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shepuryov [24]
3 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]3 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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3 years ago
A student titrates a 20.00 mL sample of an aqueous borax solution with 1.03 M H2SO4. If 2.07 mL of acid are needed to reach the
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Answer:

The concentration of the borax solution is 0.1066 M

Explanation:

Step 1: Dtaa given

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Volume of the H2SO4 = 2.07 mL = 0.00207 L

Step 2: The balanced equation

Na2B4O7*10H2O(borax) + H2SO4 ⇆ Na2SO4 + 4 H3BO3 + 5 H2O

Step 3: Calculate molarity of borax solution

b*Ca*Va = a * Cb*Vb

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⇒with Ca = the concentration of borax = TO BE DETERMINED

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⇒with a = the coefficient of borax = 1

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The concentration of the borax solution is 0.1066 M

6 0
3 years ago
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