Answer:
Cu(s) in Cu(NO₃)₂(aq)
Explanation:
The standard reduction potential (E°) is the energy necessary to reduce the atom in a redox reaction. When an atom reduces it gains electrons from other than oxides. As higher is E°, easily it will reduce. The substance that reduces is at the cathode of a cell, where the electrons go to, and the other that oxides are at the anode of the cell.
The standard reduction potentials from Al(s) and Cu(s) are, respectively, -1.66V and +0.15V, so the half-cell of Cu(s) in Cu(NO₃)₂(aq) is the cathode.
Cations from smallest to largest
Li⁺ ,Na⁺, K⁺ (from Periodic Table, the bigger number of period, the bigger size, of atom, so the bigger size of cation)
1) LiF smaller cation then KF
1,036 <span>853
</span><span>The lattice energy increases as cations get smaller, as shown by LiF and KF.
</span><span>I think this one should be correct answer, because the compared substances have also the same anion, and we can compare cations in them.
2) The same cation Li , so wrong statement.
3)</span>The same cation Na , so wrong statement.
4) NaCl smaller cation then KF
786 853
That is actually physics because it talks about motion.
Replacement I think, hope this helps ;)
Explanation:
