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Vlad1618 [11]
3 years ago
8

Here is a more complex redox reaction involving the dichromate ion in acidic solution: 3N O 2 − + 8H + + C r 2 O 7 2− → 3N O 3 −

+2Cr 3+ + 4 H 2 O Classify each reactant as the reducing agent, oxidizing agent, or neither.
Chemistry
1 answer:
Fantom [35]3 years ago
4 0

Answer:

NO2- is the reducing agent.

Cr2O7_2- is the oxidizing agent.

H+ is neither

Explanation:

Reduction is the gain in electron. A chemical specie that undergoes reduction is called the oxidizing agent.

Oxidation is simply the loss in electrons. A chemical specie that undergoes oxidation is called the reducing agent.

Let us look at the species.

The first specie is the NO2-. In this specie, the oxidation number of nitrogen changed from +3 to +5 in NO3-. Thus we can see that there is more loss of electron to have caused an increase in the oxidation number positively. This shows an oxidation. Hence, NO2- is the reducing agent.

Let us look at the chromium. We can see that the oxidation number of chromium changed from +7 to +3.

Now we can see that it is a decrease and hence, it is a gain of electron and thus it is reduction. This means the first chromium specie is the oxidizing agent.

The hydrogen ion is simply placed there to balance the ions and hence it is neither the oxidizing nor the reducing agent.

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Enough of a monoprotic weak acid is dissolved in water to produce a 0.0172 M solution. If the pH of the resulting solution is 2.
Mumz [18]

Answer:

9.7 x 10⁻⁴

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              HA     ⇄           H⁺     +        A⁻

C(eq)   0.0174             10⁻²·³⁹         10⁻²·³⁹

                              =0.0041M     =0.0041M

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3 years ago
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5 0
3 years ago
Given that it requires 27.9 ml of 0.270 m na2s2o3(aq) to titrate a 15.0-ml sample of i3–(aq), calculate the molarity of i3–(aq)
Digiron [165]

The molar concentration of the KI_3 solution is 0.251 mol/L.

<em>Step 1</em>. Write the <em>balanced chemical equation</em>

I_3^(-) + 2S_2O_3^(2-) → 3I^(-) + S_4O_6^(2-)

<em>Step 2</em>. Calculate the <em>moles of S_2O_3^(2-)</em>

Moles of S_2O_3^(2-)

= 27.9 mL S_2O_3^(2-) ×[0.270 mmol S_2O_3^(2-)/(1 mL S_2O_3^(2-)]

= 7.533 mmol S_2O_3^(2-)

<em>Step 3</em>. Calculate the <em>moles of I_3^(-) </em>

Moles of I_3^(-) = 7.533 mmol S_2O_3^(2-)))) × [1 mmol I_3^(-)/(2 mmol S_2O_3^(2-)] = 3.766 mmol I_3^(-)

<em>Step 4</em>. Calculate the <em>molar concentration of the I_3^(-) </em>

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6 0
3 years ago
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