Which of the following cell produces minimum
emf
1) Pt, H, (1 atm),HCI( 1M)//
HCI(0.1M)/H, (1atm), Pt
2) Pt, H2 (10atm),HCI(1M) //
HC1(0.1M)/Hz ( 1atm), Pt
3) Pt, H2 (10atm), HCI(0.1M) ||
// HCI(1M)/H(10atm), Pt
4) P1,A2 (0.1atm),HCI(0.1M) //
HCI (0.01M)/H, (10atm), Pt
Answers
Explanation:
Pt(s),H
2
(g)∣ 1bar HCl (aq) ∣ Ag(s) ∣ Pt(s) the cell potential is 0.92 when 10
−6
molal HCl solution is used. The standard electrode potential of (AgCl/Ag,Cl
−
) electrode is:
Given:
F
2.303RT
=0.06V at 298K
medium
MEDIUM
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ANSWER
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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Three metals A, B and C are arranged in increasing order of standard reduction electrode potential, hence their chemical reactivity order will be:
Explanation:
2nd option is correct or wrong