The line notation given represents a galvanic cell with magnesium as the anode and aluminum as the cathode. The cell notation indicates that magnesium metal is oxidized to Mg2+ ions at the anode, and aluminum ions are reduced to aluminum metal at the cathode.
The standard reduction potentials for the half-reactions involved are as follows:
Mg2+(aq) + 2e- → Mg(s) E° = -2.37 V
Al3+(aq) + 3e- → Al(s) E° = -1.66 V
To calculate the standard cell potential (E°cell) for the galvanic cell, we can use the formula:
E°cell = E°cathode - E°anode
E°cell = -1.66 V - (-2.37 V)
E°cell = 0.71 V
The positive value of E°cell indicates that the cell is spontaneous and can produce an electrical current. The higher the positive value of E°cell, the greater the driving force for the cell reaction, and the higher the efficiency of the cell in producing electrical energy. In this case, the E°cell value of 0.71 V suggests that the galvanic cell is quite efficient in converting chemical energy into electrical energy.
Consider the line notation of a galvanic cell describe Mg(s)|Mg2+(aq)||Al3+|Al(s)
Calculate the E°cell for the galvanic cell and comment on its value
1 answer
1 answer