + Cl}}}^{\mathbf{\text{ - }}}\mathbf{\text{(aq) [basic]}} \\ \mathbf{\text{(b) Fe(CN}}{{\mathbf{\text{)}}}_{\mathbf{\text{6}}}}^{\mathbf{\text{3 - }}}\mathbf{\text{(aq) + Re(s)}}\mathbf{\to }\mathbf{\text{Fe(CN}}{{\mathbf{\text{)}}}_{\mathbf{\text{6}}}}^{\mathbf{\text{4 - }}}\mathbf{\text{(aq) + Re}}{{\mathbf{\text{O}}}_{\mathbf{\text{4}}}}^{\mathbf{\text{ - }}}\mathbf{\text{(aq) [basic]}} \end{gathered}$$

(C) ${\mathbf{\text{MnO}}}_{\mathbf{\text{4}}}^{\mathbf{\text{ - }}}\mathbf{\text{(aq) + HCOOH(aq)}}\mathbf{\to }{\mathbf{\text{Mn}}}^{\mathbf{\text{2 + }}}{\mathbf{\text{(aq) + CO}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}$

Step-by-Step Solution

Answer

(a). ${SO}_3^{2-}\left(aq\right)+C{l}_2\left(g\right)+2O{H}^{-}\to {\text{SO}}_{\text{4}}^{\text{2 - }}{\text{(aq) + 2Cl}}^{\text{ - }}{\text{(aq) + H}}_{\text{2}}\text{O}$

(b). ${\text{7Fe(CN)}}_{\text{6}}^{\text{3 - }}{\text{(aq) + Re(s) + 8OH}}^{\text{ - }}\to {\text{7Fe(CN)}}_{\text{6}}^{\text{4 - }}{\text{(aq) + ReO}}_{\text{4}}^{\text{ - }}{\text{(aq) + 4H}}_{\text{2}}\text{O}$
(c). ${\text{2MnO}}_{\text{4}}^{\text{ - }}{\text{(aq) + 5HCOOH(aq) + 6H}}^{\text{ + }}\to {\text{2Mn}}^{\text{2 + }}{\text{(aq) + 5CO}}_{\text{2}}{\text{(g) + 8H}}_{\text{2}}\text{O}$

Definition of Chemical Change and Electrical Work

Electrochemistry is a discipline of physical chemistry concerned with the relationship between electrical potential as a quantifiable and quantitative phenomena and recognisable chemical change, with electrical potential as an outcome of a specific chemical change or vice versa.

Find the oxidizing and reducing agents

${\text{(a)SO}}_{\text{3}}^{\text{2 - }}{\text{(aq) + Cl}}_{\text{2}}\text{(g)}\to {\text{SO}}_{\text{4}}^{\text{2 - }}{\text{(aq) + Cl}}^{\text{ - }}\text{(aq)[basic]}$

We'll take the following steps to balance the equation:

Separate the half-reactions in the first step.

 $$\begin{gathered} {\text{(i)Cl}}_{\text{2}}\text{(g)(aq)}\to {\text{Cl}}^{\text{ - }}\text{(aq)} \\ {\text{(ii)SO}}_{\text{3}}^{\text{2 - }}\text{(aq)}\to {\text{SO}}_{\text{4}}^{\text{2 - }}\text{(aq)} \end{gathered}$$

Step 2: All elements except oxygen and hydrogen must be in balance.

$$\begin{gathered} {\text{(i)Cl}}_{\text{2}}\text{(g)}\to {\text{2Cl}}^{\text{ - }}\text{(aq)} \\ {\text{(ii)SO}}_{\text{3}}^{\text{2 - }}\text{(aq)}\to {\text{SO}}_{\text{4}}^{\text{2 - }}\text{(aq)} \end{gathered}$$.

