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shepuryov [24]
4 years ago
11

Given the balanced ionic equation representing the reaction in an operating voltaic cell: zn(s) + cu2+(aq) → zn2+(aq) + cu(s) th

e flow of electrons through the external circuit in this cell is from the
Chemistry
1 answer:
soldi70 [24.7K]4 years ago
3 0

Answer: from the Zn anode to the Cu cathode


Justification:


1) The reaction given is: Zn(s) + Cu₂⁺ (aq) -> Zn²⁺ (aq) +Cu(s)


2) From that, you can see the Zn(s) is losing electrons, since it is being oxidized (from 0 to 2⁺), while Cu²⁺, is gaining electrons, since it is being reduced (from 2⁺ to 0).


3) Then, you can already tell that electrons go from Zn to Cu.


4) The plate where oxidation occurs is called anode, and the plate where reduction occus is called cathode.


So you get that the electrons flow from the anode (Zn) to the cathode (Cu).


Always oxidation occurs at the anode, and reduction occurs at the cathode.

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4. We can set up a proportion given that 14.7 PSI = 101 KPa. This ratio should hold for 23.6 PSI. In other words, 14.7/101 = 23.6/x; to solve for x, which would be your answer, we compute 23.6 PSI × 101 kPa ÷ 14.7 PSI = 162 kPa.

5. We are told that 1.00 atm = 760. mmHg, and we want to know how many atm are equal to 854 mmHg. As we did with question 4, we set up a proportion: 1/760. = x/854, and solve for x. 854 mmHg × 1.00 atm ÷ 760. mmHg = 1.12 atm.

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