a) running with 0.1M ZnSO4 and 0.5M CuSO4 at 25°C?
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Answer:
E
cell
=E
cell
o
−
2
0.059
log
[Cu
2+
]
[Zn
2+
]
E
cell
=1.1−
2
0.059
log
[Cu
2+
]
[Zn
2+
]
Now,
E
cell
o
=E
Cu
2+
/Cu
o
−E
Zn
2+
/Zn
o
=1.1 V
Again
E
cell
=1.1−
2
0.059
log
[0.01]
[0.1]
E
cell
=1.1−0.029 log [10]
E
cell
=1.1−0.029 (as log [10]=1)
E
cell
=1.1−0.029=1.07 V
Explanation:
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