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Consider the reaction: Cr_(2)O_(2)^(2-)+14H^(+)+6e^(-) to 2Cr^(3+)+7H_(2)O What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr_(2)O_(7)^(2-) ? |
Answer» Solution :* Reaction: * Therefore, to reduce 1 mole of `Cr_(2)O_(7)^(2-),` 6 mole electrons will be USED. So, the REQUIRED quantity of electricity will be: So, 6 mole electron=6F electricity is used. 6 F electricity=6`xx96500` COULOMB =579000 Coulomb |
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