# Electrode potentials

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#1
A student warms a mixture of methanal HCHO and acidified potassium manganate. The student observes gas bubbles. Explain this observation in terms of electrode potentials and equilibria. Include overall equations in your answer.

1 = CO2 + 2H+ + 2e- —>HCOOH (EV =-0.11)
2 HCOOH + 2H+ + 2e- —> HCHO + H2O (EV= -0.03)
3 Ag+ + e- —-> Ag (EV= +0.80)
4 MnO4- + 8H+ + 5e- —> Mn2+ + 4H2O (EV= +1.51)

Can someone help me I have no clue what to do
Last edited by Unknownn1.x; 1 month ago
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#2
Methanoic acid HCOOH can be used in a fuel cell. The fuel HCOOH is supplied at one electrode and the oxidant (oxygen) at the other electrode.

The standard cell potential for this fuel cell is 1.34V.

The overall reaction is shown below:

HCOOH + 1/2O2 -> H2O + CO2

Deduce the half equation for the reaction at the oxygen electrode and calculate the standard electrode potential for the oxygen half cell.

Can someone help me idk where to start on this question
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#3
(Original post by Unknownn1.x)
Methanoic acid HCOOH can be used in a fuel cell. The fuel HCOOH is supplied at one electrode and the oxidant (oxygen) at the other electrode.

The standard cell potential for this fuel cell is 1.34V.

The overall reaction is shown below:

HCOOH + 1/2O2 -> H2O + CO2

Deduce the half equation for the reaction at the oxygen electrode and calculate the standard electrode potential for the oxygen half cell.

Can someone help me idk where to start on this question
0
1 month ago
#4
Well first of all, the gas/bubbles formed are definitely the CO2, so that's gotta be part of the last equation, And since we're starting by oxidizing methanal (an aldehyde) to methanoic acid using manganate ions, that's gotta be first.
So the first equation after balancing everything out is (5HCHO)+(2MnO4-)+(6H+)-> (2Mn2+)+(3H2O)+(5HCOOH) with change in voltage being (1.51+0.03=+1.54V) (as methanal is getting oxidized and manganate reduced, and reduce V- oxidize V) positive so thermodynamically feasible.
With methanoic acid obtained, next we use the silver ions to get CO2 (gas).
(HCOOH)+(2Ag+)-> (2Ag)+(CO2)+(2H+) with change in V being (0.8+0.11=+0.91V) (reduce V- oxidize V) Positive so thermodynamically feasible and the gas is CO2.
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#5
(Original post by Yun0)
Well first of all, the gas/bubbles formed are definitely the CO2, so that's gotta be part of the last equation, And since we're starting by oxidizing methanal (an aldehyde) to methanoic acid using manganate ions, that's gotta be first.
So the first equation after balancing everything out is (5HCHO)+(2MnO4-)+(6H+)-> (2Mn2+)+(3H2O)+(5HCOOH) with change in voltage being (1.51+0.03=+1.54V) (as methanal is getting oxidized and manganate reduced, and reduce V- oxidize V) positive so thermodynamically feasible.
With methanoic acid obtained, next we use the silver ions to get CO2 (gas).
(HCOOH)+(2Ag+)-> (2Ag)+(CO2)+(2H+) with change in V being (0.8+0.11=+0.91V) (reduce V- oxidize V) Positive so thermodynamically feasible and the gas is CO2.
Sorry how do u know that methanal is being oxidised? And also thank you so much, this is so helpful
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#6
(Original post by Yun0)
Well first of all, the gas/bubbles formed are definitely the CO2, so that's gotta be part of the last equation, And since we're starting by oxidizing methanal (an aldehyde) to methanoic acid using manganate ions, that's gotta be first.
So the first equation after balancing everything out is (5HCHO)+(2MnO4-)+(6H+)-> (2Mn2+)+(3H2O)+(5HCOOH) with change in voltage being (1.51+0.03=+1.54V) (as methanal is getting oxidized and manganate reduced, and reduce V- oxidize V) positive so thermodynamically feasible.
With methanoic acid obtained, next we use the silver ions to get CO2 (gas).
(HCOOH)+(2Ag+)-> (2Ag)+(CO2)+(2H+) with change in V being (0.8+0.11=+0.91V) (reduce V- oxidize V) Positive so thermodynamically feasible and the gas is CO2.
Also if you can, would you be able to help me with this question too:

Methanoic acid, HCOOH, can be used in a fuel cell. The fuel, HCOOH, is supplied at one electrode and the oxidant (oxygen) at the other electrode. The standard cell potential for this fuel cell is 1.34V.

