1C An electrochemical cell consists of Cu electrode dipped in 0.5M CuSO4 solution and Ag
electrode dipped in 0.25M AgNO3 solution. Write the cell representation, cell reactions
& calculate the EMF. Given SEP's of Cu & Ag are 0.34V & 0.8V respectively.
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Answer:
Given, E
Cu
2+
/Cu
∘
=0.337 volt and E
Ag
+
/Ag
∘
=0.799 volt. The standard emf will be positive if Cu/Cu
2+
is anode and Ag
+
/Ag is cathode. The cell can be represented as:
Cu∣Cu
2+
∥Ag
+
∣Ag
The cell reaction is,
Cu+2Ag
+
→Cu
2+
+2Ag
E
cell
∘
= Oxid. potential of anode + Red. potential of cathode
=−0.337+0.799
=0.462 volt
Applying the Nernst equation,
E
cell
=E
cell
∘
−
2
0.0591
log
[Ag
+
]
2
[Cu
2+
]
When, E
cell
=0
E
cell
∘
=
2
0,0591
log
[Ag
+
]
2
[Cu
2+
]
or log
[Ag
+
]
2
[Cu
2+
]
=
0.0591
0.462×2
=15.6345
[Ag
+
]
2
[Cu
2+
]
=4.3102×10
15
[Ag
+
]
2
=
4.3102×10
15
0.01
=0.2320×10
−17
=2.320×10
−18
[Ag
+
]=1.523×10
−9
M.
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