Step 3: Add $H_{2}O$ molecules to balance the oxygen atoms.

$$\begin{gathered} {\text{(i)Cl}}_{\text{2}}\text{(g)}\to {\text{2Cl}}^{\text{ - }}\text{(aq)} \\ {\text{(ii)SO}}_{\text{3}}^{\text{2 - }}{\text{(aq) + H}}_{\text{2}}\text{O}\to {\text{SO}}_{\text{4}}^{\text{2 - }}\text{(aq)} \end{gathered}$$

Step 4: We add protons, $H^{+}$, to balance hydrogen atoms.

$$\begin{gathered} {\text{(i)Cl}}_{\text{2}}\text{(g)}\to {\text{2Cl}}^{\text{ - }}\text{(aq)} \\ {\text{(ii)SO}}_{\text{3}}^{\text{2 - }}{\text{(aq) + H}}_{\text{2}}\text{O}\to {\text{SO}}_{\text{4}}^{\text{2 - }}{\text{(aq) + 2H}}^{\text{ + }} \end{gathered}$$

Step 5: Now, using electrons to balance the charge, $e^{-}$.

 $$\begin{gathered} {\text{(i)Cl}}_{\text{2}}{\text{(g) + 2e}}^{\text{ - }}\to {\text{2Cl}}^{\text{ - }}\text{(aq)} \\ {\text{(ii)SO}}_{\text{3}}^{\text{2 - }}{\text{(aq) + H}}_{\text{2}}\text{O}\to {\text{SO}}_{\text{4}}^{\text{2 - }}{\text{(aq) + 2H}}^{\text{ + }}{\text{ + 2e}}^{\text{ - }} \end{gathered}$$

Step 6: Scale the reactions to make the electron count equal.

$$\begin{gathered} {\text{(i)Cl}}_{\text{2}}{\text{(g) + 2e}}^{\text{ - }}\to {\text{2Cl}}^{\text{ - }}\text{(aq)} \\ {\text{(ii)SO}}_{\text{3}}^{\text{2 - }}{\text{(aq) + H}}_{\text{2}}\text{O}\to {\text{SO}}_{\text{4}}^{\text{2 - }}{\text{(aq) + 2H}}^{\text{ + }}{\text{ + + 2e}}^{\text{ - }} \end{gathered}$$

Step 7: Add the reactions.

${\text{Cl}}_{\text{2}}{\text{(g) + 2e}}^{\text{ - }}{\text{ + SO}}_{\text{3}}^{\text{2 - }}{\text{(aq) + H}}_{\text{2}}\text{O}\to {\text{2Cl}}^{\text{ - }}{\text{(aq) + SO}}_{\text{4}}^{\text{2 - }}{\text{(aq) + 2H}}^{\text{ + }}{\text{ + 2e}}^{\text{ - }}$

Step 8: Cancel out common terms.

${\text{Cl}}_{\text{2}}{\text{(g) + SO}}_{\text{3}}^{\text{2 - }}{\text{(aq) + H}}_{\text{2}}\text{O}\to {\text{2Cl}}^{\text{ - }}{\text{(aq) + SO}}_{\text{4}}^{\text{2 - }}{\text{(aq) + 2H}}^{\text{ + }}$

Step 9: We inject $O{H}^{-}$ to balance the $H^{+}$ ions because the process occurs in a basic media.

${\text{Cl}}_{\text{2}}{\text{(g) + SO}}_{\text{3}}^{\text{2 - }}{\text{(aq) + H}}_{\text{2}}{\text{O + 2OH}}^{\text{ - }}\to {\text{2Cl}}^{\text{ - }}{\text{(aq) + SO}}_{\text{4}}^{\text{2 - }}{\text{(aq) + 2H}}^{\text{ + }}{\text{ + 2OH}}^{\text{ - }}$

Step 10: To make a water molecule, combine $O{H}^{-}$ ions and $H^{+}$ ions on the same side.

${\text{Cl}}_{\text{2}}{\text{(g) + SO}}_{\text{3}}^{\text{2 - }}{\text{(aq) + H}}_{\text{2}}{\text{O + 2OH}}^{\text{ - }}\to {\text{2Cl}}^{\text{ - }}{\text{(aq) + SO}}_{\text{4}}^{\text{2 - }}{\text{(aq) + 2H}}_{\text{2}}\text{O}$

Step 11: Cancel out common terms.

