Answer:
The half-cell that produces this cell potential is reduction cell potential; x(s) | x2+(1.0 M)
Explanation:
[tex]E_{Cell} =E_{Cathode} -E_{Anode}[/tex]
Positive potential is associated with reduction or cathodic half reaction:
x(s) | x2+(1.0 M): [tex]= +E{Cell}[/tex]
Negative potential is associated with oxidation or Anodic half reaction:
Cu2+(1.0 M) | Cu(s): = [tex]-E_{Cell}[/tex]
The given cell potential is measured as +1.97 V,
Therefore, the half-cell that produces this cell potential is reduction cell potential; x(s) | x2+(1.0 M).
