The standard electrodes potential for pb'2+ ! Pb and zn2+ are-0.126 and-0.763 v respectively. The e.m.f. of the cell
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Answer:
We know that E
cell
is given by the equation,
E
cell
=E
cell
0
−
n
0.059
log(K)
Here for the given reaction,
Zn+Pb
2+
→Zn
2+
+Pb
n=2 for this reaction.
[Zn
2+
]=0.1M and [Pb
2+
]=1M
E
cell
0
=E
cathode
−E
anode
=0.126−(−0.763)=0.889V
Hence,
E
cell
=0.889−
2
0.059
log
1
0.1
=0.889V
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