Respuesta :
Answer:
Cell potential = +1.09V
Explanation:
Given:
[Zn2+] = 0.25 M
[Cu2+] = 0.15 M
In this electrochemical cell, Zn/Zn2+ half cell will serve as the anode and Cu/Cu2+ will act as the cathode.
The two half reactions and the reduction potentials based on the standard reduction potential chart are:
Anode (Oxidation):
[tex]Zn(s)\rightarrow Zn^{2+}(aq)+2e^{-}[/tex].......E°(anode) = -0.76 V
Cathode(Reduction):
[tex]Cu^{2+}(aq)+2e^{-}\rightarrow Cu(s)[/tex]........E°(cathode)= +0.34 V
Overall reaction:
[tex]Zn(s) + Cu^{2+}(aq)\rightarrow Cu(s) + Zn^{2+}(aq)[/tex]
[tex]E_{cell}^{0} =E_{cathode}^{0}-E_{anode}^{0}=0.34-(-0.76)=1.10V[/tex]
Based on the Nernst equation:
[tex]Ecell = E_{cell}^{0}-\frac{0.0592V}{n}log\frac{[Zn2+]}{[Cu2+]}[/tex]
[tex]Ecell = 1.10-\frac{0.0592V}{2}log\frac{[0.25]}{[0.15]}=1.09V[/tex]
