Question

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An electrochemical cell consists of a standard hydrogen electrode and a copper metal electrode. a. What is the potential of the cell at 25°C if the copper electrode is placed in a solution in which 9.0 x 10-5 M? Ecell = v b. The copper electrode is placed in a solution of unknown The measured potential at 25°C is 0.177 V. What is ? (Assume Cu2+ is reduced.)

Answer 1

The reaction of copper that takes place is

Cu_­­⁺² +2e^-------> Cu         E­_o­ = 0.34 V

In order to calculate the potential of the cell given the [Cu⁺²] concentration . we are going to apply this formula

E­_cell­ = E ­_cell­ -(RT/nF)ln ([reduction half ] / [oxidation half])

a) E= E­_0­ + (0.059/2)(log([Cu⁺²]))

thus we get

E= 0.34 + (0.059/2)(log([9.0 X 10^-5]))

E=0.220 V

b) now were asked to find the concentraction given E= 0.177

once again we use the same equation

E­_cell­ = E ­_cell­ -(RT/nF)ln ([reduction half ] / [oxidation half])

0.177= 0.34_­ + (0.059/2)(log([Cu⁺²]))

-5.525 = (log([Cu⁺²]))

thus [Cu⁺²] = 2.985 X 10^-6 M

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