Chemistry - Electrochemistry Question with Solution | TestHub

The cell representation is given as ${\text{Pb}}_{(s)}|{\text{Pb}}_{(aq)}^{+2}||{\text{Ag}}_{(aq)}^{+}|{\text{Ag}}_{(s)}$.

Here, Pb acts as the anode and Ag acts as the cathode.

The standard cell potential is given by the formula:

$E_{cell}^o=E_{cathode}^o-E_{anode}^o$

Substituting the given standard reduction potentials:

$E_{cell}^o=E_{{\text{Ag}}^{+}/\text{Ag}}^o-E_{{\text{Pb}}^{+2}/\text{Pb}}^o$

$E_{cell}^o=0.799-(-0.126)$

$E_{cell}^o=0.799+0.126=0.925$ V

Answer: $0.925$ V