The cell Pt (H2) (1 atm) | (pH = ?) || (a = 1) ||
Agl(s), Ag has emf, E298K = 0. The standard
electrode potential for the reaction Agl + e
- Ag + IQ is - 0.151 volt.
Calculate the pH value.
Me
5
th
= =
--
in
.
L
TE
=
-
-
HA
3.37
.
Elu
"
5.26
ht.
he
hu
La
2.56
4.62
Answers
Answered by
0
Answer:
:):):):):(:(:(
Explanation:
The half-cell reactions are,
At Anode:
2
1
H
2
(g)→H
+
(aq)+e
−
At Cathode:AgCl(s)+e
−
→Ag(s)+Cl
−
(aq)
Complete reaction:AgCl(s)+
2
1
H
2
(g)→Ag(s)+Cl
−
(aq)+H
+
(aq)
We know,
E
cell
0
=E
cathode
0
−E
anode
0
=(SRP)
cathode
−(SRP)
anode
We know standard hydrogen potential is assumed to be zero.
So,(SRP)
anode
=0
Let, (SRP)
cathode
=x
So,
E
cell
0
=x
Now we use Nernst equation,
E
cell
=E
cell
0
−
nF
2.303RT
log(Q)
⟹E
cell
=E
cell
0
−0.06×log([Cl
−
][H
+
])
n=1;
0.92=x−
1
0.06
log(10
−6
×10
−6
)
⟹x=0.20V
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