. Calculate the potential of the following cell at 298 K.
Sn(S)|Sn2+(0.05M)||H+(0.02M)|H2(g)(1bar)|Pt(S) ; E0
Sn²⁺/Sn =-0.14V
Answers
Answered by
1
Explanation:
Given Zn∣Zn
2+
(0.001M)∣∣Cu
2+
(0.1M)∣Cu
Overall cell reaction:
Zn⟶Zn
2+
+2e
−
Cu
2+
+2e
−
⟶Cu
Zn+Cu
2+
⟶Zn
2+
+Cu
E
cell
o
=standard reduction potential of cathode + standard oxidation potential of anode
E
cell
o
=0.34 to 0.76 V
E
cell
o
=1.1 V
K
C
=
[Cu
2+
]
[Zn
2+
]
=
10
−1
10
−3
=10
−2
EMF of the cell at any electrode concentration is:
E=E
o
−
n
0.059
log(K
C
)=1.1−
2
0.059
log(10
−2
)=1.1−
2
0.059
×(2)=1.1−0.059=1.041V
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