Electrode potentials and Bond energy

In summary, both b and c are incorrect. Adding water has no effect on the emf of a voltaic cell. Increasing the temperature does not have an effect on the cell emf.
  • #1
24
0

Homework Statement


1st question:A voltaic cell is made up of mg2+/mg half-cell and fe3+/fe2+ half-cell. Which of the following statement are correct?
a.addition of water to the fe3+/fe2+ half-cell has no effect on cell emf.
b.increasing the temperature has no effect on the cell emf.
c.Addition of the aqueous naoh to mg2+/mg half-cell decreases the cell emf.

2nd question:Which of the enthalpy change of the following reactions can only be obtained by application of Hess' Law.
a.The hydration of anhydrous copper sulphate to form crystal of CuSo4.5H2O
b.The formation of methane from its elements
c.The combustion of glucose c6h12o6.


Homework Equations





The Attempt at a Solution


1st question:I understand a is correct as water decreases concentration of both to the same extent.But i have no idea why b and c are wrong.

2nd question:The answer is both a and b are correct.But i know that enthalpy change of formation of methane can be obtained by calculating bond energy,so formation of methane can be obtained by other methods and not just Hess law!?
 
Physics news on Phys.org
  • #2
1. The EMF of cell depends on the reaction quotient of the redox reaction, the overall cell reaction. (Read: Chemical Equilibrium).

Now, changing temperature definitely changes the reaction quotient of the reaction for the same concentrations.

Please first determine which would be the anode and which would be the cathode (Look for electrochemical series), write down the half-cell reaction, try adding NaOH to the Magnesium half cell, and figure out what happened. Post here what you thought.

2. You may find out the enthalpy by individual bonds, but that's a tight situation, but gives answer in most the cases. What I means is you would like to calculate the energy required to break C-C bonds from graphite, H-H bonds from H2 gas, calculate the energy released in forming C-H bonds in methane. That's OK, until you know how many or which bond (Single or double, Partial, etc.) to break.

Hess Law reduces the load to only knowledge required that methane has 4 C-H bonds.
 
  • #3
AGNuke said:
1. The EMF of cell depends on the reaction quotient of the redox reaction, the overall cell reaction. (Read: Chemical Equilibrium).

Now, changing temperature definitely changes the reaction quotient of the reaction for the same concentrations.

Please first determine which would be the anode and which would be the cathode (Look for electrochemical series), write down the half-cell reaction, try adding NaOH to the Magnesium half cell, and figure out what happened. Post here what you thought.

2. You may find out the enthalpy by individual bonds, but that's a tight situation, but gives answer in most the cases. What I means is you would like to calculate the energy required to break C-C bonds from graphite, H-H bonds from H2 gas, calculate the energy released in forming C-H bonds in methane. That's OK, until you know how many or which bond (Single or double, Partial, etc.) to break.

Hess Law reduces the load to only knowledge required that methane has 4 C-H bonds.
For question 1: Mg2+ +2electrons -> Mg E“ =-2.38v
If we were to add Naoh, E“ becomes even more negative? As magnesium hydroxide is formed and mg2+ has a even lower tendency to be reduced. Thus Ecell increases and not decreases?

Fro question 2: The idea is that i am able to calculate enthalpy change by taking total bond formed minus total bond broken.But the answer is asking 'ONLY' be obtained by application of Hess law,which in this case (B) can be obtained by both hess law and calculating Bond energy?
 
  • #4
chewchun said:
Mg2+ +2electrons -> Mg E“ =-2.38v

Actually, its the other way round.[tex]Anode:M\!g\rightarrow M\! g^{2+}+2e^-; E^{\circ}=-2.36\; V[/tex][tex]Cathode:F\!e^{3+}+e^-\rightarrow F\!e^{2+};E^{\circ}=0.77\;V[/tex][tex]E^{\circ}_{cell}=E^{\circ}_{cathode}-E^{\circ}_{anode}=3.15\;V[/tex]

Now, adding aqueous NaOH, things changes. Now, [tex]Anode:N\!a\rightarrow N\! a^{+}+e^-; E^{\circ}=-2.71\; V[/tex][tex]E^{\circ}_{cell}=E^{\circ}_{cathode}-E^{\circ}_{anode}=3.48\;V[/tex]

See, EMF has increased. The problem arose from choosing the wrong electrodes. See to it that the cell type is Voltaic, which produces electricity, hence have positive EMF. Thus couple more attuned to oxidation is negative terminal and reduction one is positive terminal. And Oxidation always takes place at anode, irrespective of cell type.
 
