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fomenos
3 years ago
14

Write a chemical equation for the reaction that occurs in the following cell: Cu|Cu2+(aq)||Ag+(aq)|Ag Express your answer as a b

alanced net ionic equation. Identify all of the phases in your answer.
Chemistry
1 answer:
Alinara [238K]3 years ago
6 0

Answer:

Explanation:

The cell reaction properly written is shown below:

              Cu|Cu²⁺_{aq} || Ag⁺_{aq} | Ag

From this cell reaction, to get the net ionic equation, we have to split the reaction into their proper oxidation and reduction halves. This way, we can know that is happening at the electrodes and derive the overall net equation.

  Oxidation half:

                  Cu_{s}  ⇄ Cu²⁺_{aq} + 2e⁻

At the anode, oxidation occurs.

  Reduction half:

                  Ag⁺_{aq} + 2e⁻ ⇄ Ag_{s}

At the cathode, reduction occurs.

To derive the overall reaction, we must balance the atoms and charges:

             Cu_{s}  ⇄ Cu²⁺_{aq} + 2e⁻

              Ag⁺_{aq} + e⁻ ⇄ Ag_{s}

  we multiply the second reaction by 2 to balance up:

         2Ag⁺_{aq} + 2e⁻ ⇄ 2Ag_{s}

The net reaction equation:

Cu_{s} + 2Ag⁺_{aq} + 2e⁻⇄ Cu²⁺_{aq} + 2e⁻ + 2Ag_{s}

We then cancel out the electrons from both sides since they appear on both the reactant and product side:

  Cu_{s} + 2Ag⁺_{aq} ⇄ Cu²⁺_{aq} + 2Ag_{s}

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Answer- 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.

Given - Number of moles of Al(NO3)3 - 4 moles
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Solution - The complete reaction is - Al(NO3)3 + 3NaCl --> 3NaNO3 + AlCl3
To find the maximum amount of AlCl3 produced during the reaction, we need to find the limiting reagent.
Mole ratio Al(NO3)3 - 4/1 - 4
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Weight of AlCl3 - 3*133.34 - 400 grams
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Answer:

144 g

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