'A' Level Chemistry Problem Analysis: Electrolysis of Metal Sulfate Solution |
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QN: When a dilute sulfate solution of a metal J is electrolysed, metal J and a diatomic gas K are produced at the cathode and the anode respectively in the molar ratio 2:1. In another experiment, the same quantity of electricity is used to electrolyse a saturated sodium chloride solution and a gas L is evolved at the anode. What is the molar ratio of J : K : L? Thought process: At the anode, H2O is oxidized to O2 gas, requiring 4 mol of electrons per mol of O2 generated. At the cathode, J^x+ is reduced to J metal, requiring x mol of electrons per mol of J generated. Given ratio of cathodic to anodic product is 2:1, therefore 2 mol of J is generated per mol of O2. Since electrons transferred are the same, ie. 4 mol of electrons reduce 2 mol of J, thus J is J^2+, or dipositive. At the other anode, Cl- is oxidized to Cl2 gas, requiring 2 mol of electrons to be transferred per mol of Cl2 gas generated. Since 4 mol of electrons were transferred in the previous setup (per mol of O2 generated), hence 2 mol of Cl2 gas is generated per 4 mol of electrons. Molar ratio of J : K : L is therefore 2 : 1 : 2 (shown) ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ The above content is contributed by Mr Heng, owner and 'A' Level Chemistry tutor at Bedok Funland JC. He also goes by the handle UltimaOnline on various online popular homework forums. |
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