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| 1. |
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 mol of Cr_(2)O_(7)^(2-) ? |
| Answer» Solution :One `Cr_(2)O_(7)^(2-)`ion requires 6 electrons for REDUCTION as shown by the above CHEMICAL equation. THUS, 1 mol of `Cr_(2)O_(7)^(2-)` ions require 6 F= 6 x 96500 C = 579000 C of electricity for reduction to `Cr^(3+)`. | |