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Hunter-Best [27]

2 years ago

4

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

) rightarrow 2CrO42- (aq) + 3Cl2(g) + 4OH-(aq) Which species is being oxidized? Which species is being reduced? Which species is the oxidizing agent?

1 answer:

SashulF [63]2 years ago

8 0

Answer:

CrO₂⁻

ClO⁻

ClO⁻

Explanation:

Let's consider the following redox reaction.

2 CrO₂⁻(aq) + 2 H₂O(l) + 6 ClO⁻(aq) → 2 CrO₄²⁻(aq) + 3 Cl₂(g) + 4 OH⁻(aq)

We can identify both half-reactions.

Reduction: 6 H₂O(l) + 6 ClO⁻(aq) + 6 e⁻ → 3 Cl₂(g) + 12 OH⁻(aq)

Oxidation: 8 OH⁻(aq) + 2 CrO₂⁻(aq) → 2 CrO₄²⁻(aq) + 4 H₂O(l) + 6 e⁻

<em>Which species is being oxidized? </em>

CrO₂⁻ is being oxidized. Its oxidation number increases from +3 to +6.

<em>Which species is being reduced?</em>

ClO⁻ is being reduced. Its oxidation number decreases from +1 to 0.

<em>Which species is the oxidizing agent?</em>

The species being reduced is the oxidizing agent, that is, ClO⁻.

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from equation 3 mol 1 mol

given x mol 0.10 mol

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3 0

2 years ago

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A bottle containing 1,665 g of sulfuric acid (H2SO4, 98.08 g/mol) was spilled in a laboratory. The emergency spill kit contained

Tamiku [17]

Answer:

A. Yes, there is more than enough sodium carbonate.

Explanation:

Hello,

In this case, based on the given reaction which is:

$H_2SO_4(aq) + Na_2CO_3(s) \rightarrow Na_2SO_4(aq) + CO_2(g) + H_2O(l)$

By stoichiometry, one computes the grams of sodium carbonate that will neutralize 1,665 g of sulfuric acid as shown below:

$1,665gH_2SO_4*\frac{1molH_2SO_4}{98.08gH_2SO_4}*\frac{1molNa_2CO_3}{1molH_2SO_4} *\frac{105.99gNa_2CO_3}{1molNa_2CO_3}*\frac{1kgNa_2CO_3}{1000gNa_2CO_3}\\m_{Na_2CO_3}=1.80gNa_2CO_3$

Thus, the available mass is 2.0 kg so 0.2 kg are in excess, therefore: A. Yes, there is more than enough sodium carbonate.

Best regards.

5 0

2 years ago

Jesse travels 3.0 meters east and then turns and travels 4.0 meters north. The trip requires 35 seconds. What is his velocity? 0

melisa1 [442]

Displacement = √(3² + 4²)
Displacement = 5 meters north east

Velocity = displacement / time
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2 years ago

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If 25 g of NH3, and 96 g of H2S react according to the following reaction, what is the

jeyben [28]

25 g of NH₃ will produce 47.8 g of (NH₄)₂S

<u>Explanation:</u>

2 NH₃ + H₂S ----> (NH₄)₂S

Molecular weight of NH₃ = 17 g/mol

Molecular weight of (NH₄)₂S = 68 g/mol

According to the balanced reaction:

2 X 17 g of NH₃ produces 68 g of (NH₄)₂S

1 g of NH₃ will produce $\frac{68}{34}$ g of (NH₄)₂S

25g of NH₃ will produce $\frac{65}{34} X 25 g$ of (NH₄)₂S

= 47.8 g of (NH₄)₂S

Therefore, 25 g of NH₃ will produce 47.8 g of (NH₄)₂S

4 0

2 years ago

What would be the composition and ph of an ideal buffer prepared from lactic acid (ch3chohco2h), where the hydrogen atom highlig

Mashutka [201]

Answer:

$P_H =2.86$

$c=1.4\times 10^{-4}$

Explanation:

first write the equilibrium equaion ,

$C_3H_6O_3$ ⇄ $C_3H_5O_3^{-} +H^{+}$

assuming degree of dissociation $\alpha$ =1/10;

and initial concentraion of $C_3H_6O_3$ =c;

At equlibrium ;

concentration of $C_3H_6O_3 = c-c\alpha$

$[C_3H_5O_3^{-} ]= c\alpha$

$[H^{+}] = c\alpha$

$K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}$

$\alpha$ is very small so $1-\alpha$ can be neglected

and equation is;

$K_a = {c\alpha \times \alpha}$

$[H^{+}] = c\alpha$ = $\frac{K_a}{\alpha}$

$P_H =- log[H^{+} ]$

$P_H =-logK_a + log\alpha$

$K_a =1.38\times10^{-4}$

$\alpha = \frac{1}{10}$

$P_H= 3.86-1$

$P_H =2.86$

composiion ;

$c=\frac{1}{\alpha} \times [H^{+}]$

$[H^{+}] =antilog(-P_H)$

$[H^{+} ] =0.0014$

$c=0.0014\times \frac{1}{10}$

$c=1.4\times 10^{-4}$

6 0

2 years ago