The cell reaction for the Zn-H, cell is Zn(s)+ 2H* (aq) → Zn* (aq) +H, (g). Standard cell potential is +0.76 V and the cell potential at 298K is 0.45 V. Calculate the pH of the cell if [Zn*]=1.0 M and P, =1.0 atm at equilibrium.
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1
Answer:
Zn+2H
+
⟶Zn
2+
+H
2
E
cell
o
=E
H
+
/H
2
−E
Zn
2+
/Zn
o
=0.76 V
From nernst equation,
E
cell
=E
cell
o
−
2
0.0591
log[
(H
+
)
2
(Zn
2+
)
]
⟹0.28=0.76−
2
0.0591
log[
(H
+
)
2
0.1
]
⟹
0.0591
0.48×2
=log[
(H
+
)
2
0.1
]
⟹1.86×10
16
=
(H
+
)
2
0.1
⟹1.86×10
16
=
(H
+
)
2
0.1
⟹(H
+
)
+2
=5.37×10
−18
⟹(H
+
)=2.31×10
−9
⟹pH=8.62
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