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cestrela7 [59]
1 year ago
5

Write the cell notation for the voltaic cell that incorporates each of the following redox reactions:(b) Cu²⁺(aq) + SO₂(g) + 2H₂

O}(l) →Cu(s) + SO₄²⁻(aq) + 4H⁺(a q)
Chemistry
1 answer:
yawa3891 [41]1 year ago
8 0

The cell notation for the given voltaic cell is

SO₂(g) | SO₄²⁻(aq) || Cu²⁺(aq) | Cu (s)

<h3>What is a voltaic cell?</h3>

A voltaic cell often called a galvanic cell, is a type of electrochemical cell.

It is a combination of two metal rods called electrodes.

A half cell is formed when both electrodes are submerged in a solution containing a particular mix of ions.

The electrolyte is the liquid in which both electrodes are submerged and acts as a salt bridge connecting the half cells.

The cell undergoes a chemical reaction known as a redox reaction.

One of the electrodes serves as an anode and undergoes oxidation, while the other electrode serves as a cathode and undergoes reduction.

To write the cell notation, write the half-reaction of the cells.

At the anode, oxidation occurs

SO_2(g) \longrightarrow SO_4^{2-}(aq) + 4H^+ (aq) + 2e^-

At cathode

Cu^{2+}(aq) + 2e^- \longrightarrow Cu(s)

In the cell notation, the anode is written on the left side, and the cathode is written on the right side. A vertical line differentiates the phases and the double vertical line represents the salt bridge.

The cell notation for the given voltaic cell is

SO₂(g) | SO₄²⁻(aq) || Cu²⁺(aq) | Cu (s)

Learn more about cell notation:

brainly.com/question/17218591

#SPJ4

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What volume (mL) of 0.135 M NaOH is required to neutralize 13.7 mL of 0.129 M HCl? a: 0.24 b: 13.1 c: 0.076 d: 6.55 e: 14.3
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Answer:

The volume (mL) of 0.135 M NaOH that is required to neutralize 13.7 mL of 0.129 M HCl is 13.1 mL (option b).

Explanation:

The reaction between an acid and a base is called neutralization, forming a salt and water.

Salt is an ionic compound made up of an anion (positively charged ion) from the base and a cation (negatively charged ion) from the acid.

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In this case:

  • V acid= 13.7 mL= 0.0137 L (being 1,000 mL= 1 L)
  • M acid= 0.129 M
  • V base= ?
  • M base= 0.135 M

Replacing:

0.0137 L* 0.129 M= V base* 0.135 M

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V base=\frac{0.0137 L*0.129 M}{0.135 M}

V base=0.0131 L = 13.1 mL

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