Consider the following galvanic cell at 2 5 ∘ C Pt | H₂(g) | (CH₃COOH) || (H₂CO₃ + NaHCO₃) | H₂(g) | Pt (1 bar) (100 ml, 0.1M) (100 ml 0.2M) (100 ml 0.2M) (1 bar) For H₂CO₃: K a 1 = 1 0 − 7 , K a

Question

Consider the following galvanic cell at $2 5^{\circ} C$

Pt | H₂(g) | (CH₃COOH) || (H₂CO₃ + NaHCO₃) | H₂(g) | Pt (1 bar) (100 ml, 0.1M) (100 ml 0.2M) (100 ml 0.2M) (1 bar)

For H₂CO₃: $K_{a_{1}} = 1 0^{- 7}$ , $K_{a_{2}} = 1 0^{- 11}$

For CH₃COOH: $K_{a} = 1 0^{- 5}$

Now, some electrolyte is added to one of the compartments of the given cell in List-I, and the corresponding cell potential in List-II is measured. Select the correct option matched correctly:

$T ak e \frac{2.303 RT}{F} = 0.06$ Volts

List - I	List - II
(P) If no electrolyte is added in either of the compartment of the cell, then	(1) -0.36 Volts
(Q) If 100 ml 0.2 M NaOH is added in the cathode compartment of the original cell	(2) -0.24 Volts
(R) If 50 ml 0.1 M NaOH is added in the anode compartment of the original cell then	(3) -0.3 Volts
(S) If 100 mL 0.02 M HCl is added in the anode compartment of the original cell then	(4) -0.12 Volts
	(5) E.m.f. of the cell decreases after addition of electrolyte with respect to initial condition

TestHub · Accepted Answer

(P) E cell ​ = 0.06 ( P H anode − P H cathode ) P H anode = 3 for Cathode H 2 ​ CO 3 ​ + Na 2 ​ CO 3 ​ → 2 NaHCO 3 ​ 20 10 0 10 0 10 P H = PK a 1 ​ = 7 E cell ​ = 0.06 ( 3 − 7 ) = − 0.24 (Q) for Cathode H 2 ​ CO 3 ​ + NaOH → NaHCO 3 ​ 0 10 10 0 0 20 P H = 9 E cell ​ = 0.06 ( 3 − 9 ) = − 0.36 (R) for Cathode CH 3 ​ COOH + NaOH → CH 3 ​ COOH 10 5 0 5 0 5 P H = 5 E cell ​ = 0.06 ( 5 − 7 ) = − 0.12 (S) P H = 2 E cell ​ = 0.06 ( 2 − 7 ) = − 0.3

Chemistry - Electrochemistry Question with Solution | TestHub

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