Answer: The potential of the following electrochemical cell is 1.08 V.
Explanation:
[tex]E^0_(Cr^{3+}/Cr)[/tex]=-0.74V[/tex]
[tex]E^0_(Cu^{2+}/Cu)[/tex]=0.34V[/tex]
The element with negative reduction potential will lose electrons undergo oxidation and thus act as anode.The element with positive reduction potential will gain electrons undergo reduction and thus acts as cathode.
Here Cr undergoes oxidation by loss of electrons, thus act as anode. copper undergoes reduction by gain of electrons and thus act as cathode.
[tex]2Cr+3Cu^{2+}\rightarrow 2Cr^{3+}+3Cu[/tex]
[tex]E^0=E^0_{cathode}- E^0_{anode}[/tex]
Where both [tex]E^0[/tex] are standard reduction potentials, when concentration is 1M.
[tex]E^0=E^0_{[Cu^{2+}/Ni]}- E^0_{[Cr^{3+}/Cr]}[/tex]
[tex]E^0=0.34-(-0.74)=1.08V[/tex]
Thus the potential of the following electrochemical cell is 1.08 V.
