1 Faraday electricity was passed through Cu ²⁺(1.5 M , 1 ~L ) / Cu and 0.1 Faraday was passed through 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- Given: E _ Cu ²⁺ / Cu ^ o =0.34 ~V E _ Ag ⁺ / Ag ^0=0.8 ~V 2.303 RT F =0.06 ~V

Step-by-step solution

* Cu ⁺²+2 e ⁻ Cu aligned & (1 faraday = charge on 1 mole electron ) & t =0 1.5 1 ~mole & t = t 1 - 0.5 mole aligned [ Cu ⁺² ]=1 M after electrolysis * Ag + e ⁻ Ag array lll t =0 & 0.2 & 0.1 mole t = t & 0.1 & - - array [ Ag ⁺ ]=0.1 M after electrolysis Cell Cu _ ( s ) +2 Ag _ ( aq ) ⁺ Cu _ ( aq ) ⁺²+2 Ag _ ( s ) reaction aligned & E = E ^ - 0.06 n [ Cu ⁺² ] [ Ag ⁺ ]^2 & E =(0.8-0.34)- 0.06 2 1 (0.1)^2 =0.4 ~V aligned Correct answer =400 mV