Consider the cell: Zn|Zn2+(aq)(1.0M)||Cu2+(aq)(1.0M)||Cu The standard reduction potential are 0.350 V for Cu2+(aq)+2e−→Cu and -0.763 V for Zn2+(aq)+2e−→Zn.
The EMF (in V) of the cell is ( write your answer as nearest integer) :
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Solution
EMF of the cell = E⊝cathode−E⊝anode =E⊝Cu2+|Cu−E⊝Zn2+|Zn =0.350−(−0.763)=1.113V