The reaction between Mg and Al can be represented as:
Mg (s) + 2Al3+ (aq) -> Mg2+ (aq) + 2Al (s)
The half reactions are:
Mg (s) -> Mg2+ (aq) + 2e- E° = -2.37 V
Al3+ (aq) + 3e- -> Al (s) E° = -1.66 V
The overall cell reaction is the sum of the half reactions:
Mg (s) + 2Al3+ (aq) -> Mg2+ (aq) + 2Al (s)
The overall cell balance is:
Anode: Mg (s) -> Mg2+ (aq) + 2e-
Cathode: Al3+ (aq) + 3e- -> Al (s)
The flow of electrons in the cell is from the anode to the cathode.
To calculate the Ecell, we use the formula:
Ecell = Ecathode - Eanode
Ecell = (-1.66 V) - (-2.37 V)
Ecell = 0.71 V
Therefore, the Ecell for the reaction between Mg and Al is 0.71 V.
Sketch the galvanic cell of reaction between Mg and Al and flow of electron, overall cell balance , half reaction and calculate the Ecell ?
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