Answer: The
is 0.63 V
Step-by-step explanation:
In the given reaction :

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

Where both
are standard reduction potentials.
![E^0_([Pb^(2+)/Pb])= -0.13V](https://img.qammunity.org/2021/formulas/chemistry/high-school/unz0ouavml3f17wozshuzxsr9cqyl1eflb.png)
![E^0_([Zn^(2+)/Zn])=-0.76V](https://img.qammunity.org/2021/formulas/chemistry/high-school/olc4lf0brskzrvcslfp2n6xddpo3pietrg.png)
![E^0_(cell)=E^0_([Pb^(2+)/Pb])- E^0_([Zn^(2+)/Zn])](https://img.qammunity.org/2021/formulas/chemistry/high-school/bkwlhu70w76jmuwaleg599k30xfo9t02zt.png)

Thus the
is 0.63 V