${\text{Cl}}_{\text{2}}{\text{(g) + SO}}_{\text{3}}^{\text{2 - }}{\text{(aq) + 2OH}}^{\text{ - }}\to {\text{2Cl}}^{\text{ - }}{\text{(aq) + SO}}_{\text{4}}^{\text{2 - }}{\text{(aq) + H}}_{\text{2}}\text{O}$

As a result, we receive the following balanced response:

${\text{SO}}_{\text{3}}^{\text{2 - }}{\text{(aq) + Cl}}_{\text{2}}{\text{(g) + 2OH}}^{\text{ - }}\to {\text{SO}}_{\text{4}}^{\text{2 - }}{\text{(aq) + 2Cl}}^{\text{ - }}{\text{(aq) + H}}_{\text{2}}\text{O}$

Therefore, the oxidizing and reducing agents

Oxidizing agent: ${\text{Cl}}_{\text{2}}\left(\text{g}\right)$

Reducing agent: ${\text{SO}}_{\text{3}}^{\text{2 - }}\text{(aq)}$

Find the oxidizing and reducing agents

$$\begin{gathered} \left(i\right)7\times Fe{{\left(CN\right)}^{3-}}_6\left(aq\right)+7\times e^{-}\to 7\times Fe{{\left(CN\right)}^{4-}}_6\left(aq\right) \\ \left(ii\right)Re\left(s\right)+4H_{2}O\to Re{O^{-}}_4\left(aq\right)+8H^{+}+7e^{-} \end{gathered}$$(b) ${\text{Fe(CN)}}_{\text{6}}^{\text{3 - }}\text{(aq) + Re(s)}\to {\text{Fe(CN)}}_{\text{6}}^{\text{4 - }}{\text{(aq) + ReO}}_{\text{4}}^{\text{ - }}\text{(aq)[basic]}$

We'll take the following steps to balance the equation:

Step 1: Separate the half-reactions in the first step.

$$\begin{gathered} {\text{(i)Fe(CN)}}_{\text{6}}^{\text{3 - }}\text{(aq)}\to {\text{Fe(CN)}}_{\text{6}}^{\text{4 - }}\text{(aq)} \\ \text{(ii)Re(s)}\to {\text{ReO}}_{\text{4}}^{\text{ - }}\text{(aq)} \end{gathered}$$

Step 2: Ensure that all elements other than oxygen and hydrogen are in balance.

$$\begin{gathered} {\text{(i)Fe(CN)}}_{\text{6}}^{\text{3 - }}\text{(aq)}\to {\text{Fe(CN)}}_{\text{6}}^{\text{4 - }}\text{64(aq}) \\ \text{(ii)Re(s)}\to {\text{ReO}}_{\text{4}}^{\text{ - }}\text{4(aq)} \end{gathered}$$

 Step 3: To balance the oxygen atoms, add $H_{2}O$ molecules.

$$\begin{gathered} {\text{(i)Fe(CN)}}_{\text{6}}^{\text{3 - }}\text{(aq)}\to {\text{Fe(CN)}}_{\text{6}}^{\text{4 - }}\text{(aq)} \\ {\text{(ii)Re(s) + 4H}}_{\text{2}}\text{O}\to {\text{ReO}}_{\text{4}}^{\text{ - }}\text{(aq)} \end{gathered}$$

Step 4: To balance hydrogen atoms we add protons, $H^{+}$.

$$\begin{gathered} {\text{(i)Fe(CN)}}_{\text{6}}^{\text{3 - }}\text{(aq)}\to {\text{Fe(CN)}}_{\text{6}}^{\text{4 - }}\text{(aq)} \\ {\text{(ii)Re(s) + 4H}}_{\text{2}}\text{O}\to {\text{ReO}}_{\text{4}}^{\text{ - }}{\text{(aq) + 8H}}^{\text{ + }} \end{gathered}$$ 

Step 5: Now balance the charge with electrons, $e^{-}$.

 $$\begin{gathered} {\text{(i)Fe(CN)}}_{\text{6}}^{\text{3 - }}{\text{(aq) + e}}^{\text{ - }}\to {\text{Fe(CN)}}_{\text{6}}^{\text{4 - }}\text{(aq)} \\ {\text{(ii)Re(s) + 4H}}_{\text{2}}\text{O}\to {\text{ReO}}_{\text{4}}^{\text{ - }}{\text{(aq) + 8H}}^{\text{ + }}{\text{ + 7e}}^{\text{ - }} \end{gathered}$$

Step 6: Equalize the electron counts by scaling the processes.

$$\begin{gathered} \left(i\right)7\times Fe{{\left(CN\right)}_6}^{3-}\left(aq\right)+7\times e^{-}\to 7\times Fe{{\left(CN\right)}_6}^{4-}\left(aq\right) \\ \left(ii\right)Re\left(s\right)+4H_{2}O\to Re{O_4}^{-}\left(aq\right)+8H^{+}+7e^{-} \end{gathered}$$

Step 7: Add the reactions.

