My questions are as follows: For the Daniell cell, we assume that ZnSO4 and CuSO4 are 1M aqueous solutions. What if we use pure molten ZnSO4 and CuSO4 instead?

Assume that we prepare an isolated half-cell containing just a zinc metal bar immersed in a ZnSO4 solution, without any connection to any other half-cells (no salt bridge or wire). Do you think the Statement below is true? In the microscopic scale, a few number of Zn atoms of the metal bar are first oxidized to Zn2+ ions and while migrating to the aqueous solution leave their electron on the electrode. But since there is no way for the electrons of the electrode to migrate to, the accumulation of negative charge on the electrode and positive charge in the solution, force the Zn2+ ions to reduce back to Zn on the electrode. This process also occurs for an isolated half-cells of Cu-CuSO4. Is it true? Is it right to say that here, the first few Cu atoms oxide to Cu2+ ions and leave the electrode negatively charged.

What is the reason for the separation of the first few Zn atoms from the electrode and oxidizing as Zn2+? (What is the driving force?) Why doesn’t just "neutral" Zn atoms separate from the electrode and migrate to the solution? Is it necessary for the Zn atoms to leave their electrons on the electrode and migrate to the solution as Zn2+ ions?

Is it possible that the Zn electrode, instead of releasing Zn2+ ions to the solution, adsorb Zn2+ of the ZnSO4 solution and accordingly the electrode get positively charged and the solution get negatively charged? Does it occurs for Cu bar and the CuSO4 solution?

  • $\begingroup$ When you have melted CuSO4, let me now. :-) Many of heavy metal sulphates decompose to metal oxides before melting. $\endgroup$ – Poutnik Jun 30 '19 at 18:16
  • $\begingroup$ Note that metals are not formed by neutral atoms, but rather there is a lattice of metal ions with delicalized valence electrons, forming conducting electron energy band. $\endgroup$ – Poutnik Jun 30 '19 at 18:19

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