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  4. The e.m.f. of a cell, whose half-cells are given below, is: Mg2++2e- xrightarrowMg(s); E° =- 2.37 V Cu2++2e- xrightarrowCu(s); E° =+ 0.34 V

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Q. The e.m.f. of a cell, whose half-cells are given below, is: $ M{{g}^{2+}}+2{{e}^{-}}\xrightarrow{{}}Mg(s); $ $ E{}^\circ =-\,2.37\,\,V $ $ C{{u}^{2+}}+2{{e}^{-}}\xrightarrow{{}}Cu(s); $ $ E{}^\circ =+\,0.34\,\,V $

JIPMERJIPMER 2001Electrochemistry

Solution:

In the cell Mg will act as anode while Cu will act as cathode as the reduction potential of Mg is less Hence, $ E_{cell}^{\text{o}}=E_{cathode}^{\text{o}}-E_{anode}^{\text{o}} $ $ =\left( 0.34 \right)-\left( -\,2.37 \right) $ $ =0.34+2.37=2.71\,\,V $