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1 year ago

In a voltaic (galvanic) cell, oxidation occurs at the _ and is where _ in the salt bridge moves toward.

1 answer:

4 0

In a voltaic (galvanic) cell, oxidation occurs at the anode and is where anions in the salt bridge moves toward.

<h3>What is Galvanic Cell ?</h3>

Galvanic Cell or Voltaic Cell is an electrochemical cell that converts the energy of spontaneous redox reactions into electrical energy. In galvanic cell oxidation occurs at the anode and reduction occurs at the cathode. The anode is positive and cathode is negative, anode attracts anions from solution in an electrolytic cell.

Thus from the above conclusion we can say that In a voltaic (galvanic) cell, oxidation occurs at the anode and is where anions in the salt bridge moves toward.

Learn more about the Galvanic Cell here: brainly.com/question/15096829

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A 3.452 g sample containing an unknown amount of a Ce(IV) salt is dissolved in 250.0-mL of 1 M H2SO4. A 25.00 mL aliquot is anal

SOVA2 [1]

Answer:

1,812 wt%

Explanation:

The reactions for this titration are:

2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻

I₃⁻ + 2S₂O₃⁻ → 3I⁻ + S₄O₆²⁻

The moles in the end point of S₂O₃⁻ are:

0,01302L×0,03428M Na₂S₂O₃ = 4,463x10⁻⁴ moles of S₂O₃⁻. As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:

4,463x10⁻⁴ moles of S₂O₃⁻× $\frac{1molI_{3}^-}{2molS_{2}O_{3}^-}$ = 2,2315x10⁻⁴ moles of I₃⁻

As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:

2,2315x10⁻⁴ moles of I₃⁻× $\frac{2molCe^{4+}}{1molI_{3}^-}$ = 4,463x10⁻⁴ moles of Ce(IV). These moles are:

4,463x10⁻⁴ moles of Ce(IV)× $\frac{140,116g}{1mol}$ = 0,0625 g of Ce(IV)

As the sample has a 3,452g, the weight percent is:

0,0625g of Ce(IV) / 3,452g × 100 = 1,812 wt%

I hope it helps!

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3 years ago

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Hey there!

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3 years ago

In a voltaic (galvanic) cell, oxidation occurs at the _____ and is where _____ in the salt bridge moves toward.

In a voltaic (galvanic) cell, oxidation occurs at the _ and is where _ in the salt bridge moves toward.