Question

If the Ecell of the following cell is 0.945 V, what is the pH in the anode compartment? Pt(s) | H2 (1.00 atm) | H+(aq) || Ag+(0.100 M) | Ag(s)

1.245 is NOT the correct answer

Answer 1

Consider reactions taking place at anode & cathode.

At anode : 1 /2 H ₂ (g) H ⁺ (aq) + e ^-

At cathode : Ag ⁺ (aq) + e ^- Ag (s)

overall Reaction : 1 /2 H ₂ (g) + Ag ⁺ (aq) H ⁺ (aq) + Ag (s)  

Q value of above reaction is Q = [ H ⁺ ] / [ Ag ⁺ ] ( P H ₂ ) ^1/2

In overall reaction, electrons are transferred are .

Standard emf of the cell is E ⁰ cell = E ⁰ cathode - E ⁰ anode

E ⁰ cell = 0.799 V - 0.000 V  = 0.799 V

The emf of the cell is calculated by using Nernst equation E = E ⁰ cell - 2.303 R T / n F log Q

Where, n is no. of electrons transferred in the overall reaction.

0.945 = 0.799 - 2.303 R T / n F log [ H ⁺ ] / [ Ag ⁺ ] ( P H ₂ ) ^1/2

We know that , at 298 K 2.303 R T / F = 0.0592 .

0.945 = 0.799 - 0.0592 / n log [ H ⁺ ] / [ Ag ⁺ ] ( P H ₂ ) ^1/2

Put n = 1 , [ Ag ⁺ ] = 0.100 M & P H ₂ = 1.00 atm in above equation.

0.945 = 0.799 - 0.0592 / 1  log [ H ⁺ ] / ( 0.100 ) ( 1.00 ) 1/2

0.945 = 0.799 - 0.0592 log [ H ⁺ ] /0.100

- 0.0592 log [ H ⁺ ] /0.100 = 0.945 - 0.799

- 0.0592 log [ H ⁺ ] /0.100 = 0.146

- log H ⁺ ] /0.100 = 0.146 / 0.0592

- log [ H ⁺ ] /0.100 = 2.466

log [ H ⁺ ] /0.100 = - 2.466

[ H ⁺ ] /0.100 = 10 ^-2.466 = 3.420 10 ^-03

[ H ⁺ ] = 0.100 ( 3.420 10 ^-03 )

[ H ⁺ ] = ( 3.420 10 ^-04) M

We have relation, pH = - log [ H ⁺ ]

pH = - log 3.420 10 ^-04

pH = 3.466

ANSWER : pH in the anode compartment = 3.466