Mg2+ displaces hydrogen from acids but copper does not. A galvanic cell prepared by combining CU/Cu2+ and Mg/Mg2+ has and EMF of 2.71 V at 298 K. If the potential of copper electrode is 0.34V, what is the reduction potential of Mg electrode?
The galvanic cell can be represented as: E‌cell ‌∘=E‌cathode ‌∘−E‌anode ‌∘=ECu2+∘−EMg2+∕Mg∘ ⇒EMg2+∘∕Mg=ECu2+∘∕Cu∘−ECell∘(Reduction potential) =0.34−2.71V=−2.37V =(0.34−2.71)V=−2.37V