Assuming these half reactions make a galvanic cell, what is the cell potential in volts.
πΉ2(π)+2πββπΉβ(ππ)
πΈ0πππ=2.87
π΄π3++3πββπ΄π(π )
πΈππππ=β1.66
F2 + 2e ==> 2F^- .......E = 2.87 v
Al ==> Al^3+ + 3e.......E = 1.61
Add the half cells to get the cell reaction. Add the E values to obtain the voltage of the cell.
Notes: 1. Note I changed the sign of the Al half cell because I wrote it as an oxidation and not a reduction.
2. When you add the two half cells you should multiply eqn 1 by 3 and multiply eqn 2 by 2 in order to make the electron loss equal to the electron gain.
3. When you multiply the equations to make the electrons equal you do NOT (repeat, NOT) multiply the voltages.