Question

the following overall reaction: a cell that employs 22. ConsidcAl () 3Zn (ag) 2A (ag)32n (s) (ag) 3Zn (s) 22. Co What is ΔG° for the cell? Eored =-0.76 V Eoed =-1.66 V Zn2+ + 2e-→ Zn Al3+ + 3e. → Al a) + 627 kJ/mol b) -648 kJ/ mol c)-216 kJ/ mol d)-627 kJ/ mol Ve)-521 kJ/mol 12 23. Given,; Ti2++ 2e-→ Ti Eoredー-1.6284 Calculate E° for the half reaction a) -3.608 v b) -2.75 V c) +2.75 V 1.208 V e +1.94V

Can u solve 22 and 23 separately thank u
Best

Answer 1

22. As we know ?G° = - n FE°cell

F- Faraday's constant = 96485 C/mol, n- number of moles of electrons, = 6

E°cell = E°cathode - E°anode = -.76 -( - 1.66 )=0.90 V

*(right hand side half cell acts as cathode and left side anode)

So, ?G = - 6 mol×96485C/mol×0.97 V=521.019 C.V

= - 521,019 J= - 521.019 kJ ~ -521 kJ

So, the correct option is e)

23. The desires equation can be obtained by adding the given equation. When equation are added their electrode potential value are also added. So, the E° = -0.369 - 1.628

= -1.997 V ~1.94 V

So , option e) is correct.