Answer:
1.52V
Explanation:
Oxidation half equation:
2Al(s)−→2Al^3+(aq) + 6e
Reduction half equation
3Sn2^+(aq) + 6e−→3Sn(s)
E°cell= E°cathode - E°anode
E°cathode= −0.140 V
E°anode= −1.66 V
E°cell=-0.140-(-1.66)
E°cell= 1.52V
Reorder 4Fe and 3O2.
3O2 + 4Fe
First you have to divide it by seven and then with the numbers do you have to put in the first quiz and then you have to add it on
If the value of H is positive, it means you have to add that much heat to complete the reaction. If H is negative, it means that much heat is released during the chemical process. Because it is -73 kJ, 73 kJ of heat are released in the reaction.
