Question

Consider the cell: CulCu.(0.00774 M)11Ag.(0.00639 M)|Ag Ecu./C'u = 0.521 V EAg.Ag-0.799 V Cu" /Cu Calculate Ecathode (Eright) at 298 K Number Calculate Eanode (Etef) at 298 K. Number 囗 Calculate Ecell at 298 K. Number

Answer 1

Consider the cell Cu| Cu (0.0074 M) I Ag (0.00639M) Ag The half-reaction at anode (oxidation) is as follows Cu ( s) → Cu" (aq)+ e- The half-reaction at cathode (reduction) is as follows Ag" (aq)+e-→Ag (s Consider the following equation for calculating the Ethode E-0.521V Cu/ cu Ag, Ag-0799 V 0.0591 -log | Ag Here, Etoda is cathode potential, E is standard cathode potential, n is the number of electron, and Ag is concentration. Ag /Ag Substitute the value of E follows (0.799V), n (2), and [Ag 0.00639) in equation 1 as Ag /Ag cathode = 0.799 V-0.0591 log 0.00639 -0.99-(-0.13) =0.929 V Therefore, the E at 298 K is 0.929 V catiode

Consider the following equation for calculating the Ee 0.0591 Here, is anode potential, E is standard anode potential, n is the number of electron, and Cu is concentration. Cu Cu Substitute the value of Ecucu (-0.521 V), n (2), and Lor」(0.00774) in equation 2 as follows 0.0591 =-0.521V log 0,00774 anode -0.521V- 0.125V =-0.646 V Therefore, the Em, at 298 K is 1-0.646V Calculate the Eis as follows = E anode-Ecathode -0.646V0.929 V 0.283V cell Therefore, the Ee at 298 K is 0.283 V

Consider the formula for calculating the initial potential V - IR Here, V is the potential, I is the current and R is the resistance. Substitute the value of 1 (0.600 A) and R (4.60 Ω) as follows: V 0.600A 4.60 Ω Therefore, the initial potential is 2.76 V