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Can anybody plz explain to me the difference between the anode and cathode in electrolysis, voltaic cell amd diode ??

2007-07-25 22:42:06 · 2 answers · asked by Sherlynn W 1 in Science & Mathematics Physics

2 answers

Yes. the ANODE is the POSITIVE electrode while the CATHODE is the NEGATIVE electrode.

As a result if you put a salt such as Na Cl in water and you use a constant current , the ion Cl- which is negative goes to the anode Whereas the ion Na+ which is positive goes to the cathode

Cl- is called anion, Na+ is called cation

2007-07-25 22:49:35 · answer #1 · answered by maussy 7 · 0 0

electrolyis is forcing a non-spontaneous (negative E-cell) reaction to occur by applying electrical energy: 2 H2O(l) + 2e– → H2(g) + 2 OH–(aq), E0 =−0.83V Cl2(g) + 2e− → 2Cl−(aq), E0 = +1.36V since the reduction potential for Cl2 is higher, if the components were all mixed together it would be reduced - but you want to produce Cl2. the overall reaction for this is 2 H2O(l) + 2Cl−(aq) → Cl2(g) + H2(g) + 2 OH–(aq), E0 = -2.19V here water is being reduced (so it's the oxidising agent), and chloride ion (NOT chlorine!) is being oxidised (so it's the reducing agent). For the second part you need to find the number of moles of chlorine produced, and thus the number of moles of electrons that must be passed through the solution to produce that amount of chlorine. the current is in amps (coulombs / second), and you can find out Faraday's constant, the charge of a mole of electrons (coulombs / mole). From that and the number of moles of electrons calculated earlier you can find how long you need to run the current for. then you can use E = V*I*t to find the amount of electrical energy it corresponds to. edit: steve_geo1 is correct i think, the terms 'oxidisng agent' and 'reducing agent' really only make proper sense when the reaction is spontaneous. when electrical energy is being provided, as here, that's the proper cause of the reaction, not the properties of the chemicals.

2016-05-18 23:20:11 · answer #2 · answered by ? 3 · 0 0

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