Select 1 reduction half reaction choices - Fe2+(aq) + 2e- -> Fe(s), Fe(s) -> Fe2+(aq)+2e- , Au3+ (aq) + 3e- -> Au(s) ,

[answerhappygod]

Select 1 reduction half reaction choices - Fe2+(aq) + 2e- -> Fe(s), Fe(s) -> Fe2+(aq)+2e- , Au3+ (aq) + 3e- -> Au(s) , Au(s) -> Au3 +(aq) + 3e-
Select 2 oxidation half reaction choices - Fe2+(aq) + 2e- -> Fe(s), Fe(s) -> Fe2+(aq)+2e- , Au3+ (aq) + 3e- -> Au(s) , Au(s) -> Au3 +(aq) + 3e-
Select 3 Number of electrons in balanced reaction choices - 1, 2, 3, 4
Select 4 standard cell potential, choices- 0.44 V, 1,50 V, 1.89 V , 1.94 V
Select 5 emf of the cell - 1.83 V , 1.89 V, 1.94 V, 1.99 V

Select 1 Reduction Half Reaction Choices Fe2 Aq 2e Fe S Fe S Fe2 Aq 2e Au3 Aq 3e Au S 1 (59.92 KiB) Viewed 15 times

Select 1 Reduction Half Reaction Choices Fe2 Aq 2e Fe S Fe S Fe2 Aq 2e Au3 Aq 3e Au S 2 (59.92 KiB) Viewed 15 times

Given the cell notation: Fe(s) | Fe2+ (aq) || Au3+ | Au Identify the ff. 1. Reduction Half reaction: [Select] < 2. Oxidation Half reaction: (Select] 3. Number of electrons in balanced reaction: Select) 4. Standard cell potential, Eºcell: (Select) 5. emf of the cell if [ Fe2+ ] = 1.500 M and [ Aus+ 1 = 0.0040 M: Select) Reduction Hall Reactions EV Fe2+ + 2e -> Fe -0.44 AU3+ + 3e - Au +1.50