1. Sketch of galvanic cell:
Al(s) | Al^3+(aq) || Mg^2+(aq) | Mg(s)
Electrons flow from Al to Mg, so the anode is Al(s) and the cathode is Mg(s).
2. Overall balanced cell notation:
Al(s) + 3Mg^2+(aq) → Al^3+(aq) + 3Mg(s)
3. Half-reactions:
Anode (oxidation): Al(s) → Al^3+(aq) + 3e^-
Cathode (reduction): Mg^2+(aq) + 2e^- → Mg(s)
4. E°cell calculation:
E°cell = E°cathode - E°anode
E°cell = E°Mg^2+/Mg - E°Al^3+/Al
E°cell = -2.37 V - (-1.66 V)
E°cell = -0.71 V
The negative E°cell value indicates that the reaction is not spontaneous under standard conditions. However, under non-standard conditions, the cell may still be able to produce a potential difference and do work.
Consider the line notation of a galvanic cell describe below.
Mg|Mg^2+||Al^3+|Al
1. Sketch the galvanic cell based on the above half-reactions and shows the direction of the electrons flow, and identify the cathode and the anode?
2. Give the overall balance cell notation?
3. Provide the half-reactions for the galvanic cell line notation?
4. Calculate E°cell for the galvanic cell and comment on its value?
1 answer