Which of the following energy sources obtains its energy from gravitational potential energy?

Barium (Ba) has two valence electrons, and these electrons are located in the 6s subshell. Without using the periodic table, in

Katena32 [7]

Barium is located in Group 2A, Period 6

7 0

3 years ago

Read 2 more answers

Consider the reaction of metallic copper with iron(!!) to give copper(ll) and ironin 0.77V Fe* (aq) + e-Fe (aq) Cup (aq) + 2e --

frosja888 [35]

Answer :

(a) The anode and cathode will be $E^o_{(Cu^{2+}/Cu)}$

and $E^o_{(Fe^{3+}/Fe^{2+})}$ respectively.

(b) The emf of cell potential is 1.022 V

Explanation :

(a) The standard reduction potentials for iron and copper are:

$E^o_{(Fe^{3+}/Fe^{2+})}=0.77V\\E^o_{(Cu^{2+}/Cu)}=0.34V$

In the voltaic cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.

From the standard reduction potentials we conclude that, the substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction.

So, iron will undergo reduction reaction will get reduced. Copper will undergo oxidation reaction and will get oxidized.

The given cell reactions are:

Oxidation half reaction (anode): $Cu\rightarrow Cu^{2+}+2e^-$

Reduction half reaction (cathode): $Fe^{3+}+1e^-\rightarrow Fe^{2+}$

Thus, the anode and cathode will be $E^o_{(Cu^{2+}/Cu)}$

and $E^o_{(Fe^{3+}/Fe^{2+})}$ respectively.

(b) Now we have to calculate the potential of a cell.

Oxidation half reaction (anode): $Cu\rightarrow Cu^{2+}+2e^-$

Reduction half reaction (cathode): $Fe^{3+}+1e^-\rightarrow Fe^{2+}$

In order to balance that electrons, we will multiple the reduction reaction by 2, we get:

Oxidation half reaction (anode): $Cu\rightarrow Cu^{2+}+2e^-$

Reduction half reaction (cathode): $2Fe^{3+}+2e^-\rightarrow 2Fe^{2+}$

The overall cell reaction will be,

$2Fe^{3+}+Cu\rightarrow Cu^{2+}+2Fe^{2+}$

$E^o_{[Fe^{3+}/Fe^{2+}]}=2\times 0.77V=1.54V$

$E^o_{[Cu^{2+}/Cu]}=0.34V$

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$E^o=E^o_{[Fe^{3+}/Fe^{2+}]}-E^o_{[Cu^{2+}/Cu]}$

$E^o=1.54V-(0.34V)=1.20V$

Now we have to calculate the cell potential.

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Fe^{2+}]^2[Cu^{2+}]}{[Fe^{3+}]^2}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = emf of the cell = ?

Now put all the given values in the above equation, we get:

$E_{cell}=1.20-\frac{0.0592}{2}\log \frac{(0.20)^2(0.25)}{(0.0001)^2}$

$E_{cell}=1.022V$

Therefore, the emf of cell potential is 1.022 V

5 0

3 years ago

What is the electron shielding effect? What is the trend for it? How and why does it happen?

andriy [413]

Answer:

See explanation

Explanation:

In an atom, the inner electrons may shield the outer electrons from the attractive force of the nucleus. We, refer to this phenomenon as the <u><em>shielding effect</em></u>, It is defined as a decrease in the magnitude of attraction between an electron and the nucleus of an atom having more than one electron shell (energy level).

Shielding effect increases down the group due to addition of more shells but decreases across the period due to the increase in the size of the nuclear charge.

As the magnitude of shielding increases down the group, ionization of electrons becomes easier and the first ionization energies of elements decreases as we move down the group. Since shielding effect decreases across the period, the first ionization energies of elements increases across the period.

6 0

2 years ago

Perform the following operation and express the answer in scienfific notation 7.5x1-^9 - 2.5 x 10^8

asambeis [7]

34^67 42’063 1220 thank you

6 0

1 year ago

What type of bond connects the fatty acid?

scoray [572]

The type of bond connects the fatty acid will be " an ester bond".

Fatty acids could be joined to each of the three carbons of the glycerol molecule in a fat molecule by an ester bond that passes through into the oxygen atom. Three molecules are emitted during the establishment of the ester bond. Fats were also known as triacylglycerols as well as triglycerides because they are made consisting of three fatty acids and glycerol.

The three major nutrients in a diet may be carbohydrates, proteins, and fats. According to how many hydrogen bonds they contain, fats are classified as either saturated as well as unsaturated.

The type of bond connects the fatty acid will be " an ester bond".

Hence, the correct answer will be option (c)

To know more about fatty acid

brainly.com/question/13528495

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7 0

1 year ago