Question

Using the following data, determine the standard cell potential E^o cell for the electrochemical cell constructed using the following reaction, where zinc is the anode and lead is the cathode.

Zn(s) + Pb2+(aq) -> Zn2+(aq) + Pb(s)

Half-reaction: Standard Reduction Potential:

Zn2+(aq) + 2e- -> Zn(s)= -0.763

Pb2+(aq) + 2e- -> Pb(s)= -0.126

a. -0.889 V

b. +0.889 V

c. +0.637 V

d. +1.274 V

e. -0.637 V

Answer 1

Answer: C)  0.637 V

Explanation:

The balanced redox reaction is:

$Zn(s)+Pb^{2+}(aq)\rightarrow Zn^{2+}(aq)+Pb(s)$

Here Zn undergoes oxidation by loss of electrons, thus act as anode. Lead undergoes reduction by gain of electrons and thus act as cathode.

$E^0=E^0_{cathode}- E^0_{anode}$

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

$Zn^{2+}(aq)+2e^-\rightarrow Zn(s)$= -0.763

$Pb^{2+}(aq)+2e^-\rightarrow Pb(s)$= -0.126

$E^0_{[Zn^{2+}/Zn]}=-0.763V$

$E^0_{[Pb^{2+}/Pb]}=-0.126V$

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

$E^0=-0.126-(-0.763V)=0.637V$

The standard emf of a cell is 0.637 V