Question

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

(a)$\mathit{A}{\mathit{s}}_{\mathbf{4}}{\mathit{O}}_{\mathbf{6}}\mathbf{(}\mathbf{}\mathit{s}\mathbf{)}\mathbf{+}\mathit{M}\mathit{n}{\mathit{O}}_{\mathbf{4}}^{\mathbf{-}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{\to }\mathit{A}\mathit{s}{\mathit{O}}_{\mathbf{4}}^{\mathbf{3}\mathbf{-}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}\mathit{M}{\mathit{n}}^{\mathbf{2}\mathbf{+}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{\left[}\mathit{a}\mathit{c}\mathit{i}\mathit{d}\mathit{i}\mathit{c}\mathbf{\right]}$

(b) ${\mathit{P}}_{\mathbf{4}}\mathbf{(}\mathbf{}\mathit{s}\mathbf{)}\mathbf{\to }\mathit{H}\mathit{P}{\mathit{O}}_{\mathbf{3}}^{\mathbf{2}\mathbf{-}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}\mathit{P}{\mathit{H}}_{\mathbf{3}}\mathbf{(}\mathbf{}\mathit{g}\mathbf{)}\mathbf{\left[}\mathit{a}\mathit{c}\mathit{i}\mathit{d}\mathit{i}\mathit{c}\mathbf{\right]}$

(c) $\mathit{M}\mathit{n}{\mathit{O}}_{\mathbf{4}}^{\mathbf{-}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{+}\mathit{C}{\mathit{N}}^{\mathbf{-}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{\to }\mathit{M}\mathit{n}{\mathit{O}}_{\mathbf{2}}\mathbf{(}\mathbf{}\mathit{s}\mathbf{)}\mathbf{+}\mathit{C}\mathit{N}{\mathit{O}}^{\mathbf{-}}\mathbf{\left(}\mathit{a}\mathit{q}\mathbf{\right)}\mathbf{\left[}\mathit{b}\mathit{a}\mathit{s}\mathit{i}\mathit{c}\mathbf{\right]}$

Short answer

(a)$A{s}_4{O}_6(s)+Mn{O}_4^{-}\left(aq\right)\to As{O}_4^{3-}\left(aq\right)+M{n}^{2+}\left(aq\right)\left[acidic\right]$

(b)$P_4(s)\to HP{O}_3^{2-}\left(aq\right)+P{H}_3(g)\left[acidic\right]$

(c)$Mn{O}_4^{-}\left(aq\right)+C{N}^{-}\left(aq\right)\to Mn{O}_2(s)+CN{O}^{-}\left(aq\right)\left[basic\right]$

Step-by-step solution

Definition of Oxidising Agent

A substance that obtains or "accepts"/"receives" an electron from a reducing agent is known as an oxidizing agent in a redox chemical process. To put it another way, an oxidizer is anything that oxidizes anything else.

Balancing of the given reaction in part a.

(a) ${\text{As}}_4{\text{O}}_6(\text{s})+{\text{MnO}}_4^{-}\left(\text{aq}\right)\to {\text{AsO}}_4^{3-}\left(\text{aq}\right)+{\text{Mn}}^{2+}\left(\text{aq}\right)\left[acidic\right]$

To balance the equation, we will follow the following steps:

1: Separate the half-reactions.

$$\begin{gathered} {\text{MnO}}_4^{-}\left(aq\right)\to {\text{Mn}}^{2+}\left(aq\right) \\ {\text{As}}_4{\text{O}}_6\left(s\right)\to {\text{AsO}}_4^{3-}\left(aq\right) \end{gathered}$$

2: Balance all the elements other than Oxygen and Hydrogen.

$$\begin{gathered} {\text{MnO}}_4^{-}\left(\text{aq}\right)\to {\text{Mn}}^{2+}\left(\text{aq}\right) \\ {\text{As}}_4{\text{O}}_6\left(s\right)\to 4{\text{AsO}}_4^{3-}\left(aq\right) \end{gathered}$$

3: Now add $H_{2}O$molecules to balance oxygen atoms.

