ElectrochemistryHard
Question
In which of the following cells, EMF is greater than E-cell ?
Options
A.Pt, H2(g) | H+(pH = 5) || H+(pH = 3) | H2(g), Pt
B.Zn(s) | Zn2+(0.2 M) || Cu2+(0.1 M) | Cu(s)
C.Cr(s) | Cr3+(0.1 M) || Cu2+(0.2 M) | Cu(s)
D.Pt, H2(g) | H+ (pH = 4) || H+(pH = 6) | H2(g) | Pt
Solution
Use: Ecell = E-cell - 
log Qcell
For Ecell > E-cell, Qcell < 1 (i.e., log Qcell < 0)
a. H2(1 atm) + 2H+ (10-3 M) → 2H+ (10-5 M) + H2 (1 atm)
Qcell =
= 10 - 4 < 1
b . Zn(s) + Cu2+(0.1 M) → Zn2+ (0.2 M) + Cu(s)
Qcell =
= 2 > 1
c. 2Cr(s) + 3Cu2+(0.2 M) → 2Cr3+ (0.1 M) + 3Cu(s)
Qcell =
> 1
d. H2(1 atm) + 2H+ (10-6 M) → 2H+(10-4 M) + H2 (1 atm)
Qcell =
= 104 > 1
log QcellFor Ecell > E-cell, Qcell < 1 (i.e., log Qcell < 0)
a. H2(1 atm) + 2H+ (10-3 M) → 2H+ (10-5 M) + H2 (1 atm)
Qcell =
= 10 - 4 < 1b . Zn(s) + Cu2+(0.1 M) → Zn2+ (0.2 M) + Cu(s)
Qcell =
= 2 > 1c. 2Cr(s) + 3Cu2+(0.2 M) → 2Cr3+ (0.1 M) + 3Cu(s)
Qcell =
> 1d. H2(1 atm) + 2H+ (10-6 M) → 2H+(10-4 M) + H2 (1 atm)
Qcell =
= 104 > 1Create a free account to view solution
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