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MHT CET · Chemistry · Electrochemistry

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. A \(6.369 \times 10^5 \mathrm{C}\)
  2. B \(1.462 \times 10^5 \mathrm{C}\)
  3. C \(4.839 \times 10^5 \mathrm{C}\)
  4. D \(3.419 \times 10^5 \mathrm{C}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(6.369 \times 10^5 \mathrm{C}\)

Step-by-step Solution

Detailed explanation

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