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A solution of Ni(NO_(3))_(2) is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode ? |
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Answer» Solution :* Mass of electricity `Q=5Axx20xx60s` =6000C * By the following reaction NI is formed. Ionization : `Ni(NO_(3))_(2(aq)) to Ni_((aq))^(2+)+2NO_(3(aq))^(-)` REDUCTION on cathode: `Ni_((aq))^(2+) + underset("2 mol "e^(-))(2e^(-)) to underset("1 mol Ni")(Ni_((l)))` * According to this reaction, 2 mol `e^(-)`, 2F electricity 1 mol `Ni=58.7` g Ni So, `2xx96500` COULOMB electricity produce 58.79 gm Ni and so mass of Ni obtained by using 6000 coulomb electricity `=(6000xx58.79)/(2xx96500)` `=1.8277g` Ni |
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