Calculate the halfcell potential at 298k for the r^n. cu2+ + 2e- -------->cu . if concentration of cu2+ is 5mol and standard electrode potential of cu is 0.34v.
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Explanation:
We know that, Ered=Ered∘+n0.0591log10[ion]
Putting the values of Ered∘=0.34 V,n=2 and [Cu2+]=0.1M
Ered=0.34+20.0591log10[0.1]
=0.34+0.02955×(−1)
=0.34−0.02955=0.31045 volt.
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