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Consider the reaction : Cr(2)O(7)^(2-)...

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

Text Solution

Verified by Experts

6 mol of electrons are required to reduce 1 mol of `Cr_(2) O_(7)^(2-)`
Now 1 mol of electrons = 96500 C
6 mol of electrons = `96500 xx 6 = 5.79 xx 10^(5) C `
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