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प्रश्न
In the electrochemical cell: Zn|ZnSO4 (0.01 M)||CuSO4 (1.0 M)|Cu, the emf of this Daniel cell is E1. When the concentration of ZnSO4 is changed to 1.0 M and that CuSO4 changed to 0.01 M, the emf changes to E2. From the above, which one is the relationship between E1 and E2?
पर्याय
E1 < E2
E1 > E2
E2 ≥ E1
E1 = E2
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उत्तर
E1 > E2
Explanation:
`"E"_"cell" = "E"_"cell"^0 - 0.0591/2 log (["Zn"^(2+)])/(["Cu"^(2+)])`
E1 = `"E"_"cell"^0 - 0.0591/2 log 10^-2/1`
E1 = `"E"_"cell"^0 + 0.0591` .....................(1)
\[\ce{Zn_{(s)} -> Zn^{2+}_{( aq)} + 2e^-}\]
\[\ce{Cu^{2+}_{( aq)} + 2e^- -> Cu_{(s)}}\]
\[\ce{Zn_{(s)} + Cu^{2+}_{( aq)} -> Zn^{2+}_{( aq)} + Cu_{(s)}}\]
E2 = `"E"_"cell"^0 - 0.0591/2 log 1/10^-2`
E2 = `"E"_"cell"^0 - 0.0591` .....................(2)
∴ E1 > E2
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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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