PLEASE HELP!! LAST QUESTION IM RUNNING OUT OF TIME!!

Under standard-state conditions, which of the following species is the best reducing agent? a. Ag+ b. Pb c. H2 d. Ag e. Mg2+

eimsori [14]

Answer: The correct answer is Option b.

Explanation:

Reducing agents are defined as the agents which help the other substance to get reduced and itself gets oxidized. They undergo oxidation reaction.

$X\rightarrow X^{n+}+ne^-$

For determination of reducing agents, we will look at the oxidation potentials of the substance. Oxidation potentials can be determined by reversing the standard reduction potentials.

For the given options:

Option a: $Ag^+$

This ion cannot be further oxidized because +1 is the most stable oxidation state of silver.

Option b: $Pb$

This metal can easily get oxidized to $Pb^{2+}$ ion and the standard oxidation potential for this is 0.13 V

$Pb\rightarrow Pb^{2+}+2e^-;E^o_{(Pb/Pb^{2+})}=+0.13V$

Option c: $H_2$

This metal can easily get oxidized to $H^{+}$ ion and the standard oxidation potential for this is 0.0 V

$H_2\rightarrow 2H^++2e^-;E^o_{(H_2/H^{+})}=0.0V$

Option d: $Ag$

This metal can easily get oxidized to $Ag^{+}$ ion and the standard oxidation potential for this is -0.80 V

$Ag\rightarrow Ag^{+}+e^-;E^o_{(Ag/Ag^{+})}=-0.80V$

Option e: $Mg^{2+}$

This ion cannot be further oxidized because +2 is the most stable oxidation state of magnesium.

By looking at the standard oxidation potential of the substances, the substance having highest positive $E^o$ potential will always get oxidized and will undergo oxidation reaction. Thus, considered as strong reducing agent.

From the above values, the correct answer is Option b.

8 0

3 years ago

Name the two possible products in the precipitation reaction of copper (II) chloride and sodium phosphate. Use the charges on th

satela [25.4K]

Answer:

General equation for a double-displacement reaction:

AB + CD --> AC + BD

• sodium chloride – NaCl copper sulfate – CuSO₄

NaCl + CuSO₄ --> Na₂SO₄ + CuCl₂

The products formed are sodium sulfate and copper (II) chloride.

Copper (II) chloride forms a blue colored solution.

• sodium hydroxide – NaOH copper sulfate – CuSO₄

NaOH + CuSO₄ --> Na₂SO₄ + Cu(OH)₂

The products formed are sodium sulfate and copper (II) hydroxide.

Copper (II) hydroxide forms a blue colored solution.

• sodium phosphate – Na₂HPO₂ copper sulfate – CuSO₄

Na₂HPO₄ + CuSO₄ --> Na₂SO₄ + CuHPO₄

The products formed are sodium sulfate and copper (II) hydrogen phosphate.

Copper (II) hydrogen phosphate forms a blue colored solution.

• sodium chloride – NaCl silver nitrate – AgNO₃

NaCl + AgNO₃--> AgCl + NaNO₃

The products formed are silver chloride and sodium nitrate.

Silver chloride forms a white precipitate.

• sodium hydroxide – NaOH silver nitrate – AgNO₃

NaOH + AgNO₃ --> NaNO₃ + AgOH

The products formed are silver hydroxide and sodium nitrate.

Silver hydroxide forms a white precipitate.

• sodium phosphate – Na₂HPO₄ silver nitrate – AgNO₃

Na₂HPO₄ + AgNO₃ --> NaNO₃ + Ag₂HPO₄

The products formed are sodium nitrate and silver hydrogen phosphate.

Silver hydrogen phosphate forms a colorless solution.

Explanation:

5 0

3 years ago

The periodic table

astra-53 [7]

it's Lithium or Li

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5 0

3 years ago

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Consider this equilibrium: COCl2(g) CO(g) + Cl2(g) Keq = 8.1 x 10-4. What is true for this system?

fiasKO [112]

Answer is: [COCl₂] > [CO][Cl₂]
Chemical reaction: COCl₂(g) ⇄ CO(g) + Cl₂(g); Keq = 8.1 x 10⁻⁴.
Keq = [CO] · [Cl₂] / [COCl₂]; equilibrrium constant of chemical reaction.
[CO] · [Cl₂] / [COCl₂] = 0,00081.
Equilibrium product concentration is much more less than equilibrium concentration of reactant.

6 0

3 years ago

Which ones better, old youngboy or new youngboy??

Sholpan [36]

old youngboy for sure

7 0

3 years ago

Read 2 more answers