Question

Exercise 20.105

Part A

The cell potential of the following electrochemical cell depends on the pH of the solution in the anode half-cell:
Pt(s)|H2(g,1atm)|H+(aq,?M)||Cu2+(aq,1.0M)|Cu(s).
What is the pH of the solution if Ecell is 370 mV ?

Express your answer using two significant figures.

pH =

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Answer 1

First, lets write the half cell reactions:

2H⁺ + 2 e^- --------> H₂ ......(I)

Cu²⁺ + 2 e^- -------------> 2 Cu (s) .........(II)

From Nernst equation, we know

E⁰_cell = E^-_cell - RT/nF ln Q

where R =8.314 J K⁻¹ mol⁻¹

T =298 K (assuming standard temperature)

F =9.65×10⁴ C mol⁻¹

n = 2 ( from equation II)

Q = rate constant of the reaction

The reaction of the cell is:

Cu²⁺ + H₂ ---------> Cu + 2 H⁺

So, Q = [Cu][H⁺]²/ [Cu²⁺]

Cu being solid , its concentration is unity. Concentration of Cu²⁺ is 1.0 moles given

For this cell,

E^-_cell = 0.34 V (From literature)

Hence, putting all the values we get,

0.37 = 0.34 - 0.059 log₁₀ [ H⁺]

or, -log₁₀ [ H⁺] = 0.51

or, pH = 0.51 ( pH = -log₁₀ [ H⁺])