Question

1 mole of $C{{r}_{2}}{{O}_{7}}^{2-}$ oxidises _____ moles of \[Fe{{C}_{2}}{{O}_{4}}\] in acidic medium.

Answer 1

Hint: This is a redox reaction. Use the half equation method and write both the oxidation half and reduction half of the reaction. Balance them separately and then add them to get the balanced equation. Balancing the equation will give the number of moles of \[Fe{{C}_{2}}{{O}_{4}}\] oxidised.

Complete step by step answer:
- So, to find the number of moles of \[Fe{{C}_{2}}{{O}_{4}}\] oxidised, we first have to write the chemical reaction that occurs.
\[{{K}_{2}}C{{r}_{2}}{{O}_{7}}+Fe{{C}_{2}}{{O}_{4}}\overrightarrow{\text{ acidic medium}}\text{ }2C{{r}^{3+}}+F{{e}^{3+}}+C{{O}_{2}}\]

-The method used to balance the reaction is called the half equation method. In this method, we write the reduction reaction and oxidation reaction occurring separately and then balance them. The reduction half and oxidation half can be written as,
\[\begin{align}
 & \text{Oxidation half : }F{{e}^{2+}}+{{C}_{2}}{{O}_{4}}^{2-}\to F{{e}^{3+}}+C{{O}_{2}} \\
 & \operatorname{Re}\text{duction half : C}{{\text{r}}_{2}}{{O}_{7}}^{2-}\to C{{r}^{3+}} \\
\end{align}\]

- After writing the oxidation and reduction halves, we have to balance the number of the oxidised and reduced elements in the half equations. Here, we have to make the number of Cr atoms equal in the reduction half.
\[\begin{align}
 & \text{Oxidation half : }F{{e}^{2+}}+{{C}_{2}}{{O}_{4}}^{2-}\to F{{e}^{3+}}+C{{O}_{2}} \\
 & \operatorname{Re}\text{duction half : C}{{\text{r}}_{2}}{{O}_{7}}^{2-}\to 2C{{r}^{3+}} \\
\end{align}\]

- The next step is to balance the elements other than O and H. We have to balance C in the oxidation half in this case.
\[\begin{align}
 & \text{Oxidation half : }F{{e}^{2+}}+{{C}_{2}}{{O}_{4}}^{2-}\to F{{e}^{3+}}+2C{{O}_{2}} \\
 & \operatorname{Re}\text{duction half : C}{{\text{r}}_{2}}{{O}_{7}}^{2-}\to 2C{{r}^{3+}} \\
\end{align}\]

- Next, we have to balance the number of oxygen atoms. For doing this, we add ${{H}_{2}}O$ to the side lacking oxygen atoms.
\[\begin{align}
 & \text{Oxidation half : }F{{e}^{2+}}+{{C}_{2}}{{O}_{4}}^{2-}\to F{{e}^{3+}}+2C{{O}_{2}} \\
 & \operatorname{Re}\text{duction half : C}{{\text{r}}_{2}}{{O}_{7}}^{2-}\to 2C{{r}^{3+}}+7{{H}_{2}}O \\
\end{align}\]

- Next, we have to balance the number of H atoms in both half equations. As it is an acidic medium, we add ${{H}^{+}}$ ions to the side lacking H atoms.
\[\begin{align}
  & \text{Oxidation half : }F{{e}^{2+}}+{{C}_{2}}{{O}_{4}}^{2-}\to F{{e}^{3+}}+2C{{O}_{2}} \\
 & \operatorname{Re}\text{duction half : C}{{\text{r}}_{2}}{{O}_{7}}^{2-}+14{{H}^{+}}\to 2C{{r}^{3+}}+7{{H}_{2}}O \\
\end{align}\]


- Next, we have to balance the charge. This is done simply by adding electrons in required amount to the side having positive charge.
\[\begin{align}
  & \text{Oxidation half : }F{{e}^{2+}}+{{C}_{2}}{{O}_{4}}^{2-}\to F{{e}^{3+}}+2C{{O}_{2}}+3{{e}^{-}} \\
 & \operatorname{Re}\text{duction half : C}{{\text{r}}_{2}}{{O}_{7}}^{2-}+14{{H}^{+}}+6{{e}^{-}}\to 2C{{r}^{3+}}+7{{H}_{2}}O \\
\end{align}\]

- Now, we have to make the number of electrons equal in both half equations and then add them. In this case, we have to multiply the oxidation half by 2 and add to the reduction half.
\[\text{ C}{{\text{r}}_{2}}{{O}_{7}}^{2-}+14{{H}^{+}}+2F{{e}^{2+}}+2{{C}_{2}}{{O}_{4}}^{2-}+6{{e}^{-}}\to 2F{{e}^{3+}}+4C{{O}_{2}}+2C{{r}^{3+}}+7{{H}_{2}}O+6{{e}^{-}}\]

- Now, we can cancel out the common terms to get the final balanced equation.
\[\text{ C}{{\text{r}}_{2}}{{O}_{7}}^{2-}+14{{H}^{+}}+2Fe{{C}_{2}}{{O}_{4}}\to 2F{{e}^{3+}}+4C{{O}_{2}}+2C{{r}^{3+}}+7{{H}_{2}}O\]

- From the balanced equation, it is clear that 1 mole of $C{{r}_{2}}{{O}_{7}}^{2-}$ oxidises 2 moles of \[Fe{{C}_{2}}{{O}_{4}}\] in acidic medium.

Note: You can also use trial and error methods for balancing the equation. But you have to be very careful while doing so. In half equation method, make sure that you make the number of electrons in the reduction half and oxidation half equal before adding the half reactions.