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Chemistry
Chemistry
Problem
The voltage of the cell given below is
−
0.46
V
-0.46\text{ V}
−
0.46
V
:
Pt
H
2
∣
NaHSO
3
(
0.4
M
)
∣
Na
2
SO
3
(
6.44
×
10
−
3
M
)
∣
Zn
2
+
(
0.3
M
)
∣
Zn
(
s
)
\text{Pt}_{\text{H}_2}\,|\,\text{NaHSO}_3\ (0.4\text{ M})\,|\,\text{Na}_2\text{SO}_3\ (6.44\times 10^{-3}\text{ M})\,|\,\text{Zn}^{2+}\ (0.3\text{ M})\,|\,\text{Zn}(s)
Pt
H
2
∣
NaHSO
3
(
0.4
M
)
∣
Na
2
SO
3
(
6.44
×
1
0
−
3
M
)
∣
Zn
2
+
(
0.3
M
)
∣
Zn
(
s
)
Also,
Zn
2
+
+
2
e
⟶
Zn
(
s
)
\text{Zn}^{2+} + 2e \longrightarrow \text{Zn}(s)
Zn
2
+
+
2
e
⟶
Zn
(
s
)
,
E
∘
=
−
0.763
V
E^\circ = -0.763\text{ V}
E
∘
=
−
0.763
V
. Calculate the value of
K
2
K_2
K
2
, where
K
2
=
[
H
+
]
[
SO
3
2
−
]
[
HSO
3
−
]
K_2 = \frac{[\text{H}^+][\text{SO}_3^{2-}]}{[\text{HSO}_3^-]}
K
2
=
[
HSO
3
−
]
[
H
+
]
[
SO
3
2
−
]
.
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