Answer: C) 0.637 V
Explanation:
The balanced redox reaction is:
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.
= -0.763
= -0.126
![E^0_{[Zn^{2+}/Zn]}=-0.763V](https://tex.z-dn.net/?f=E%5E0_%7B%5BZn%5E%7B2%2B%7D%2FZn%5D%7D%3D-0.763V)
![E^0_{[Pb^{2+}/Pb]}=-0.126V](https://tex.z-dn.net/?f=E%5E0_%7B%5BPb%5E%7B2%2B%7D%2FPb%5D%7D%3D-0.126V)
![E^0=E^0_{[Pb^{2+}/Pb]}- E^0_{[Zn^{2+}/Zn]}](https://tex.z-dn.net/?f=E%5E0%3DE%5E0_%7B%5BPb%5E%7B2%2B%7D%2FPb%5D%7D-%20E%5E0_%7B%5BZn%5E%7B2%2B%7D%2FZn%5D%7D)
The standard emf of a cell is 0.637 V
The chemical equation needs to be balanced so that it follows the law of conservation of mass.
<span>The best answer is B. ICl experiences induced dipole-induced dipole interactions. Both iodine and chlorine belongs to the same group of the periodic table. Electronegativity decreases as you go down a group therefore Cl will have a greater attraction with the bond it forms with another atom. Dipole-dipole interactions form between I and Cl. For the Br2 molecule, no dipole occurs because they are two identical atoms. Therefore we will be expecting ICl will have a higher boiling point due to higher binding energy it forms.</span>
The amount of oxygen bound to hemoglobin is 98.5%
