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Nikolay [14]
3 years ago
5

rank these species by their ability to act as an oxidizing agent from best oxidizing agent to best reducing agent. Ag+ cr3+ sn2+

Na+
Chemistry
2 answers:
jolli1 [7]3 years ago
7 0

Answer: Order of the species in their decreasing oxidizing agent ability is Ag^+>Sn^{2+}>Cr^{3+}>Na^+

Explanation: Oxidizing agent is defined as the agent which helps the other element to get oxidized and it itself gets reduced. These agents should have a positive value of reduction potentials. More is the positive value, best is the oxidizing agent.

Reducing agents are defined as the agents which helps the other element to get reduced and itself gets oxidized. These agents have a negative value of reduction potentials. The more the negative value, best is the reducing agent.

Reduction potentials of the following elements are:

Ag^+:0.80V\\Cr^{3+}:-0.74V\\Sn^{2+}:-0.14V\\Na^+:-2.71V

From the given reduction potentials, the best oxidizing agent is Ag^+ and the best reducing agent is Na^+

Order in the decreasing oxidizing agent ability of the species is Ag^+>Sn^{2+}>Cr^{3+}>Na^+

swat323 years ago
3 0

From best oxidizing agent to best reducing agent : <u>Ag⁺, Sn²⁺, Cr³⁺, Na⁺</u>

<h3>Further explanation</h3>

The oxidation-reduction reaction or abbreviated as Redox is a chemical reaction in which there is a change in oxidation number

3 basic theories explain this Redox concept:

  • 1. Binding / release of oxygen

The oxidation reaction is the binding of a substance with oxygen. (O₂)

For example:

2SO₂ + O₂ ----> 2SO₃

The reduction reaction is the release of oxygen from a substance.

For example:

2CuO → 2Cu + O₂

  • 2. Electron release / binding reaction

Oxidation is an electron release event

Example:

2F ---> 2Fe³⁺ + 6e⁻

The reduction is an electron capture event

Example:

3O₂ + 6e⁻ ---> 3O²⁻

  • 3. The reaction of addition / reduction of oxidation number

Oxidation is an increase  in oxidation number, while reduction is a decrease in oxidation number.

In the redox reaction, it is also known

Reducing agents are substances that experience oxidation

Oxidizing agents are substances that experience reduction

The metal activity series is expressed in voltaic series

Li-K-Ba-Ca-Na-Mg-Al-Mn- (H2O) -Zn-Cr-Fe-Cd-Co-Ni-Sn-Pb- (H) -Cu-Hg-Ag-Pt-Au

The more to the left, the metal is more reactive (easily release electrons) and the stronger reducing agent

The more to the right, the metal is less reactive (harder to release electrons) and the stronger oxidizing  agent

So that the metal located on the left can push the metal on the right in the redox reaction

From the ion ion below:

 Ag⁺ ;Cr³⁺ ;Sn²⁺; Na⁺

If we sort according to the voltaic series: (from left to right)

Na, Cr, Sn, Ag

Then:

Na, the most reactive, strong reducing agent

Ag, less reactive, strong oxidizing agents

So that the order from best oxidizing agent to best reducing agent.

Ag⁺, Sn²⁺, Cr³⁺, Na⁺

<h3>Learn more </h3>

an oxidation-reduction reaction

ainly.com/question/2973661

a reducing agent during a redox reaction

brainly.com/question/2890416

loses electrons in a chemical reaction

brainly.com/question/2278247

Keywords: reduction, oxidation

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3 years ago
A buffer solution is made that is 0.347 M in H2C2O4 and 0.347 M KHC2O4.
irga5000 [103]

Answer:

1. pH = 1.23.

2. H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)

Explanation:

Hello!

1. In this case, for the ionization of H2C2O4, we can write:

H_2C_2O_4\rightleftharpoons HC_2O_4^-+H^+

It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:

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pKa=-log(Ka)=-log(5.90x10^{-2})=1.23

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pH=1.23+log(\frac{0.347M}{0.347M} )\\\\pH=1.23+0\\\\pH=1.23

2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:

n_{acid}=0.347mol/L*1.00L=0.347mol\\\\n_{acid}^{remaining}=0.347mol-0.070mol=0.277mol

It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:

H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)

Which is also shown in net ionic notation.

Best regards!

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3 years ago
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