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The reaction of copper that takes place is
Cu+2 +2e-------> Cu Eo = 0.34 V
In order to calculate the potential of the cell given the [Cu+2] concentration . we are going to apply this formula
Ecell = E cell -(RT/nF)ln ([reduction half ] / [oxidation half])
a) E= E0 + (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
Ecell = 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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