ElectrochemistryHard
Question
What would be the magnitude of the EMF of the following cell:
Pt(H2) =
Pt(H2)
at 25oC ? The ionization constant of acetic acid, Ka ~ 10-5, while that of formic acid, Ka ~ 10-4.
Pt(H2) =
Pt(H2) at 25oC ? The ionization constant of acetic acid, Ka ~ 10-5, while that of formic acid, Ka ~ 10-4.
Options
A. 0.0295 V
B.0.059 V
C.-0.059 V
D.-0.0295 V
Solution
At anode:
0.1 M acetic acid + 0.1 M sodium acetate ⇒ Acidic buffer.
∴ pHa = pKa + log
(pKa(CH3COOH) = 5)
pHc = 4
Ecell = -0.059(pHc - pHa) = -0.059(4 - 5) = 0.059 V
0.1 M acetic acid + 0.1 M sodium acetate ⇒ Acidic buffer.
∴ pHa = pKa + log
(pKa(CH3COOH) = 5)pHc = 4
Ecell = -0.059(pHc - pHa) = -0.059(4 - 5) = 0.059 V
Create a free account to view solution
View Solution FreeMore Electrochemistry Questions
The specific conductance of AgCl solution in water was determined to be 1.8 × 10−6 Ω−1 cm−1 at 298 K. The molar conducta...Calculate the cell EMF in mV for Pt | H2 (1atm) | HCl (0.01 M) | AgCl(s) | Ag(s) at 298 Kif ᐃGfo values are at 25o...The chemical reaction,2AgCl(s) + H2(g) → 2HCl(aq) = 2Ag(s)Taking place in a galvanic cell is represented by the no...The correct order of equivalent conductance at infinte dilution of LiCl, NaCl is :...The EMFs of the cell obtained by combining separately Zn and Cu electrodes of a Daniel cell with normal calomel electrod...