$$\begin{gathered} {\text{MnO}}_4^{-}\left(aq\right)\to {\text{Mn}}^{2+}\left(\text{aq}\right)+4{\text{H}}_2\text{O} \\ {\text{As}}_4{\text{O}}_6(\text{s})+10{\text{H}}_2\text{O}\to 4{\text{AsO}}_4^{3-}\left(aq\right) \end{gathered}$$

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

$$\begin{gathered} {\text{MnO}}_4^{-}\left(\text{aq}\right)+8{\text{H}}^{+}\to {\text{Mn}}^{2+}\left(\text{aq}\right)+4{\text{H}}_2\text{O} \\ {\text{As}}_4{\text{O}}_6(\text{s})+10{\text{H}}_2\text{O}\to 4{\text{AsO}}_4^{3-}\left(aq\right)+2{\text{OH}}^{+} \end{gathered}$$

Balancing of the given reaction in part a.

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

$$\begin{gathered} {\text{MnO}}_4^{-}\left(aq\right)+8{\text{H}}^{+}+5{\text{e}}^{-}\to {\text{Mn}}^{2+}\left(\text{aq}\right)+4{\text{H}}_2\text{O} \\ {\text{As}}_4{\text{O}}_6(\text{s})+10{\text{H}}_2\text{O}\to 4{\text{AsO}}_4^{3-}\left(\text{aq}\right)+20{\text{H}}^{+}+8{\text{e}}^{-} \end{gathered}$$

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

$$\begin{gathered} 8{\text{MnO}}_4^{-}\left(aq\right)+64H^{+}+40e^{-}\to 8\times {\text{Mn}}^{2+}\left(aq\right)+32H_{2}O \\ 5A{s}_4{O}_6\left(s\right)+50H_{2}O\to 20As{O}_4^{3-}\left(aq\right)+100H^{+}+40e^{-} \end{gathered}$$

7: Add the reactions.

$8{\text{MnO}}_4^{-}\left(\text{aq}\right)+64{\text{H}}^{+}+40{\text{e}}^{-}+5{\text{As}}_4{\text{O}}_6(\text{s})+50{\text{H}}_2\text{O}\to 8{\text{Mn}}^{2+}\left(\text{aq}\right)+32{\text{H}}_2\text{O}+20{\text{AsO}}_4^{3-}\left(\text{aq}\right)+100{\text{H}}^{+}+40{\text{e}}^{-}$

Balancing of the given reaction in part a.

8: Cancel out common terms.

$8{\text{MnO}}_4^{-}\left(\text{aq}\right)+5{\text{As}}_4{\text{O}}_6(\text{s})+18{\text{H}}_2\text{O}\to 8{\text{Mn}}^{2+}\left(\text{aq}\right)+20{\text{AsO}}_4^{3-}\left(\text{aq}\right)+36{\text{H}}^{+}$

Hence, we get the following balanced reaction:

Oxidizing agent: ${\text{MnO}}_4^{-}\left(\text{aq}\right)$

Reducing agent:${\text{As}}_4{\text{O}}_6(\text{s})$

Balancing of the given reaction in part b.

(b) $P_4(\text{s})\to {\text{HPO}}_3^{2-}\left(\text{aq}\right)+P{H}_3(\text{g})\left[acidic\right]$

To balance the equation, we will follow the following steps:

1: Separate the half-reactions.

$$\begin{gathered} {\text{P}}_4(\text{s})\to {\text{PH}}_3\left(g\right) \\ {\text{P}}_4(\text{s})\to {\text{HPO}}_3^{2-}\left(aq\right) \end{gathered}$$

2: Balance all the elements other than Oxygen and Hydrogen.

$$\begin{gathered} {\text{P}}_4(\text{s})\to 4{\text{PH}}_3\left(g\right) \\ {\text{P}}_4(\text{s})\to 4{\text{HPO}}_3^{2-}\left(aq\right) \end{gathered}$$

3: Now add molecules to balance oxygen atoms.

$$\begin{gathered} {\text{P}}_4(\text{s})\to 4{\text{PH}}_3(\text{g}) \\ {\text{P}}_4(\text{s})+12{\text{H}}_2\text{O}\to 4{\text{HPO}}_3^{2-}\left(aq\right) \end{gathered}$$

Balancing of the given reaction in part b.