${\text{7Fe(CN)}}_{\text{6}}^{\text{3 - }}{\text{(aq) + 7e}}^{\text{ - }}{\text{ + Re(s) + 4H}}_{\text{2}}\text{O}\to {\text{7Fe(CN)}}_{\text{6}}^{\text{4 - }}{\text{(aq) + ReO}}_{\text{4}}^{\text{ - }}{\text{(aq) + 8H}}^{\text{ + }}{\text{ + 7e}}^{\text{ - }}$

Step 8: Cancel out common terms.

${\text{7Fe(CN)}}_{\text{6}}^{\text{3 - }}{\text{(aq) + Re(s) + 4H}}_{\text{2}}\text{O}\to {\text{7Fe(CN)}}_{\text{6}}^{\text{4 - }}{\text{64(aq) + ReO}}_{\text{4}}^{\text{ - }}{\text{4(aq) + 8H}}^{\text{ + }}$

Step 9: Because the reaction is taking place in a basic media, we must inject $O{H}^{-}$to balance the $H^{+}$ ions.

${\text{7Fe(CN)}}_{\text{6}}^{\text{3 - }}{\text{(aq) + Re(s) + 4H}}_{\text{2}}{\text{O + 8OH}}^{\text{ - }}\to {\text{7Fe(CN)}}_{\text{6}}^{\text{4 - }}{\text{(aq) + ReO}}_{\text{4}}^{\text{ - }}{\text{(aq) + 8H}}^{\text{ + }}{\text{ + 8OH}}^{\text{ - }}$

Step 10: Combine $O{H}^{-}$ ions and$H^{+}$ ions present on the same side to form water molecule.

${\text{7Fe(CN)}}_{\text{6}}^{\text{3 - }}{\text{(aq) + Re(s) + 4H}}_{\text{2}}{\text{O + 8OH}}^{\text{ - }}\to {\text{7Fe(CN)}}_{\text{6}}^{\text{4 - }}{\text{(aq) + ReO}}_{\text{4}}^{\text{ - }}{\text{(aq) + 8H}}_{\text{2}}\text{O}$

Step 11: Cancel out common terms.

${\text{7Fe(CN)}}_{\text{6}}^{\text{3 - }}{\text{(aq) + Re(s) + 8OH}}^{\text{ - }}\to {\text{7Fe(CN)}}_{\text{6}}^{\text{4 - }}{\text{(aq) + ReO}}_{\text{4}}^{\text{ - }}{\text{(aq) + 4H}}_{\text{2}}\text{O}$

Hence, we get the following balanced reaction:

${\text{7Fe(CN)}}_{\text{6}}^{\text{3 - }}{\text{(aq) + Re(s) + 8OH}}^{\text{ - }}\to {\text{7Fe(CN)}}_{\text{6}}^{\text{4 - }}{\text{(aq) + ReO}}_{\text{4}}^{\text{ - }}{\text{(aq) + 4H}}_{\text{2}}\text{O}$

Therefore, the oxidizing and reducing agents

Oxidizing agent: ${\text{Fe(CN)}}_{\text{6}}^{\text{3 - }}$

Reducing agent: $\text{Re}\left(\text{s}\right)$

Find the oxidizing and reducing agents

${\text{(c)MnO}}_{\text{4}}^{\text{ - }}\text{(aq) + HCOOH(aq)}\to {\text{Mn}}^{\text{2 + }}{\text{(aq) + CO}}_{\text{2}}\text{(g)[acidic]}$

We'll take the following steps to balance the equation:

Step 1: Separate the half-reactions.

