EMF of a cell whose half cells are given below is Mg^(2+)+2e^(-)toMg( – InterviewSolution

1.	EMF of a cell whose half cells are given below is Mg^(2+)+2e^(-)toMg(s),E=-2.37V Cu^(2+)+2e^(-)toCu(s),E=+0.33V
Answer» SOLUTION :`E_(CELL)^(o)=E_("CATHODE")^(o)-E_("ANODE"),""E_(cell)^(o)=0.34-(-2.37)` `E_(cell)^(o)=2.71V`

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