The potential of the following given cell is 1.02 v at 298 k temperature calculate the PH of HCl solution
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Answer:
Using Nernst equation for the cell reaction:
Anode:Sn→Sn
2+
+2e; E
Sn
2+
/Sn
0
=−0.14 V
Cathode:2H
+
+2e→H
2
; E
H
+
/H
2
0
=0 V
cell reaction:
Sn+2H
+
→Sn
2+
+H
2
E
cell
=E
0
−
nF
RT
ln(
[H
+
]
2
[Sn
2+
]
)
Given: E
cell
=0.092 V and E
0
=E
H
+
/H
2
0
−E
Sn
2+
/Sn
0
E
0
=0−(−0.14)=0.14 V
E
cell
=E
0
−
nF
RT
ln(
[H
+
]
2
[Sn
2+
]
)
0.092=0.14−
2
0.059
ln(
x
2
0.05
)
x=0.1
Calculation of pH:
pH=−log([H
+
])
pH=−log(0.1)
pH=1
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