Question

Question: Balance the following equations by the half –reaction method

Short answer

Answer

All the given equations are balanced using half-reaction method.

Step-by-step solution

Subpart a)

Split the whole reaction into half-reactions.

$$\begin{gathered} \text{Reduction} : H^{+}\to H_2 \\ \text{Oxidation} : C{l}^{-}+Fe\to HFeC{l}_4 \end{gathered}$$

In this equation disregarding the atoms

$$\begin{gathered} \text{Reduction} : H\to H_2 \\ \text{Oxidation} : 4C{l}^{-}+Fe\to HFeC{l}_4 \end{gathered}$$

Balancing ofH-atoms by adding the H-atom:

$$\begin{gathered} \text{Reduction} : 2e^{-}+H^{+}\to H_2 \\ \text{Oxidation} : H^{+}+4C{l}^{-}+Fe\to HFeC{l}_4+3e \end{gathered}$$

Subpart b)

$F{e}_{\left(s\right)}+HC{l}_{\left(aq\right)}\to HFeC{l}_{4\left(aq\right)}+H_{2\left(g\right)}$

Balancing the equation by adding :

$$\begin{gathered} \text{Reduction} : 2e^{-}+2H^{+}\to H_2 \\ \text{Oxidation} : H^{+}+4C{l}^{-}+Fe\to HFeC{l}_4+3e^{-} \end{gathered}$$

For removing the electrons, one should multiply the reduction reaction by 3 and the oxidation reaction by 2.

 Step 3: Subpart c)

$$\begin{gathered} \text{Oxidation} : 3I^{-}\to {I_3}^{-}+2e^{-} \\ \text{Reduction} : 3I{O_3}^{-}+16e^{-}+18H\to {I_3}^{-}+9H_{2}O \end{gathered}$$

The final equation will be

${O_3}^{-}+ 8I^{-}+6H\to 3{I_3}^{-}+3H_{2}O$

Subpart d)

$$\begin{gathered} \text{Oxidation}: Cr{(NCS{)}_6}^{4-}+54H_{2}O-97e^{-}\to C{r}^{3+}+6N{O}_3^{-}+6C{O}_2+6S{O}_4^{2-}+108H^{+}\backslash \mathrm{hfill} \\ \text{Rediction} : C{e}^{4+}+e^{-} \to C{e}^{3+} \end{gathered}$$

The final equation will be,

Subpart e)

$$\begin{gathered} \text{Oxidation} : Cr{I}_3+16H_{2}O+32O{H}^{-}\to Cr{O}_4^{2-}+27e^{-}+3I{O}_4^{-}+32H_{2}O \\ Reduction : C{l}_2+2e^{-}\to 2Cl \end{gathered}$$

The final equation will be,

$2Cr{I}_3+27C{l}_2+64O{H}^{-}\to 2Cr{O}_4^{2-}+54C{l}^{-}+6I{O}_4^{-}+32H_{2}O$

Subpart f)

$$\begin{gathered} \text{Oxidation} : Fe(CN{)}_6^{4-}+21H_{2}O+39O{H}^{-}\to Fe(OH{)}_3+ 6C{O}_3^{2-}+ 39H_{2}O+31e^{-} \\ \text{Reduction} : C{e}^{4+}+3H_{2}O+3O{H}^{-}+e^{-}\to Ce(OH{)}_3+3H_{2}O \end{gathered}$$

The final equation will be,

$Fe(CN{)}_6^{4-}+31C{e}^{4+}+32O{H}^{-}\to Fe(OH{)}_3+31Ce(OH{)}_3+6C{O}_3^{2-}+18H_{2}O$