Zn (s)? + CuSO4?(aq)→ Cu (s)? + ZnSO4?(aq)
Zn (s)?????Zn2+?(aq)?+ 2e–
When a metal is dipped into a solution containing its ions an equilibrium is established between the metal and its ions. This is the basis of a half cell in an electrochemical cell.
Zn(s) ? Zn2+(aq) + 2e–
Zn2+(aq) + 2e–??? Zn(s)
Cu2+(aq) + 2e–??? Cu(s)???– reduction (rod becomes positive)
Cu(s) ? Cu2+(aq) + 2e–??? – oxidation (rod becomes negative)
Oxidation of copper atoms
Reduction of copper(II) ions
Br2(l) + 2e-?? 2Br-(aq)? ? ? ??E?= +1.09 V
Na+(aq) + e-?? Na(s)? ? ? ??E?= -2.71 V
Zn (s)∣Zn2+?(aq) ∥Cu2+?(aq)∣Cu (s)??????????????????E?cell?= +1.10 V
Cu (s)∣Cu2+?(aq) ∥Zn2+?(aq)∣Zn (s)??????????????????E?cell?= -1.10 V
Writing a cell diagram
If you connect an aluminium electrode to a zinc electrode, the voltmeter reads 0.94V and the aluminium is the negative. Write the conventional cell diagram to the reaction.
Answer
Al (s)∣Al3+?(aq) ∥ Zn2+?(aq)∣Zn (s)??????????????????E?cell?= +0.94 V
It is also acceptable to include phase boundaries on the outside of cells as well:
∣ Al (s)∣Al3+?(aq) ∥ Zn2+?(aq)∣Zn (s) ∣ ? ? ? ? ? ? ? ???E?cell?= +0.94 V
Students often confuse the redox processes that take place in electrochemical cells.
Remember, oxidation is the loss of electrons, so you are losing electrons at the negative.
∣ Al (s)∣Al3+?(aq) ∥Zn2+?(aq)∣Zn (s) ∣ ? ? ? ? ? ? ? ???Ecell?= +0.94 V
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