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KUET_14-15
HSC - রসায়ন ২য় পত্র
→
অধ্যায়-০৪ঃ তড়িৎ রসায়ন
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P
b
∣
P
b
2
+
(
1.0
M
)
∣
∣
H
+
(
0.4
M
)
∣
H
2
(
1
a
t
m
)
∣
P
t
\mathrm{Pb}\left|\mathrm{Pb}^{2+}(1.0 \mathrm{M})\right|\left|\mathrm{H}^{+}(0.4 \mathrm{M})\right| \mathrm{H}_{2}(1 \mathrm{~atm}) \mid \mathrm{Pt}
Pb
Pb
2
+
(
1.0
M
)
∣
H
+
(
0.4
M
)
∣
H
2
(
1
atm
)
∣
Pt
[দেওয়া আছে,
E
P
b
2
+
/
P
b
0
=
−
0.14
V
]
\left.\mathrm{E}_{\mathrm{Pb}^{2+} / \mathrm{Pb}}^{0}=-0.14 \mathrm{~V}\right]
E
Pb
2
+
/
Pb
0
=
−
0.14
V
]
A.
0.1165V
B.
0.170V
C.
0.1155V
D.
0.1161V
E.
0.1175 V
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