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Answer:
Anode:
3Mg(s) ----------> 3Mg2+(aq) + 6e
Cathode:
2Al3+(aq) +6e ---------> 2Al(s)
Explanation:
Anode:
3Mg(s) ----------> 3Mg2+(aq) + 6e
Cathode:
2Al3+(aq) +6e ---------> 2Al(s)
Magnesium is more electro positive than aluminum hence it functions as the anode. Six electrons are lost/gained in the redox process as shown in the oxidation and reduction half reaction equations above. Magnesium is oxidized to magnesium ion while aluminum is reduced to elemental aluminum.
Answer:
It emits 1.64 x 10⁻¹⁸J of energy
Explanation:
The n = 1 is a lower quantum level compared to n = 2.
When a hydrogen atom moves from a higher level to a lower one, it simply emits the energy difference between the two levels.
- If a hydrogen atom moves from a lower energy level to a higher one such as from 1 to 2, they absorb the energy difference to attain the new excited state.
- So, for an electron in the hydrogen atom to move from a higher energy level to a lower one, it must emit 1.64 x 10⁻¹⁸J of energy.