For the given cell Fe2+(aq) + Ag+(aq) Fe3+(aq) + Ag(s) The standard cell potential of the above reaction is Given: Ag+ + e- Ag E = xV Fe2+ + 2e- Fe E = yV Fe3+ + 3e- Fe E = zV
- x + 2y - 3z
- x + 2y
- y - 2x
- x + y - z
Answer
(1) x + 2y - 3z
Solution
E is not additive; G =-nFE is.
Fe3++e- Fe2+ = (3e⁻ eq.) − (2e⁻ eq.): -FE =-3Fz+2Fy E =3z-2y
E cell=E Ag+/Ag-E Fe3+/Fe2+=x-(3z-2y)=x+2y-3z
Answer: (1)