Solve
Guides
Join / Login
Use app
Login
0
You visited us
0
times! Enjoying our articles?
Unlock Full Access!
Standard XII
Chemistry
Gibb's Energy and Nernst Equation
Question
The half-cell reaction for the corrosion,
2
H
+
+
1
2
O
2
+
2
e
−
→
H
2
O
;
E
∘
=
1.23
V
F
e
2
+
+
2
e
−
→
F
e
(
s
)
;
E
∘
=
−
0.44
V
Find the
△
G
∘
(in kJ) for the overall reaction.
−
322
k
J
−
76
k
J
−
161
k
J
−
152
k
J
A
−
322
k
J
B
−
161
k
J
C
−
76
k
J
D
−
152
k
J
Open in App
Solution
Verified by Toppr
F
e
(
s
)
→
F
e
2
+
+
2
e
−
;
△
G
∘
1
2
H
+
+
2
e
−
+
1
2
O
2
→
H
2
O
(
l
)
;
△
G
∘
2
–
––––––––––––––––––––––––––––––––––––
–
F
e
(
s
)
+
2
H
+
+
1
2
O
2
→
F
e
2
+
+
H
2
O
;
△
G
∘
3
–
–––––––––––––––––––––––––––––––––––––––––––––
–
Applying
△
G
∘
1
+
△
G
∘
2
=
△
G
∘
3
△
G
∘
3
=
(
−
2
F
×
0.44
)
+
(
−
2
F
×
1.23
)
=
(
−
2
×
96500
×
0.44
)
+
(
−
2
×
96500
×
1.23
)
= -322310\ J $
=
−
322
k
J
.
Was this answer helpful?
4
Similar Questions
Q1
The half-cell reaction for the corrosion,
2
H
+
+
1
2
O
2
+
2
e
−
→
H
2
O
;
E
∘
=
1.23
V
F
e
2
+
+
2
e
−
→
F
e
(
s
)
;
E
∘
=
−
0.44
V
Find the
△
G
∘
(in kJ) for the overall reaction.
View Solution
Q2
The half cell reactions for the corrosion are
2
H
+
+
1
/
2
O
2
+
2
e
−
→
H
2
O
;
E
o
=
1.23
V
F
e
2
+
+
2
e
−
→
F
e
(
s
)
;
E
o
=
−
0.44
V
. Find teh
Δ
G
o
(in kJ) for the overall reaction:
View Solution
Q3
The half cell reactions for rusting of iron are :
2
H
+
+
1
2
O
2
+
2
e
−
→
H
2
O
;
E
o
=
1.23
V
F
e
2
+
+
2
e
−
→
F
e
;
E
o
=
−
0.44
V
Δ
G
o
(in kJ) for the complete cell reaction is :
View Solution
Q4
The half-cell reactions for rusting of iron are,
2
H
+
+
1
2
O
2
+
2
e
−
→
2
H
2
O
,
E
∘
=
+
1.23
V
,
F
e
2
+
+
2
e
−
→
F
e
(
s
)
;
E
∘
=
−
0.44
V
.
Δ
G
∘
(
i
n
k
J
)
for the reaction is :
View Solution
Q5
The rusting of iron takes place as follows :
2
H
⊕
+
2
e
−
+
1
2
O
2
⟶
H
2
O
(
l
)
;
E
⊙
=
+
1.23
V
F
e
2
+
+
2
e
−
⟶
F
e
(
s
)
;
E
⊙
=
−
0.44
V
Calculate
Δ
G
⊙
for the net process.
View Solution