Mg +2 Ag + (0.0001M) → Mg 2+(0.130M) + 2Ag, E° =3.17V
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ANSWER
Mg+2Ag+(0.0001M)⟶Mg
2+
(0.130M)+2Ag
E
cell
=E
o
−
2
0.0591
log[
(Ag
+
)
2
(Mg
2+
)
]
E
cell
=8.17−
2
0.0591
log[
10
−8
0.130
]=2.959 V
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