Redox Reactions and Electrochemistry: JEE Main 2025 Chemistry question

1 Faraday electricity was passed through $\mathrm{Cu^{2+}}$ (1.5 M, 1 L)/Cu and 0.1 Faraday was passed through $\mathrm{Ag^+}$ (0.2 M, 1 L)/Ag electrolytic cells. After this the two cells were connected as shown below to make an electrochemical cell. The emf of the cell thus formed at 298 K is ______ mV (nearest integer). Given: $E^\circ_{\mathrm{Cu^{2+}/Cu}}=0.34$ V, $E^\circ_{\mathrm{Ag^+/Ag}}=0.8$ V, $\frac{2.303RT}{F}=0.06$ V