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Q. Consider the reaction,
$\text{Cr}_{2} \text{O}_{7}^{2 -} + 1 4 \text{H}^{+} + 6 \text{e}^{-} \rightarrow 2 \text{Cr}^{3 +} + 7 \text{H}_{2} \text{O}$
What is the quantity of electricity in coulombs needed to reduce 1 mole of $\text{Cr}_{2} \text{O}_{7}^{2 -}$ ?
(Given: 1F ≈ 96500 C)

NTA AbhyasNTA Abhyas 2022Electrochemistry

Solution:

1 mole $\text{Cr}_{2} \text{O}_{7}^{2 -}$ require 6 moles of electrons to completely reduce into $Cr^{3 +}$ . That means 6F electricity is required. (One Faraday = 96500C)
Hence, the charge in coulombs $=6x96500C=5.79x10^{5}C$