Consider the Galvanic cell
 $Ag\left|AgN{O}_3\left(0.001M\right)\right|\left|AgN{O}_3\left(0.1M\right)\right|Ag,E_{A{g}^{+},Ag}^0=0.80volt$
Choose the CORRECT options when both half cells are connected by an electronic conductor.
 

 $E_{cell}=E_2-E_1$
= $\{E_{A{g}^{+},Ag}^0+\frac{0.0591}{n}\mathcal{l}og{\left[A{g}^{+}\right]}_{right}\}-\{E_{A{g}^{+}Ag}^0+\frac{0.0591}{n}\mathcal{l}og{\left[A{g}^{+}\right]}_{left}\}$
= $+\frac{0.0591}{n}\mathcal{l}og{\left[A{g}^{+}\right]}_{right}-\frac{0.0591}{n}\mathcal{l}og{\left[A{g}^{+}\right]}_{left}$
= $\frac{0.0591}{n}\mathcal{l}og\left\{\frac{{\left[A{g}^{+}\right]}_{right}}{{\left[A{g}^{+}\right]}_{left}}\right\}$ = $\frac{0.0591}{1}\mathcal{l}og\frac{0.1}{0.001}=0.0591\mathcal{l}og100$
= $\frac{0.0591}{1}\mathcal{l}og\frac{0.1}{0.001}=0.0591\mathcal{l}og100$  = 0.1182 volt
Since $E_{cell}$   is positive, the right electrode will act as cathode (where reduction occurs) and the left electron will act as anode (where oxidation occurs). The electrons will thus travel from left to right in external circuit.