Question

# Exercise 20.105 Part A The cell potential of the following electrochemical cell depends on the pH...

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 ?

pH =

First, lets write the half cell reactions:

2H+ + 2 e- --------> H2 ......(I)

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

From Nernst equation, we know

E0cell = 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+])

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