How would you draw a galvanic cell for Aluminum and Charcoal? Anode: Al(s) + 3OH-(aq) >>>> Al(OH)3(s) + 3e-
E cell/V: -2.31
Cathode: O2(g) + 2H2O(l) +4e- >>>> 4OH-(aq)
E cell/V: 0.401
Overall: 4Al(s) + 3O2(g) + 6H2O(l) >>>> 4Al(OH)3(s)

Please just write it out because I can not see the files you send put into a link thank you

Answerthe half-reaction having the greatest tendency to occur at the anode:     i₂(aq) + 2e⁻  ⇄ 2i⁻  ;   e = +0.53 v======oxidation occurs at the anode. we must therefore find the half-reaction with the greatest tendency to oxidize.the given values of e are reduction potentials (the form of the half-reactions are of reduction).the lower the reduction potential, the higher the potential to oxidize. the half-cell with the lowest reduction potential will have the greatest tendency to oxidize at the anode.the half-reaction having the greatest tendency must be    i₂(aq) + 2e⁻  ⇄ 2i⁻    e = +0.53 vbecause its reduction potential is the lowest out of the three.

It should be 2.8 c/m

Most elements have more than one naturally occurring isotope. as you learned previously, the atoms of those isotopes have the same atomic number (number of protons), making them belong to the same element, but they have different mass numbers (total number of protons and neutrons) giving them different atomic masses. so which mass is put on the periodic table for each element? is it the most common isotope’s mass? the heaviest mass? this activity will answer that question.