1: Sketch of galvanic cell:
Al(s) | Al3+(aq) || Mg2+(aq) | Mg(s)
2: Overall balanced cell reaction:
Al(s) + Mg2+(aq) -> Al3+(aq) + Mg(s)
3: Half-cell reactions:
Anode (oxidation): Mg(s) -> Mg2+(aq) + 2e-
Cathode (reduction): Al3+(aq) + 3e- -> Al(s)
4: To calculate Ecell, we need to look up the standard reduction potentials for each half-reaction:
Mg2+ + 2e- -> Mg : E° = -2.37 V
Al3+ + 3e- -> Al : E° = -1.66 V
Ecell = E°cathode - E°anode
Ecell = -1.66 V - (-2.37 V)
Ecell = 0.71 V
The positive value for Ecell indicates that the cell reaction is spontaneous and can produce electrical energy. The fact that the Ecell value is relatively high suggests that this galvanic cell has a high potential for generating electricity.
Consider the line notation of galvanic cell of,
Mg (s)/Mg2+(aq)//Al3+(aq)/Al(s).
1: sketch the galvanic cell
2: give overall balance cell reaction
3: half cell reaction of galvanic cell above
4: calculate Ecell and comments on its value.
1 answer
1 answer