The standard electrode potential of Pb ions is 0.126V. What will be the potential of a 0.5M solution at 298 K?
Answers
Answered by
0
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
Hence,option D is correct answer
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