  • #5
AGNuke said:
Actually, its the other way round.[tex]Anode:M\!g\rightarrow M\! g^{2+}+2e^-; E^{\circ}=-2.36\; V[/tex][tex]Cathode:F\!e^{3+}+e^-\rightarrow F\!e^{2+};E^{\circ}=0.77\;V[/tex][tex]E^{\circ}_{cell}=E^{\circ}_{cathode}-E^{\circ}_{anode}=3.15\;V[/tex]

Now, adding aqueous NaOH, things changes. Now, [tex]Anode:N\!a\rightarrow N\! a^{+}+e^-; E^{\circ}=-2.71\; V[/tex][tex]E^{\circ}_{cell}=E^{\circ}_{cathode}-E^{\circ}_{anode}=3.48\;V[/tex]

See, EMF has increased. The problem arose from choosing the wrong electrodes. See to it that the cell type is Voltaic, which produces electricity, hence have positive EMF. Thus couple more attuned to oxidation is negative terminal and reduction one is positive terminal. And Oxidation always takes place at anode, irrespective of cell type.
I don't think so...sodium should be oxidised,but naoh contains sodium ion...
 
  • #6
Sorry I oxidised Na, even though it wasn't present. Sodium Hydroxide was only added to provide Hydroxide ions.

E will increase, hope I am right this time.[tex]M\!g^{2+}+2O\!H^-\rightarrow M\!g(O\!H)_2;\;K_{sp}=1.5\times 10^{-11}[/tex]
Magnesium Hydroxide formed is sparingly soluble in water and hence the concentration of Mg2+ will decrease drastically. Now, for any non-standard cell,[tex]E_{cell}=E^{\circ}_{cell}-\frac{0.0591}{n}\log_{10}Q[/tex]
Write the Reaction Quotient of the overall cell reaction. See, reducing Magnesium ion concentration will decrease Q, If you decrease Q, you subtract less from 3.15 V, effectively, Ecell increases.

Alternatively, you can find out the standard potential of [itex]E_{OH^-|Mg(OH)_2|Mg}[/itex], which is Metal-insoluble salt electrode.[tex]E_{OH^-|Mg(OH)_2|Mg}=E^{\circ}_{Mg^{2+}|Mg}+ \frac{0.0591}{2}\log_{10}K_{sp}[/tex]
It is even less than standard electrode, which is good.
 
Last edited:
  • #7
agnuke said:
sorry i oxidised na, even though it wasn't present. Sodium hydroxide was only added to provide hydroxide ions.

E will increase, hope i am right this time.[tex]m\!g^{2+}+2o\!h^-\rightarrow m\!g(o\!h)_2;\;k_{sp}=1.5\times 10^{-11}[/tex]
magnesium hydroxide formed is sparingly soluble in water and hence the concentration of mg2+ will decrease drastically. Now, for any non-standard cell,[tex]e_{cell}=e^{\circ}_{cell}-\frac{0.0591}{n}\log_{10}q[/tex]
write the reaction quotient of the overall cell reaction. See, reducing magnesium ion concentration will decrease q, if you decrease q, you subtract less from 3.15 v, effectively, ecell increases.

Alternatively, you can find out the standard potential of [itex]e_{oh^-|mg(oh)_2|mg}[/itex], which is metal-insoluble salt electrode.[tex]e_{oh^-|mg(oh)_2|mg}=e^{\circ}_{mg^{2+}|mg}+ \frac{0.0591}{2}\log_{10}k_{sp}[/tex]
it is even less than standard electrode, which is good.
thanks a lot,totally got it!
 

1. What is an electrode potential?

An electrode potential is a measure of the tendency of an electrode to lose or gain electrons when it is in contact with an electrolyte solution. It is a measure of the strength of the electrochemical reaction at the electrode.

2. How is electrode potential measured?

Electrode potential is measured using a standard hydrogen electrode (SHE) as a reference. The SHE has a defined potential of 0 volts and is used to compare the potential of other electrodes.

3. What is bond energy?

Bond energy is the amount of energy required to break a chemical bond between two atoms. It is a measure of the strength of the bond and is typically expressed in units of kilojoules per mole (kJ/mol).

4. How does bond energy affect electrode potential?

The bond energy between atoms in an electrode affects its electrode potential. A stronger bond will result in a more negative electrode potential, indicating a greater tendency for the electrode to gain electrons. Conversely, a weaker bond will result in a more positive electrode potential, indicating a greater tendency for the electrode to lose electrons.

5. What is the relationship between electrode potential and bond energy?

Electrode potential and bond energy are directly related. As the bond energy increases, the electrode potential becomes more negative. This is because a stronger bond requires more energy to be broken, resulting in a greater tendency for the electrode to gain electrons. Conversely, as the bond energy decreases, the electrode potential becomes more positive.

Suggested for: Electrode potentials and Bond energy

Replies
6
Views
1K
Replies
4
Views
1K
Replies
7
Views
3K
Replies
2
Views
1K
Replies
3
Views
1K
Replies
7
Views
2K
Replies
3
Views
2K
Back
Top