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

$$\begin{gathered} {\text{P}}_4(\text{s})+12{\text{H}}^{+}\to 4{\text{PH}}_3(\text{g}) \\ {\text{P}}_4(\text{s})+12{\text{H}}_2\text{O}\to 4{\text{HPO}}_3^{2-}\left(aq\right)+20{\text{OH}}^{+} \end{gathered}$$

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

$$\begin{gathered} {\text{P}}_4(\text{s})+12{\text{H}}^{+}+12{\text{e}}^{-}\to 4{\text{PH}}_3(\text{g}) \\ {\text{P}}_4(\text{s})+12{\text{H}}_2\text{O}\to 4{\text{HPO}}_3^{2-}\left(aq\right)+2{\text{OH}}^{-}+12{\text{e}}^{-} \end{gathered}$$

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

$$\begin{gathered} {\text{P}}_4(\text{s})+12{\text{H}}^{+}+12{\text{e}}^{-}\to 4{\text{PH}}_3(\text{g}) \\ {\text{P}}_4(\text{s})+12{\text{H}}_2\text{O}\to 4{\text{HPO}}_3^{2-}\left(aq\right)+2{\text{OH}}^{+}+12{\text{e}}^{-} \end{gathered}$$

7: Add the reactions.

${\text{P}}_4(\text{s})+12{\text{H}}^{+}+12{\text{e}}^{-}+{\text{P}}_4(\text{s})+12{\text{H}}_2\text{O}\to 4{\text{PH}}_3(\text{g})+4{\text{HPO}}_3^{2-}\left(\text{aq}\right)+20{\text{H}}^{+}+12{\text{e}}^{-}$

Balancing of the given reaction in part b.

8: Cancel out common terms.

$2{\text{P}}_4(\text{s})+12{\text{H}}_2\text{O}\to 4{\text{PH}}_3(\text{g})+4{\text{HPO}}_3^{2-}\left(\text{aq}\right)+8{\text{H}}^{+}$

Hence, we get the following balanced reaction:

$2{\text{P}}_4(\text{s})+12{\text{H}}_2\text{O}\to 4{\text{HPO}}_3^{2-}\left(\text{aq}\right)+4{\text{PH}}_3(\text{g})+8{\text{H}}^{+}$

Oxidizing agent: ${\text{P}}_4(\text{s})$

Reducing agent:${\text{P}}_4(\text{s})$

Balancing of the given reaction in part c.

$\left(\text{c}\right){\text{MnO}}_4^{-}\left(\text{aq}\right)+{\text{CN}}^{-}\left(\text{aq}\right)\to {\text{MnO}}_2(\text{s})+{\text{CNO}}^{-}\left(\text{aq}\right)\left[\text{basic}\right]$

To balance the equation, we will follow the following steps:

1: Separate the half-reactions.

$$\begin{gathered} {\text{MnO}}_4^{-}\left(aq\right)\to {\text{MnO}}_2\left(s\right) \\ {\text{CN}}^{-}\left(aq\right)\to {\text{CNO}}^{-}\left(aq\right) \end{gathered}$$

2: Balance all the elements other than Oxygen and Hydrogen.

$$\begin{gathered} {\text{MnO}}_4^{-}\left(aq\right)\to {\text{MnO}}_2\left(s\right) \\ {\text{CN}}^{-}\left(aq\right)\to {\text{CNO}}^{-}\left(aq\right) \end{gathered}$$

3: Now add$H_{2}O$ molecules to balance oxygen atoms.

$$\begin{gathered} {\text{MnO}}_4^{-}\left(aq\right)\to {\text{MnO}}_2\left(s\right)+2{\text{H}}_2\text{O} \\ {\text{CN}}^{-}\left(aq\right)+{\text{H}}_2\text{O}\to {\text{CNO}}^{-}\left(aq\right) \end{gathered}$$

 Balancing of the given reaction in part c.

