Exercise 6L Question 7

[Vivian Leung 1C]

Hi,

For this question (Exercise 6L Question 7), I am confused as to tell which substances are the reactants and which substances are the products when writing the 1/2 reactions in part b and part c.

Also, a specific question for:

Part c: Where did the KOH (aq) come from (see galvanic cell).

[Sophia Hu 1A]

For parts b and c, determining the reactants and products for each half reaction is based on the overall reaction.

For instance, for part b:
We find the half reactions for H+(aq) and OH-(aq)
O2(g) + 4H+(aq) + 4 e- --> 2H2O(l)
4OH-(aq) --> O2(g) + 2H2O(l) + 4e- (I flipped this half-reaction because this will get us the overall reaction we want)
overall: 4H+ (aq) + 4OH-(aq) --> 4H2O(l) which can simplify to H+ (aq) + OH-(aq) --> H2O(l) like the reaction given in the problem
We can also verify this is the correct orientation by calculating the Enaught_cell and it does end up giving + 0.83V.

A similar process can be followed for step 3 where we find the half-reactions that contain Cd(s) + 2 Ni(OH)3

I think what might be confusing is finding the half-reaction for the oxidation reaction since Appendix 2 gives all the reduction half-reactions. For part c determining this is easier. For instance Cd goes from Cd(s) with an oxidation number of 0 to Cd(OH)2 with an oxidation number of + 2. Therefore, Cd was oxidized and lost 2 electrons. Therefore, we need to find the reduction half-reaction that includes Cd(OH)2 and flip the reaction to make it an oxidation half-reaction which would result in Cd(s) being a reactant and Cd(OH)2 being a product which matches up with the overall equation. We can then easily find the reduction of Ni(OH)3 by finding its reduction half-reaction and leaving it the way that it is as Ni(OH)3 is the reactant in the half-reaction and Ni(OH)2 is the product. We should then verify that this chosen reduction and oxidation half-reactions will give an overall positive Enaught_cell and it does.

I hope this helps!

[Ellison Gonzales 1H]

For parts b and c, determining the reactants and products for each half reaction is based on the overall reaction.

For instance, for part b:
We find the half reactions for H+(aq) and OH-(aq)
O2(g) + 4H+(aq) + 4 e- --> 2H2O(l)
4OH-(aq) --> O2(g) + 2H2O(l) + 4e- (I flipped this half-reaction because this will get us the overall reaction we want)
overall: 4H+ (aq) + 4OH-(aq) --> 4H2O(l) which can simplify to H+ (aq) + OH-(aq) --> H2O(l) like the reaction given in the problem
We can also verify this is the correct orientation by calculating the Enaught_cell and it does end up giving + 0.83V.

A similar process can be followed for step 3 where we find the half-reactions that contain Cd(s) + 2 Ni(OH)3

I think what might be confusing is finding the half-reaction for the oxidation reaction since Appendix 2 gives all the reduction half-reactions. For part c determining this is easier. For instance Cd goes from Cd(s) with an oxidation number of 0 to Cd(OH)2 with an oxidation number of + 2. Therefore, Cd was oxidized and lost 2 electrons. Therefore, we need to find the reduction half-reaction that includes Cd(OH)2 and flip the reaction to make it an oxidation half-reaction which would result in Cd(s) being a reactant and Cd(OH)2 being a product which matches up with the overall equation. We can then easily find the reduction of Ni(OH)3 by finding its reduction half-reaction and leaving it the way that it is as Ni(OH)3 is the reactant in the half-reaction and Ni(OH)2 is the product. We should then verify that this chosen reduction and oxidation half-reactions will give an overall positive Enaught_cell and it does.

I hope this helps!

This was super helpful, Sophia! Thank you so much!

[Ellison Gonzales 1H]

I was wondering if anyone knew where the KOH came from? I’m not sure if it’s a typo or if it has a purpose. I was also wondering where the solo Ni(s) came from in the cell diagram? Is it because it’s a solid metal and so it can be used as the electrode?

[Ellison Gonzales 1H]

This question also specifies that it is a galvanic cell, so I was wondering if we are to treat this problem any differently or if there are any differences in the process compared to the other types of cells? Thanks!

[Vivian Leung 1C]

For parts b and c, determining the reactants and products for each half reaction is based on the overall reaction.

For instance, for part b:
We find the half reactions for H+(aq) and OH-(aq)
O2(g) + 4H+(aq) + 4 e- --> 2H2O(l)
4OH-(aq) --> O2(g) + 2H2O(l) + 4e- (I flipped this half-reaction because this will get us the overall reaction we want)
overall: 4H+ (aq) + 4OH-(aq) --> 4H2O(l) which can simplify to H+ (aq) + OH-(aq) --> H2O(l) like the reaction given in the problem
We can also verify this is the correct orientation by calculating the Enaught_cell and it does end up giving + 0.83V.

A similar process can be followed for step 3 where we find the half-reactions that contain Cd(s) + 2 Ni(OH)3

I think what might be confusing is finding the half-reaction for the oxidation reaction since Appendix 2 gives all the reduction half-reactions. For part c determining this is easier. For instance Cd goes from Cd(s) with an oxidation number of 0 to Cd(OH)2 with an oxidation number of + 2. Therefore, Cd was oxidized and lost 2 electrons. Therefore, we need to find the reduction half-reaction that includes Cd(OH)2 and flip the reaction to make it an oxidation half-reaction which would result in Cd(s) being a reactant and Cd(OH)2 being a product which matches up with the overall equation. We can then easily find the reduction of Ni(OH)3 by finding its reduction half-reaction and leaving it the way that it is as Ni(OH)3 is the reactant in the half-reaction and Ni(OH)2 is the product. We should then verify that this chosen reduction and oxidation half-reactions will give an overall positive Enaught_cell and it does.

I hope this helps!

Thank you so much! This helps a lot!