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Clarify stoichiometry of reaction occurs on the electrodes of electrolytic cell and its relationship with quantity of electricity. |
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Answer» Solution :* The amount of electricity(or charge) required for oxidation or reduction depends on the stoichiometry of the electrode reaction. For example, in the reaction : `Ag_((aq))^(+)+e^(-) to Ag_((S))` * One mole of the electron is required for the reduction of one mole of silver ions. * We know that charge on one electron is equal to `1.6021xx10^(-19)C`. * So, the charge on one mole of electrons is equal to : `=N_(A)xx1.6021xx10^(-19)C` `=6.02xx10^(23)mol^(-1)xx1.6021xx10^(-19)C` `=96487C" "mol^(-1)` * This quantity of electricity is called Faraday and is represented by the symbol F. for approximate calculations, we use `1F=96500C" "mol^(-1)` * 1F=electric quantity of 1 mol electron is 96487 (0.96500) COULOMB is known as 1 Faraday. Generally, in electrolytic reactions current is measured in ampere and time in second. * So TOTAL electricity=quantity of electricity=Q=It * To calculate required quantity of electricity to produced product on electrode following STEPS should be followed: (i) WRITE oxidation/reduction reaction occurs on the electrode. (ii) Determine the required quantity of Faraday based on reaction occurs for 1 mol of product. (iii) Calculate current in ampere by using Faraday value. e.g., (i) reaction occurs for the reduction of Ag metals: Reaction : `Ag_((aq))^(+)+e^(-) to Ag_((S))` 1 mole electron=1F, 1 mol electricity will produced 1 mol of Ag. (ii) required electricity for reduction of 1 mol Mg: Reaction : `Mg_((aq))^(2+)+2e^(-) to Mg_((S))` 2 mole electron=2F , 1 mole Mg is produced by 2 F electricity. (iii) required electricity for reduction of 1 mol of Al: `Al^(3+)+3e^(-) to Al` 3 mol electron=3F, for 1 mole Al 3 F electricity is required. |
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