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2 years ago

What is the chemical formula for the binary compound composed of li+ and o2- ions?

Chemistry

2 answers:

Paraphin [41]2 years ago

5 0

Answer:

$Li_2O$

Explanation:

To determine the formula of a binary compound given two ions you need to have in mind that the global charges must be balanced.

In this case, that means that the positive charges of the Li atoms must equal the negative charges of de O ones. So, two Li ions neutralize one O ion:

$2 Li^+ + O^{-2} \longrightarrow Li_2O$

AnnyKZ [126]2 years ago

3 0

Should be Li₂O ! Hope this helped :')

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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

2 years ago

The solubility of co in water at 0c and 1 atm co pressure is 0. 0354 mg of co in 1 ml of water. Calculate the moalrity of aqueso

Dmitriy789 [7]

From all the calculations that were carried out, the concentration of 2atm is 0.0026 M.

<h3>What is molarity?</h3>

The term molarity is defined as the number of moles divided by the volume of the solution. Using the formula; C = kp where

C = concentration
k = constant
p = pressure

C = 0. 0354 * 10^-3 g/28 g/mol × 1000/1 L

C = 0.0013 M

k = C/p = 0.0013 M/1 atm

k = 0.0013 Matm-1

Now;

C = 0.0013 Matm-1 * 2 atm

C= 0.0026 M

Learn more about molarity: brainly.com/question/12127540

4 0

1 year ago