Calculate the amount of electricity required to convert 1.1 mol of $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$ to…

Calculate the amount of electricity required to convert 1.1 mol of $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$ to $\mathrm{Cr}^{+3}$ in acidic medium.
  1. $6.369 \times 10^5 \mathrm{C}$
  2. $1.462 \times 10^5 \mathrm{C}$
  3. $4.839 \times 10^5 \mathrm{C}$
  4. $3.419 \times 10^5 \mathrm{C}$

Solution

$14 \mathrm{H}^{+}+\mathrm{Cr}_2 \mathrm{O}_7^{2-}+6 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Cr}^{3-}+7 \mathrm{H}_2 \mathrm{O}$
1 mole of $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$ gets reduced by 6 moles of electrons to give 2 moles of $\mathrm{Cr}^{3+}$. $\begin{array}{ll}\therefore \quad & 1.1 \text { mole of } \mathrm{Cr}_2 \mathrm{O}_7{ }^{2-}=6 \times 1.1=6.6 \text { moles of } \mathrm{e}^{-} \\ & 1 \text { mole of electrons }=96500 \mathrm{C} \text { of electricity } \\ \therefore \quad & 6.6 \text { moles of electrons } \\ & =6.6 \times 96500 \mathrm{C}=6.369 \times 10^5 \mathrm{C}\end{array}$

Asked in: MHT CET 2024 (03 May Shift 1)

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