Answer:
[tex]E^0_{[At_2/At^-]}=0.30V[/tex]
Explanation:
[tex]Ni+At_{2}\rightarrow Ni^{2+}+2At^-[/tex]
Here Ni undergoes oxidation by loss of electrons, thus act as anode. At undergoes reduction by gain of electrons and thus act as cathode.
[tex]E^0=E^0_{cathode}- E^0_{anode}[/tex]
Where both [tex]E^0[/tex] are standard reduction potentials.
[tex]E^0_{[Ni^{2+}/Mg]}= -0.25V[/tex]
[tex]E^0_{[At_2/At^-]}=?[/tex]
[tex]E^0_{cell}=0.55V[/tex]
[tex]E^0=E^0_{[At_2/At^-]}- E^0_{[Ni^{2+}/Mg]}[/tex]
[tex]0.55V=E^0_{[At_2/At^-]-(-0.25V)[/tex]
[tex]E^0_{[At_2/At^-]}=0.30V[/tex]