$$\begin{gathered} {\text{(i)MnO}}_{\text{4}}^{\text{ - }}\text{4(aq)}\to {\text{Mn}}^{\text{2 + }}\text{(aq)} \\ \text{(ii)HCOOH(aq)}\to {\text{CO}}_{\text{2}}\text{(g)} \end{gathered}$$

Step 2: Ensure that all elements other than oxygen and hydrogen are in balance.

$$\begin{gathered} {\text{(i)MnO}}_{\text{4}}^{\text{ - }}\text{4(aq)}\to {\text{Mn}}^{\text{2 + }}\text{ + (aq)} \\ \text{(ii)HCOOH(aq)}\to {\text{CO}}_{\text{2}}\text{(g)} \end{gathered}$$

Step 3: To balance the oxygen atoms, add $H_{2}O$ molecules.

$$\begin{gathered} {\text{(i)MnO}}_{\text{4}}^{\text{ - }}\text{(aq)}\to {\text{Mn}}^{\text{2 + }}{\text{(aq) + 4H}}_{\text{2}}\text{O} \\ \text{(ii)HCOOH(aq)}\to {\text{CO}}_{\text{2}}\text{(g)} \end{gathered}$$

Step 4: We add protons, $H^{+}$, to balance hydrogen atoms.

$$\begin{gathered} {\text{(i)MnO}}_{\text{4}}^{\text{ - }}{\text{(aq) + 8H}}^{\text{ + }}\to {\text{Mn}}^{\text{2 + }}{\text{(aq) + 4H}}_{\text{2}}\text{O} \\ \text{(ii)HCOOH(aq)}\to {\text{CO}}_{\text{2}}{\text{(g) + 2H}}^{\text{ + }} \end{gathered}$$

Step 5: Now, using electrons to balance the charge, $e^{-}$.

$$\begin{gathered} {\text{(i)MnO}}_{\text{4}}^{\text{ - }}{\text{(aq) + 8H}}^{\text{ + }}{\text{ + 5e}}^{\text{ - }}\to {\text{Mn}}^{\text{2 + }}{\text{2 + (aq) + 4H}}_{\text{2}}\text{2O} \\ \text{(ii)HCOOH(aq)}\to {\text{CO}}_{\text{2}}{\text{(g) + 2H}}^{\text{ + }}{\text{ + 2e}}^{\text{ - }} \end{gathered}$$

Step 6: Equalize the electron counts by scaling the processes.

$$\begin{gathered} \left(i\right)2\times Mn{O_4}^{-}\left(aq\right)+2\times 5e^{-}\to 2\times M{n}^{2+}+\left(aq\right)+2\times H_4{H}_{2}O \\ \left(ii\right)5\times HCOOH\left(aq\right)\to 5\times C{O}_2\left(g\right)+5\times 2{H_2}^{+}+5\times 2e- \end{gathered}$$

Step 7: Add the reactions.

${\text{2MnO}}_{\text{4}}^{\text{ - }}{\text{(aq) + 16H}}^{\text{ + }}{\text{ + 10e}}^{\text{ - }}\text{ + 5HCOOH(aq)}\to {\text{2Mn}}^{\text{2 + }}{\text{(aq) + 8H}}_{\text{2}}{\text{O + 5CO}}_{\text{2}}{\text{(g) + 10H}}^{\text{ + }}{\text{ + 10e}}^{\text{ - }}$

Step 8: Cancel out common terms.

${\text{2MnO}}_{\text{4}}^{\text{ - }}{\text{(aq) + 6H}}^{\text{ + }}\text{ + 5HCOOH(aq)}\to {\text{2Mn}}^{\text{2 + }}{\text{ + (aq) + 8H}}_{\text{2}}{\text{O + 5CO}}_{\text{2}}\text{(g)}$

Hence, we get the following balanced reaction:

${\text{2MnO}}_{\text{4}}^{\text{ - }}{\text{(aq) + 5HCOOH(aq) + 6H}}^{\text{ + }}\to {\text{2Mn}}^{\text{2 + }}{\text{ + (aq) + 5CO}}_{\text{2}}{\text{(g) + 8H}}_{\text{2}}\text{O}$

Therefore, the oxidizing and reducing agents

Oxidizing agent: ${\text{MnO}}_{\text{4}}^{\text{ - }}$

Reducing agent: $\text{HCOOH}\left(\text{aq}\right)$