The overall reaction is shown below:

HCOOH + 1/2O2 -> H2O + CO2

Using the information below use the half equation for the reaction at the oxygen electrode and calculate the standard electrode potential for the oxygen half cell.

1 CO2 + 2H+ + 2e- —>HCOOH (EV =-0.11)
2 HCOOH + 2H+ + 2e- —> HCHO + H2O (EV= -0.03)
3 Ag+ + e- —-> Ag (EV= +0.80)
4 MnO4- + 8H+ + 5e- —> Mn2+ + 4H2O (EV= +1.51)
0
1 month ago
#7
(Original post by Unknownn1.x)
Sorry how do u know that methanal is being oxidised? And also thank you so much, this is so helpful
Acidified potassium manganate is an oxidising agent, and methanal is an aldehyde which can be oxidized to methanoic acid/carboxylic acid (unlike ketones)
0
1 month ago
#8
0
1 month ago
#9
(Original post by r0yyjte)
HA, being a while since that happened
0
1 month ago
#10
(Original post by Unknownn1.x)
Also if you can, would you be able to help me with this question too:

Methanoic acid, HCOOH, can be used in a fuel cell. The fuel, HCOOH, is supplied at one electrode and the oxidant (oxygen) at the other electrode. The standard cell potential for this fuel cell is 1.34V.

The overall reaction is shown below:

HCOOH + 1/2O2 -> H2O + CO2

Using the information below use the half equation for the reaction at the oxygen electrode and calculate the standard electrode potential for the oxygen half cell.

1 CO2 + 2H+ + 2e- —>HCOOH (EV =-0.11)
2 HCOOH + 2H+ + 2e- —> HCHO + H2O (EV= -0.03)
3 Ag+ + e- —-> Ag (EV= +0.80)
4 MnO4- + 8H+ + 5e- —> Mn2+ + 4H2O (EV= +1.51)
Overall charge =+1.34, and we know that the oxygen is acting as the oxidizing agent here (the oxidant), thus the HCOOH should be reduced in the other electrode, which is what is happening with equation number 2.
Let's set standard electrode potential for the oxygen half cell as X.
So oxygen is gonna get reduced (since it's oxidizing at it's electrode), so based on reduced V-oxidized V, X-(-0.03)=1.34 X+0.03=1.34 X=+1.31V
(let me know if i get it right )
0
#11
(Original post by Yun0)
Overall charge =+1.34, and we know that the oxygen is acting as the oxidizing agent here (the oxidant), thus the HCOOH should be reduced in the other electrode, which is what is happening with equation number 2.
Let's set standard electrode potential for the oxygen half cell as X.
So oxygen is gonna get reduced (since it's oxidizing at it's electrode), so based on reduced V-oxidized V, X-(-0.03)=1.34 X+0.03=1.34 X=+1.31V
(let me know if i get it right )
Thank you!! I will. Thank you so much you really helped me 0
#12
(Original post by Yun0)
Well first of all, the gas/bubbles formed are definitely the CO2, so that's gotta be part of the last equation, And since we're starting by oxidizing methanal (an aldehyde) to methanoic acid using manganate ions, that's gotta be first.
So the first equation after balancing everything out is (5HCHO)+(2MnO4-)+(6H+)-> (2Mn2+)+(3H2O)+(5HCOOH) with change in voltage being (1.51+0.03=+1.54V) (as methanal is getting oxidized and manganate reduced, and reduce V- oxidize V) positive so thermodynamically feasible.
With methanoic acid obtained, next we use the silver ions to get CO2 (gas).
(HCOOH)+(2Ag+)-> (2Ag)+(CO2)+(2H+) with change in V being (0.8+0.11=+0.91V) (reduce V- oxidize V) Positive so thermodynamically feasible and the gas is CO2.
You know where you balanced the first equation, when I balanced mine I got HCHO + MnO4- + 6H -> Mn2+ + 3H2O + HCOOH

