Chemical BondingHard
Question
Find Ecell for the following cell
Mg(s)| Mg+2 (0.10 M)|| Ag+ (0.001 M)|Ag(s)
Given : Eocell = 3.17 V
= 0.06
Mg(s)| Mg+2 (0.10 M)|| Ag+ (0.001 M)|Ag(s)
Given : Eocell = 3.17 V
= 0.06Options
A.3.17 V
B.3.02 V
C.3.32 V
D.None of these
Solution
NCERT XII part 1 ; Page No. 71
Mg(s) → Mg+2 + 2e- (at anode)
2Ag+ + 2e- → 2Ag(s) (at cathode)
Mg(s) + 2Ag+ → Mg+2 + 2Ag(s)
Q =
= 105
Ecell = Eocell -
log10105
Ecell = 3.17 -
× 5log1010 = 3.17 - 0.03 × 5
= 3.17 - 0.15
= 3.02 V
Mg(s) → Mg+2 + 2e- (at anode)
2Ag+ + 2e- → 2Ag(s) (at cathode)
Mg(s) + 2Ag+ → Mg+2 + 2Ag(s)
Q =
= 105Ecell = Eocell -
log10105Ecell = 3.17 -
× 5log1010 = 3.17 - 0.03 × 5= 3.17 - 0.15
= 3.02 V
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