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Question
For the given cell, \[\ce{Mg | Mg^{2+} || Cu^{2+} | Cu}\]
(i) \[\ce{Mg}\] is cathode
(ii) \[\ce{Cu}\] is cathode
(iii) The cell reaction is \[\ce{Mg^+ Cu^{2+} -> Mg^{2+} + Cu}\]
(iv) \[\ce{Cu}\] is the oxidising agent
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Solution
(ii) \[\ce{Cu}\] is cathode
(iii) The cell reaction is \[\ce{Mg^+ Cu^{2+} -> Mg^{2+} + Cu}\]
Explanation:
(i) Left side of cell reaction represents oxidation half-cell i.e., oxidation of \[\ce{Mg}\] and right side of cell represents reduction half-cell reactions i.e., reduction of copper.
(ii) \[\ce{Cu}\] is reduced and reduction occurs at cathode.
(iii) \[\ce{Mg}\] is oxidized and oxidation occurs at anode.
(iv) Whole cell reaction can be written as
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Oxidation-reduction reactions are commonly known as redox reactions. They involve transfer of electrons from one species to another. In a spontaneous reaction, energy is released which can be used to do useful work. The reaction is split into two half-reactions. Two different containers are used and a wire is used to drive the electrons from one side to the other and a Voltaic/Galvanic cell is created. It is an electrochemical cell that uses spontaneous redox reactions to generate electricity. A salt bridge also connects to the half-cells. The reading of the voltmeter gives the cell voltage or cell potential or electromotive force. If \[\ce{E^0_{cell}}\] is positive the reaction is spontaneous and if it is negative the reaction is non-spontaneous and is referred to as electrolytic cell. Electrolysis refers to the decomposition of a substance by an electric current. One mole of electric charge when passed through a cell will discharge half a mole of a divalent metal ion such as Cu2+. This was first formulated by Faraday in the form of laws of electrolysis.
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