what did I do wrong ?
0
1 month ago
#13
(Original post by Unknownn1.x)
You know where you balanced the first equation, when I balanced mine I got HCHO + MnO4- + 6H -> Mn2+ + 3H2O + HCOOH

what did I do wrong ?
The number of electrons in each half-equation needs to be the same.
1
1 month ago
#14
(Original post by Yun0)
Well first of all, the gas/bubbles formed are definitely the CO2, so that's gotta be part of the last equation, And since we're starting by oxidizing methanal (an aldehyde) to methanoic acid using manganate ions, that's gotta be first.
So the first equation after balancing everything out is (5HCHO)+(2MnO4-)+(6H+)-> (2Mn2+)+(3H2O)+(5HCOOH) with change in voltage being (1.51+0.03=+1.54V) (as methanal is getting oxidized and manganate reduced, and reduce V- oxidize V) positive so thermodynamically feasible.
With methanoic acid obtained, next we use the silver ions to get CO2 (gas).
(HCOOH)+(2Ag+)-> (2Ag)+(CO2)+(2H+) with change in V being (0.8+0.11=+0.91V) (reduce V- oxidize V) Positive so thermodynamically feasible and the gas is CO2.
There is no mention of using silver ions in the question.

The methanoic acid is further oxidised by excess manganate(VII) in acid.
0
1 month ago
#15
(Original post by charco)
There is no mention of using silver ions in the question.

The methanoic acid is further oxidised by excess manganate(VII) in acid.
OH, yea yea yea
That makes more sense, i thought i had to use all the equations. (don't )
So the last one is (5HCOOH)+(2MnO4-)+(6H+)-> (2Mn2+)+(8H2O)+(5CO2) and 1.51+0.11=1.62V
0
#16
(Original post by Yun0)
OH, yea yea yea
That makes more sense, i thought i had to use all the equations. (don't )
So the last one is (5HCOOH)+(2MnO4-)+(6H+)-> (2Mn2+)+(8H2O)+(5CO2) and 1.51+0.11=1.62V
Thank you!
0
#17
(Original post by Unknownn1.x)
Also if you can, would you be able to help me with this question too:

Methanoic acid, HCOOH, can be used in a fuel cell. The fuel, HCOOH, is supplied at one electrode and the oxidant (oxygen) at the other electrode. The standard cell potential for this fuel cell is 1.34V.

The overall reaction is shown below:

HCOOH + 1/2O2 -> H2O + CO2

Using the information below use the half equation for the reaction at the oxygen electrode and calculate the standard electrode potential for the oxygen half cell.

1 CO2 + 2H+ + 2e- —>HCOOH (EV =-0.11)
2 HCOOH + 2H+ + 2e- —> HCHO + H2O (EV= -0.03)
3 Ag+ + e- —-> Ag (EV= +0.80)
4 MnO4- + 8H+ + 5e- —> Mn2+ + 4H2O (EV= +1.51)
charco could you also help me with this question so I can gain a better understanding?
0
#18
(Original post by Yun0)
Well first of all, the gas/bubbles formed are definitely the CO2, so that's gotta be part of the last equation, And since we're starting by oxidizing methanal (an aldehyde) to methanoic acid using manganate ions, that's gotta be first.
So the first equation after balancing everything out is (5HCHO)+(2MnO4-)+(6H+)-> (2Mn2+)+(3H2O)+(5HCOOH) with change in voltage being (1.51+0.03=+1.54V) (as methanal is getting oxidized and manganate reduced, and reduce V- oxidize V) positive so thermodynamically feasible.
With methanoic acid obtained, next we use the silver ions to get CO2 (gas).
(HCOOH)+(2Ag+)-> (2Ag)+(CO2)+(2H+) with change in V being (0.8+0.11=+0.91V) (reduce V- oxidize V) Positive so thermodynamically feasible and the gas is CO2.
For the change in voltage of the first equation you did: 1.51+0.03 = 1.54, but 0.03 is negative so would it not be 1.51 + (-0.03) = 1.48 v

Or is that wrong ?
0
1 month ago
#19
It’s 1.51-(-0.03), double negative so it’s plus
0
#20
(Original post by Yun0)
It’s 1.51-(-0.03), double negative so it’s plus
Ohhh sorry silly mistake 😅I got it thank you!
0
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