# Electrode potentials and Bond energy

chewchun

## 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.

## 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!?

Gold Member
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.

chewchun
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???

Gold Member
Mg2+ +2electrons -> Mg E“ =-2.38v

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

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

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.

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

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

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 dont think so...sodium should be oxidised,but naoh contains sodium ion....

Gold Member
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.$$M\!g^{2+}+2O\!H^-\rightarrow M\!g(O\!H)_2;\;K_{sp}=1.5\times 10^{-11}$$
Magnesium Hydroxide formed is sparingly soluble in water and hence the concentration of Mg2+ will decrease drastically. Now, for any non-standard cell,$$E_{cell}=E^{\circ}_{cell}-\frac{0.0591}{n}\log_{10}Q$$
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 $E_{OH^-|Mg(OH)_2|Mg}$, which is Metal-insoluble salt electrode.$$E_{OH^-|Mg(OH)_2|Mg}=E^{\circ}_{Mg^{2+}|Mg}+ \frac{0.0591}{2}\log_{10}K_{sp}$$
It is even less than standard electrode, which is good.

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chewchun
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.$$m\!g^{2+}+2o\!h^-\rightarrow m\!g(o\!h)_2;\;k_{sp}=1.5\times 10^{-11}$$
magnesium hydroxide formed is sparingly soluble in water and hence the concentration of mg2+ will decrease drastically. Now, for any non-standard cell,$$e_{cell}=e^{\circ}_{cell}-\frac{0.0591}{n}\log_{10}q$$
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 $e_{oh^-|mg(oh)_2|mg}$, which is metal-insoluble salt electrode.$$e_{oh^-|mg(oh)_2|mg}=e^{\circ}_{mg^{2+}|mg}+ \frac{0.0591}{2}\log_{10}k_{sp}$$
it is even less than standard electrode, which is good.
thanks a lot,totally got it!!!!