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kogti [31]
3 years ago
11

What gets reduced in an electrolytic cell made with nickel and copper electrodes?

Chemistry
1 answer:
jekas [21]3 years ago
8 0

Answer:

D. Ni²⁺  

Explanation:

We know at once that the answer cannot be A or C, because Ni and Cu are already in their lowest oxidation states.

The correct answer must be either B or D.

An electrolytic cell is the opposite of a galvanic cell. In the former, the reaction proceeds spontaneously. In the latter, you must force the reaction to occur.  

One strategy to solve this problem is:

  1. Look up the standard reduction potentials for the half reaction·
  2. Figure out the spontaneous direction.
  3. Write the equation in the reverse direction.

1. Standard reduction potentials

                                E°/V

Cu²⁺ + 2e⁻ ⟶ Cu; 0.3419

Ni²⁺ + 2e⁻ ⟶ Ni;  -0.257

2. Galvanic Cell

We reverse the direction of the more negative half cell and add.

                                       <u>E°/V </u>

Ni ⟶ Ni²⁺ + 2e⁻;           0.257

<u>Cu²⁺ + 2e⁻ ⟶ Cu;      </u>   0.3419

Ni + Cu²⁺ ⟶ Cu + Ni²⁺; 0.599

This is the spontaneous direction.

Cu²⁺ is reduced to Cu.

3. Electrochemical cell

                                        <u>E°/V</u>

Ni²⁺ + 2e⁻ ⟶ Ni;           -0.257

<u>Cu ⟶ Cu²⁺ + 2e⁻;        </u> <u>-0.3419</u>

Cu + Ni²⁺ ⟶ Ni + Cu²⁺; -0.599

This is the non-spontaneous direction.

Ni²⁺ is reduced to Ni in the electrolytic cell.

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A soluble iodide was dissolved in water. Then an excess of silver nitrate, AgNO3, was added to precipitate all of the io- dide i
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Answer:

m_{I^-}=1.18gI^-

Explanation:

Hello,

In this case, the reaction is given as:

I^-+AgNO_3\rightarrow AgI+NO_3^-

Thus, starting by the yielded grams of silver iodide, we obtain:

n_{I^-}=2.185gAgI*\frac{1molAgI}{234.77gAgI}*\frac{1molI}{1molAgI}=9.31x10^{-3}molI^-

Which correspond to the iodide grams in the silver iodide. In such a way, by means of the law of the conservation of mass, it is known that the grams of each atom MUST remain constant before and after the chemical reaction whereas the moles do not, therefore, the mass of iodine from the silver iodide will equal the mass of iodine present in the soluble iodide, thereby:

m_{I^-}=9.31x10^{-3}molI^-\frac{127gI^-}{1molI^-} =1.18gI^-

And the rest, correspond to the iodide's metallic cation which is unknown. Such value has sense since it is lower than the initial mass of the soluble iodide which is 1.454g, so 0.272 grams correspond to the unknown cation.

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the element cesium forms a with the charge fill in the blank 2 . the symbol for this ion is , and the name is fill in the blank
yulyashka [42]

The element cesium forms a  positive ion with the charge +1 . The symbol for this ion is Ce⁺, and the name is cesium ion. The number of electrons in this ion is 54.

<h3>What group does the element cesium belong to in the periodic table?</h3>

Cesium belong to the Group 1A of the periodic table.

Group 1A elements are known as alkali metals.

The alkali metals react with water to form alkalis. They form ions with a positive charge of +1 by the loss of an electron.

In conclusion, cesium is an alkali metals it belongs to group 1A of the periodic table. It forms a positive ion with a charge of +1 by the loss of an electron.

Learn more about alkali metals at: brainly.com/question/25317545

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