Consider the reaction,
`Cr_(2)O_(7)^(2-)+ 14H^(+) + 6e^(-) to 2 Cr^(3+) + 7H_(2)O`
What is the quantity of electricity in coulombs needed to reduce 1 mole of `2 Cr_(2)O_(7)^(2-)` ?
(Given 1F= 96500C)

Options

A. `5.79 xx 10^(5)`
B. `5.69 xx 10^(5)`
C. `5.59 xx 10^(5)`
D. `5.49 xx 10^(5)`

Answer

A

Explanation

1 mole `Cr_(2)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
`6 xx 96500 C= 5.79 xx 10^(5)C`