The voltaic cell made up of cobalt, copper, and their m2+ ions, has e°cell = 0.62 v. if e° of the cathode half-cell is 0.34 v, what is e° of the anode half-cell? cu2+(aq) + co(s) → cu(s) + co2+(aq)

-0.28V

Explanation:

E°cell= E°cathode- E°anode

E°cell=0.62V

E°cathode= 0.34V

E°anode= E°cathode- E°cell

E°anode= 0.34-0.62

E°anode= -0.28V

you need to take a look at the periodic system in order to sort these elements into pairs that would most likely exhibit similar chemical properties. first, you need to find elements that are found in the same group (groups are vertical in the periodic system), and then to see whether they have the same valence number. if they meet these criteria, they will have similar properties.

the first pair: ne (neon) and kr (krypton), are noble gases found in group 8

the second pair: li (lithium) and k (potassium), found in group 1

the third pair: s (sulfur) and te (tellurium), found in group 6

i'm assuming that it's c.) red bank..