All categories

3 years ago

What element is reduced in the following reaction?

1 answer:

riadik2000 [5.3K]3 years ago

6 0

the oxidation states of the elements before and after the reaction is;
Pb oxidation state changes from 0 to +2
SO₄²⁻ ion there's no change in the oxidation state during the reaction
Au oxidation state changes from +3 to 0
reduction reactions are when there's a decrease in the oxidation state of the species
oxidation reactions are when theres an increase in the oxidation state of the species
the element where there's a decrease in oxidation state is Au.
Therefore Au gets reduced.
answer is B) Au

You might be interested in

Plzzzzzzzzzzzzzz answer fast

vfiekz [6]

Answer:

Welding and computer chips

8 0

3 years ago

Read 2 more answers

Consider the reaction

SOVA2 [1]

Answer :

(a) The average rate will be:

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

(b) The average rate will be:

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

Explanation :

The general rate of reaction is,

$aA+bB\rightarrow cC+dD$

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

$\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}$

$\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}$

$\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}$

$\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}$

$Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}$

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

$5Br^-(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 3Br_2(aq)+3H_2O(l)$

The expression for rate of reaction :

$\text{Rate of disappearance of }Br^-=-\frac{1}{5}\frac{d[Br^-]}{dt}$

$\text{Rate of disappearance of }BrO_3^-=-\frac{d[BrO_3^-]}{dt}$

$\text{Rate of disappearance of }H^+=-\frac{1}{6}\frac{d[H^+]}{dt}$

$\text{Rate of formation of }Br_2=+\frac{1}{3}\frac{d[Br_2]}{dt}$

$\text{Rate of formation of }H_2O=+\frac{1}{3}\frac{d[H_2O]}{dt}$

Thus, the rate of reaction will be:

$\text{Rate of reaction}=-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{d[BrO_3^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}=+\frac{1}{3}\frac{d[H_2O]}{dt}$

Part (a) :

Given:

$\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s$

As,

$-\frac{1}{5}\frac{d[Br^-]}{dt}=+\frac{1}{3}\frac{d[Br_2]}{dt}$

and,

$\frac{d[Br_2]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}$

$\frac{d[Br_2]}{dt}=\frac{3}{5}\times 1.56\times 10^{-4}M/s$

$\frac{d[Br_2]}{dt}=9.36\times 10^{-5}M/s$

Part (b) :

Given:

$\frac{1}{5}\frac{d[Br^-]}{dt}=1.56\times 10^{-4}M/s$

As,

$-\frac{1}{5}\frac{d[Br^-]}{dt}=-\frac{1}{6}\frac{d[H^+]}{dt}$

and,

$-\frac{1}{6}\frac{d[H^+]}{dt}=\frac{3}{5}\frac{d[Br^-]}{dt}$

$\frac{d[H^+]}{dt}=\frac{6}{5}\times 1.56\times 10^{-4}M/s$

$\frac{d[H^+]}{dt}=1.87\times 10^{-4}M/s$

5 0

3 years ago

In a 1.0× 10–6 M solution of Ba(OH)2(aq) at 25 °C, identify the relative molar amounts of these species.

Marysya12 [62]

Thank you for posting your question here. Below is the solution:

HNO3 --> H+ + NO3-
HNO3 = strong acid so 100% dissociation 
** one doesn't need to find the molarity of water since it is the solvent 

0M HNO3 
1x10^-6M H3O+ 
1x10^-6M NO3- 
1x10^-8M OH-.....the Kw = 1x10^-14 = [H+][OH-] 
you have 1x10^-6M H+ so, 1x10^-14 / 1x10^-6 = 1x10^-8M OH- 

1x10^-6 Ba(OH)2 = strong base, 100% dissociation 
1x10^-6M Ba2+ 
2x10^-6M OH- since there are 2 OH- / 1 Ba2+ 
0M Ba(OH)2 
5x10^-9M H3O+

4 0

3 years ago

State the oxidation number of S in <img src="https://tex.z-dn.net/?f=H_%7B2%7DSO_%7B3%7D" id="TexFormula1" title="H_{2}SO_{3

Nezavi [6.7K]

Taking into account the definition of oxidation number, the oxidation numbers of S in H₂SO₃ is 4.

<h3>Definition of oxidation number</h3>

The oxidation number is the charge that an atom has; is an integer that represents the number of electrons an atom puts into play when it forms a given compound.

In other words, the oxidation number of an element is a value that indicates the number of electrons that element gains or loses when it combines with another.

<h3>Oxidation number determination</h3>

To determine the oxidation state of different elements it is necessary to know that:

The oxidation number of hydrogen in a compound is +1, except in metal hydrides, where is –1.
The oxidation number of oxygen in a compound is –2, except in peroxides, where it is –1.

On the other side, the sum of the oxidation numbers of the existing elements in a chemical formula must add up to zero.

Then, considering the oxidation numbers of each element, multiplying it by the number of existing elements in the chemical formula and adding it and equaling it to zero, the value of the missing oxidation number can be obtained.

<h3>Oxidation numbers of S</h3>

In this case, the oxidation numbers of S in H₂SO₃ is calculated as:

2× (+1) + oxidation numbers of S + 3×(-2)= 0

2 + oxidation numbers of S -6= 0

oxidation numbers of S -4= 0

oxidation numbers of S= 4

Finally, the oxidation numbers of S in H₂SO₃ is 4.

Learn more about the oxidation number:

brainly.com/question/8990767

brainly.com/question/6498977

5 0

2 years ago