Hi I am Mary and I need help solving these questions

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

A 32.0-g sample of HF is dissolved in water to give 2.0 × 102 mL of solution. The concentration of the solution is:

guajiro [1.7K]

The concentration of the solution is 8 M

calculation

Concentration = moles/volume in liters

step 1: find moles of HF

moles of HF =mass/molar mass

molar mass of HF = 1+ 19 )= 20 g/mol

moles is therefore = 32.0 g/ 20 g/mol= 1.6 moles

Step 2: convert ml to L

volume in liters = 2.0 x 10^2 / 1000 =0.2 l

step 3: find the concentration

concentration = 1.6 mol / 0.2 l = 8 M

4 0

3 years ago

How would you write a balanced equation for the combustion of zinc sulfide?

lara31 [8.8K]

Answer:

The solution is: ZnS + O^2 => ZnSO

6 0

4 years ago

True or False: The only planet that has gravity is the Earth.

taurus [48]

Answer:

False.

Explanation:

all planets have gravity because if they didnt, the gasses and whatever makes up the planet wouldnt be pulled down to the core of the planet, and it would just be floating around in space.

6 0

3 years ago

How many grams of chlorine gas can be produced from the decomposition of 1.46x1023 formula units of AuCl3?

icang [17]

Answer:

c

Explanation:

4 0

3 years ago

Hi I am Mary and I need help solving these questions ​

Hi I am Mary and I need help solving these questions