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JEE Advanced 2016 Paper 2 Chemistry Q1
Chemistry · Electrochemistry
Problem
For the following electrochemical cell at 298 K,
P
t
(
s
)
∣
H
X
2
(
g
,
1
b
a
r
)
∣
H
X
+
(
a
q
,
1
M
)
∣
∣
M
X
4
+
(
a
q
)
,
M
X
2
+
(
a
q
)
∣
P
t
(
s
)
\ce{Pt(s) | H2(g,1 bar) | H+(aq,1 M) || M^{4+}(aq), M^{2+}(aq) | Pt(s)}
Pt
(
s
)
∣
H
X
2
(
g
,
1
bar
)
∣
H
X
+
(
aq
,
1
M
)
∣∣
M
X
4
+
(
aq
)
,
M
X
2
+
(
aq
)
∣
Pt
(
s
)
E
c
e
l
l
=
0.092
V
E_{\mathrm{cell}}=0.092\,\mathrm V
E
cell
=
0.092
V
when
[
M
X
2
+
(
a
q
)
]
[
M
X
4
+
(
a
q
)
]
=
10
x
\dfrac{[\ce{M^{2+}(aq)}]}{[\ce{M^{4+}(aq)}]}=10^x
[
M
X
4
+
(
aq
)
]
[
M
X
2
+
(
aq
)
]
=
1
0
x
. Given:
E
M
X
4
+
/
M
X
2
+
∘
=
0.151
V
E^\circ_{\ce{M^{4+}/M^{2+}}}=0.151\,\mathrm V
E
M
X
4
+
/
M
X
2
+
∘
=
0.151
V
;
2.303
R
T
F
=
0.059
V
2.303\dfrac{RT}{F}=0.059\,\mathrm V
2.303
F
R
T
=
0.059
V
. The value of
x
x
x
is
Related theory
Electrochemistry
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