A galvanic cell consisting of a Cu 2+ - Cu half cell versus a H + - H 2 half cel

Question

A galvanic cell consisting of a Cu2+ - Cu half cell versus a H+ - H2 half cell was used to determine the pH of an unknown solution. The unknown solution was placed in the hydrogen half cell and the pressure of the hydrogen gas was controlled at 1 atm. The concentration of Cu2+ in the Cu2+ - Cu half cell was 1.0 M and the electromotive force (emf) of the cell at 25°C was determined to be +0.48V.

Write the balanced equation for the reaction and calculate the pH of the unknown solution.

Answer for pH is 2.42.

How do I arrive at this answer. Highly confused

Explanation / Answer

The spontaneous reaction is

Cu2+    + H2 = Cu    + 2H+       (n = 2 electrons)

Eocell = 0.34 V – 0.00 = 0.34 V (see a Standard Reduction Potential Table and verify the value for Cu/Cu2+)

If the cell is not in standard condition ( here [H+] is not 1 M), then:

Ecell = Eocell – (0.0592 V/ 2) log [H+]2

0.48 V = 0.34V –0.0296 V · log [H+]2

- log [H+]2 = (0.48 -0.34)/0.0296

2 pH = 4.73

pH = 2.37

Note about Ecell = Eocell – (0.0592 V/ 2) log [H+]2

In general Ecell = Eocell – (0.0592 V/ 2) log Q

Q , reaction quotient has the expression of the equilibrium constant, but the used concentrations are the actual ones, not for equilibrium. Here [Cu] = 1 by convention (is in solid state), [Cu2+] = 1 M , pressure of H2 = 1 atm. Then

Log Q = log [H+]2

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