Calculate the single electrode potential at 298 K of the `Cu ^2 +` / Cu electrode in which the concentration of ions is 2.0M

Explanation

Ans: According to Nernst Equation
``

E . =dotE^0+0.0591/n log [M^n+]
=0.34+0.0591/2 log [Cu^2+]
=0.34 +0.0591/2 log 2=0.34+0.059 .1/2 . xx 0.3010
= 0.3489 ~V
endaligned
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