Question

How many moles of ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ is required to oxidise one mole of$\mathrm{Fe}\left({\mathrm{C}}_2{\mathrm{O}}_4\right)$ in an acidic medium.

• A. $2$
• B. $3$
• C. $1/2$
• D. $1/3$

Answer 1

The correct option is C

$1/2$

The explanation for the correct option:

Option (C):

Step 1: Given data:

${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ is given to oxidise one mole of$\mathrm{Fe}\left({\mathrm{C}}_2{\mathrm{O}}_4\right)$ in an acidic medium, for which the reaction can be written as:

${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7\left(\mathrm{aq}\right)+2{\mathrm{FeC}}_2{\mathrm{O}}_4\left(\mathrm{aq}\right)+14{\mathrm{H}}^{+}\left(\mathrm{aq}\right)\to 2{\mathrm{K}}^{+}\left(\mathrm{aq}\right)+2\mathrm{Cr}³⁺\left(\mathrm{aq}\right)+2{\mathrm{Fe}}^{+3}\left(\mathrm{aq}\right)+4{\mathrm{CO}}_2\left(\mathrm{g}\right)+7{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

Here the given ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$(Potassium dichromate) is known as a strong oxidizing agent, in the acidic medium.

In the acidic medium, Potassium dichromate helps in oxidizing the other substance while itself gets reduced, and the oxidation state of Chromium changes by $+3.$

Step 2: Oxidation and reduction taking place in the given reaction:

In the given reaction, ${\mathrm{Fe}}^{2+}$ in Ferrous oxalate$\mathrm{Fe}\left({\mathrm{C}}_2{\mathrm{O}}_4\right)$ gets oxidized by acidified${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ to form ${\mathrm{Fe}}^{3+}$.

While ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$(Potassium dichromate), acting as an oxidizing agent a gets reduced from ${\mathrm{Cr}}^{+6}\mathrm{to}{\mathrm{Cr}}^{+3}$, which can be shown as:

$\underset{\left(\mathrm{Potassium}\mathrm{dichromate}\right)}{{\mathrm{K}}_2\stackrel{\left(+6\right)}{{\mathrm{Cr}}_2}{\mathrm{O}}_7\left(\mathrm{aq}\right)}+\underset{\left(\mathrm{Ferrous}\mathrm{oxalate}\right)}{2\stackrel{(+2)}{\mathrm{Fe}}{\mathrm{C}}_2{\mathrm{O}}_4\left(\mathrm{aq}\right)}+14{\mathrm{H}}^{+}\left(\mathrm{aq}\right)\to 2{\mathrm{K}}^{+}\left(\mathrm{aq}\right)+2\stackrel{\left(+3\right)}{\mathrm{Cr}}³⁺\left(\mathrm{aq}\right)+2{\mathrm{Fe}}^{+3}\left(\mathrm{aq}\right)+4{\mathrm{CO}}_2\left(\mathrm{g}\right)+7{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

Step 3: Calculation of required moles of Potassium dichromate:

In the above reaction,$1$ mole ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ is oxidizing $2$ moles of $\mathrm{Fe}\left({\mathrm{C}}_2{\mathrm{O}}_4\right)$ .

Hence,

$2$ moles of $\mathrm{Fe}\left({\mathrm{C}}_2{\mathrm{O}}_4\right)$ = $1$ mole ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$oxidizes

To oxidize 1 mole of $\mathrm{Fe}\left({\mathrm{C}}_2{\mathrm{O}}_4\right)$ we require = $\frac{1}{2}$mole of${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$.

Thus, the number of moles of ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$required to oxidize one mole of$\mathrm{Fe}\left({\mathrm{C}}_2{\mathrm{O}}_4\right)$ in an acidic medium is calculated to be $\frac{1}{2}$mole.