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

$$\begin{gathered} {\text{MnO}}_4^{-}\left(aq\right)+4{\text{H}}^{+}\to {\text{MnO}}_2(\text{s})+2{\text{H}}_2\text{O} \\ {\text{CN}}^{-}\left(aq\right)+{\text{H}}_2\text{O}\to {\text{CNO}}^{-}\left(aq\right)+2{\text{H}}^{+} \end{gathered}$$

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

$$\begin{gathered} {\text{MnO}}_4^{-}\left(\text{aq}\right)+4{\text{H}}^{+}+3{\text{e}}^{-}\to {\text{MnO}}_2(\text{s})+2{\text{H}}_2\text{O} \\ {\text{CN}}^{-}\left(\text{aq}\right)+{\text{H}}_2\text{O}\to {\text{CNO}}^{-}\left(\text{aq}\right)+2{\text{H}}^{+}+2{\text{e}}^{-} \end{gathered}$$

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

$$\begin{gathered} 2{\text{MnO}}_4^{-}\left(aq\right)+8H^{+}+6e^{-}\to 2{\text{MnO}}_2(\text{s})+4H_{2}O \\ 3C{N}^{-}\left(aq\right)+3H_2\text{O}\to 3{\text{CNO}}^{-}\left(aq\right)+6H^{+}+6e^{-} \end{gathered}$$

7: Add the reactions.

$2{\text{MnO}}_4^{-}\left(\text{aq}\right)+8{\text{H}}^{+}+6{\text{e}}^{-}+3{\text{CN}}^{-}\left(\text{aq}\right)+3{\text{H}}_2\text{O}\to 2{\text{MnO}}_2(\text{s})+4{\text{H}}_2\text{O}+3{\text{CNO}}^{-}\left(\text{aq}\right)+6{\text{H}}^{+}+6{\text{e}}^{-}$

Balancing of the given reaction in part c.

8: Cancel out common terms.

$2{\text{MnO}}_4^{-}\left(\text{aq}\right)+2{\text{H}}^{+}+3{\text{CN}}^{-}\left(\text{aq}\right)\to 2{\text{MnO}}_2(\text{s})+{\text{H}}_2\text{O}+3{\text{CNO}}^{-}\left(\text{aq}\right)$

9: Since, the reaction is in basic medium we add $O{H}^{-}$to balance the $H^{+}$ions.

$2{\text{MnO}}_4^{-}\left(\text{aq}\right)+2{\text{H}}^{+}+2{\text{OH}}^{-}+3{\text{CN}}^{-}\left(\text{aq}\right)\to 2{\text{MnO}}_2(\text{s})+{\text{H}}_2\text{O}+3{\text{CNO}}^{-}\left(\text{aq}\right)+2{\text{OH}}^{-}$

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

$2{\text{MnO}}_4^{-}\left(\text{aq}\right)+2{\text{H}}_2\text{O}+3{\text{CN}}^{-}\left(\text{aq}\right)\to 2{\text{MnO}}_2(\text{s})+{\text{H}}_2\text{O}+3{\text{CNO}}^{-}\left(\text{aq}\right)+2{\text{OH}}^{-}$

11: Cancel out common terms.

$2{\text{MnO}}_4^{-}\left(\text{aq}\right)+{\text{H}}_2\text{O}+3{\text{CN}}^{-}\left(\text{aq}\right)\to 2{\text{MnO}}_2(\text{s})+3{\text{CNO}}^{-}\left(\text{aq}\right)+2{\text{OH}}^{-}$

Hence, we get the following balanced reaction:

$2{\text{MnO}}_4^{-}\left(\text{aq}\right)+3{\text{CN}}^{-}\left(\text{aq}\right)+{\text{H}}_2\text{O}\to 2{\text{MnO}}_2(\text{s})+3{\text{CNO}}^{-}\left(\text{aq}\right)+2{\text{OH}}^{-}$

Oxidizing agent: ${\text{MnO}}_4^{-}\left(aq\right)$

Reducing agent: ${\text{CN}}^{-}\left(aq\right)$.