Consider the cell H₂ (Pt)| H|H|H₂ (Pt) 1 bar a M b M 1 bar then the above cell is not working when a = 1 M, b = 1.5 M a = 0.5 M, b = 0.75 M a = 0.02 M, b = 0.01 M a=0.4 M, b=0.6 M Consider the cell H₂ ( Pt ) | H | H | H₂ ( Pt ) 1 bar a M b M 1 bar then the above cell is not working when
a = 1 M , b = 1.5 M
a = 0.5 M , b = 0.75 M
a = 0.02 M , b = 0.01 M
a = 0.4 M , b = 0.6 M
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Explanation:
Cathode: Ag
+
+e
−
⟶Ag E
Ag+/Ag
o
=0.799V
Anode: 1/2H
2
⟶H
+
+e
−
E
H
2
/H
+
o
=OV
Ag
+
+1/2H
2
⟶Ag+H
+
E
cell
o
=0.799−0=0.799V
Using Nernst Equation, at 25
o
C
E
cell
=E
cell
o
−
1
0.0591
log
[Ag
+
]
[H
+
]
as pH=5.5=−log[H
+
]
thus [H
+
]=10
−5.5
⇒1.023=0.799−
1
0.0591
log
[x]
[10
−5.5
]
⇒0.224=5.5×0.0591+0.0591log[x]
On solving, we get
⇒[x]=2×10
−2
M
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Explanation:
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