Question

1. The emf of the following cell at 25 C is equal to 0.34v. Calculate the reduction potential
of copper electrode .
Pt (s)| H, (g, latm)| H(aq, 1M)|| Cu?" (aq, IM) Cu (s)​

Answer 1

Answer:

We know that, Ered=Ered∘+n0.0591log10[ion]

Putting the values of Ered∘=0.34 V,n=2and [Cu2+]=0.1M

Ered=0.34+20.0591log10[0.1]

=0.34+0.02955×(−1)

=0.34−0.02955=0.31045 volt.

ok bro