Question

It is expected that a chemical reaction will occur when copper metal is combined with aqueous zinc sulfate. Explain why there will be no reaction when zinc metal and aqueous copper sulfate solution are combined. Identify the anode and the cathode, assuming a voltaic cell is constructed. Note: Be careful in the calculation of the standard cell potential ( Eo cathode - Eo anode). Do not change the sign of the given reduction potential. The sign is already taken care of using the formula for calculating the standard cell potential.

Standard reduction potential: Cu2+(aq) + 2e- -------> Cu(s) Eo = -0.34 V

Zn2+(aq) + 2e- -------> Zn(s) Eo = -0.76 V

Standard cell potential = Eo cathode - Eo anode

Although , combination of the reactants do not result in voltaic cells, it can be predicted if the reaction is spontaneous, based on the standard reduction potential.

Answer 1

Answer:

$E_{cell}$= +ve, reaction is spontaneous

$E_{cell}$= -ve, reaction is non spontaneous

$E_{cell}$= 0, reaction is in equilibrium

Case 1: when copper metal is combined with aqueous zinc sulfate.

$Cu+ZnSO_4\rightarrow Zn+CuSO_4$

Here copper is undergoing oxidation ad thus acts as anode and zinc is undergoing reduction , thus acts as cathode.

$E^o_{cell}$ = standard electrode potential =$E^0_{cathode}- E^0_{anode}$

Where both $E^0$ are standard reduction potentials.

$E^0_{[Cu^{2+}/Cu]}= +0.34V$

$E^0_{[Zn^{2+}/Zn]}= -0.76V$

$E^0=E^0_{[Zn^{2+}/Zn]}- E^0_{[Cu^{2+}/Cu]}$

$E^0=-0.76-(+0.34)=-1.10V$

Thus as $E_{cell}$ is negative , the reaction is non spontaneous.

Case 2: when zinc metal and aqueous copper sulfate solution are combined.

$Zn+CuSO_4\rightarrow Cu+ZnSO_4$

Here zinc is undergoing oxidation ad thus acts as anode and copper is undergoing reduction , thus acts as cathode.

$E^o_{cell}$ = standard electrode potential =$E^0_{cathode}- E^0_{anode}$

Where both $E^0$ are standard reduction potentials.

$E^0_{[Cu^{2+}/Cu]}= +0.34V$

$E^0_{[Zn^{2+}/Zn]}= -0.76V$

$E^0=E^0_{[Cu^{2+}/Cu]}- E^0_{[Zn^{2+}/Zn]}$

$E^0=+0.34-(-0.76)=+1.10V$

Thus as $E_{cell}$ is positive , the reaction is spontaneous.