The emf of following cell containing 2 hydrogen electrodes is pt1/2H2(g)|H2(g)|H^+(10^-8)||M^+(0.001M)|1/2H2(g)pt
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Explanation:
According to Nernst equation,
E=E
o
−0.059logQ/n
Here,
E
cell
=E
o
+
1
0.059
log
[H
+
]
LHS
[H
+
]
RHS
=0+0.059log
10
−8
10
−3
=0.059×5=0.295V (as E
o
=0 for concentration cell)
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