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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` |
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Answer» Correct Answer - C `E_(cell)^(o)=E_("cathode")^(o)-E_("anode"), " "E_(cell)^(o)=0.34-(-2.37)` `E_(cell)^(o)=2.71V` |
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