Question

The cell reaction ${Cr}_2{O}_7^{2+}\left(aq\right)+14H^{+}\left(aq\right)+6{Fe}^{2+}\left(aq\right)\to 6{Fe}^{3+}\left(aq\right)+2{Cr}^{3+}\left(aq\right)+7H_{2}O\left(l\right)$, is best represented by:

• A. $Pt\left(s\right)|{Fe}^{2+}\left(aq\right),{Fe}^{3+}\left(aq\right)\parallel {Cr}_2{O}_7^{2+}\left(aq\right),{Cr}^{3+}\left(aq\right)|Pt\left(s\right)$
• B. $Pt\left(s\right)|{Cr}_2{O}_7^{2+}\left(aq\right),{Cr}^{3+}\left(aq\right)\parallel {Fe}^{3+}\left(aq\right),{Fe}^{2+}\left(aq\right)|Pt\left(s\right)$
• C. ${Fe}^{2+}\left(aq\right)|{Fe}^{3+}\left(aq\right)\parallel {Cr}_2{O}_7^{2+}\left(aq\right)|{Cr}^{3+}\left(aq\right)$
• D. ${Cr}_2{O}_7^{2+}\left(aq\right)|{Cr}^{3+}\left(aq\right)\parallel {Fe}^{3+}\left(aq\right)|{Fe}^{2+}\left(aq\right)$

Answer 1

The correct option is A $Pt\left(s\right)|{Fe}^{2+}\left(aq\right),{Fe}^{3+}\left(aq\right)\parallel {Cr}_2{O}_7^{2+}\left(aq\right),{Cr}^{3+}\left(aq\right)|Pt\left(s\right)$

${Cr}_2{O}_7^{-2}\left(aq\right)+14H^{+}\left(aq\right)+6{Fe}^{2+}\left(aq\right)⟶6{Fe}^{+3}\left(aq\right)+2{Cr}^{+3}\left(aq\right)+7H_{2}O\left(l\right)$

Here, ${\stackrel{+6}{Cr}}_2{O}_7^{2-}\left(aq\right)$ is reduced to ${Cr}^{+3}\left(aq\right)$

& ${Fe}^{+2}\left(aq\right)$ is oxidized to ${Fe}^{+3}\left(aq\right)$

Reducing takes place at cathode & oxidation takes place at anode.

The cell is represented by anode on the left hand side of electrolyte bridge & cathode on the right hand side of electrolyte bridge.

$\therefore$ It can be represented as :

$Pt\left(s\right)\left|{Fe}^{3+}\left(aq\right){Fe}^{2+}\left(aq\right)\right||{Cr}_2{O}_7^{2-}\left(aq\right),{Cr}^{+3}\left(aq\right)|Pt\left(s\right)$