calculate the cell potential,E, at 25°c for the cell if the initial concentration of Ni(NO3)2 is 0.1 molar and the initial concentration of AgNO3 is 1.00 molar.[E°Ni 2+/Ni=-0.25v;E° ag3+/ag=0.80v]
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The overall cell reaction is given by:
Cu
2+
+H
2
→Cu+2H
+
E
cell
∘
=E
Cu
2+
/Cu
∘
−E
H
+
/H
2
∘
⟹E
cell
∘
=0.337−0=0.337V [standard hydrogen electrode potential is zero].
Now, E
cell
=0.337−
2
0.059
log[
[Cu
2+
]
[H
+
]
2
]
E
cell
=0.337−
2
0.059
log(
(0.1)
(0.01)
2
)
E
cell
=0.425V
Option C is correct.
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