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ch4aika [34]
3 years ago
5

Consider the following balanced redox reaction (do not include state of matter in your answers):2CrO2−(aq) + 2H2O(l) + 6ClO−(aq)

→2CrO42−(aq) + 3Cl2(g) + 4OH−(aq)(a) Which species is being oxidized? (b) Which species is being reduced? (c) Which species is the oxidizing agent? (d) Which species is the reducing agent? (e) From which species to which does electron transfer occur?
Chemistry
2 answers:
Valentin [98]3 years ago
6 0

Answer:

E i think

Explanation:

zhenek [66]3 years ago
5 0

Answer:

Explanation:

Hello,

At first, it is necessary to establish the oxidation states for all the species involved in the reaction:

2(Cr^{+3}O_2^{-2})^{-}(aq) + 2H_2^{+1}O^{-2}(l) + 6(Cl^{+1}O^{-2})^{-1}(aq) -->2(Cr^{+6}O_4^{-2})^{-2}(aq) + 3Cl_2^0(g) + 4(O^{-2}H^{+1})^-(aq)

In such a way, we answer to the questions as follows:

(a) Oxidized species is the one that has an increase in its oxidation state, this is chromium which passes from +3 to +6.

(b) Reduced species is the one that has a decrease in its oxidation state, this is chlorine which passes from +1 to 0.

(c) Oxidizing agent is the same reduced species, this is chlorine as it removes electrons.

(d) Reducing agent is the same oxidized species, this is chromium as it gains electrons.

(e) Electron transfer occur from chlorine to chromium as chlorine removes electrons and chromium gains them.

Best regards.

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jeyben [28]

<u>Answer:</u> The amount of barium sulfate produced in the given reaction is 0.667 grams.

<u>Explanation:</u>

To calculate the number of moles from molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Molarity of barium chloride = 0.113 M

Volume of barium chloride = 25.34 mL = 0.02534 L   (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

0.113mol/L=\frac{\text{Moles of barium chloride}}{0.02534L}\\\\\text{Moles of barium chloride}=0.00286mol

For the given chemical reaction:

BaCl_2(aq.)+Na_2SO_4(aq.)\rightarrow BaSO_4(s)+2NaCl(aq.)

By Stoichiometry of the reaction:

1 mole of barium chloride is producing 1 mole of barium sulfate.

So, 0.00286 moles of barium chloride will produce = \frac{1}{1}\times 0.00286mol=0.00286mol of barium sulfate.

Now, to calculate the mass of barium sulfate, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Molar mass of barium sulfate = 233.38 g/mol

Moles of barium sulfate = 0.00286 moles

Putting values in above equation, we get:

0.00286mol=\frac{\text{Mass of barium sulfate}}{233.38g/mol}\\\\\text{Mass of barium sulfate}=0.667g

Hence, the amount of barium sulfate produced in the given reaction is 0.667 grams

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