# Electrode potentials

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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

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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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

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

reply

**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

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#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

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) (

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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(Original post by

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

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) (

**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.
0

reply

**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.

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)

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#7

(Original post by

Sorry how do u know that methanal is being oxidised? And also thank you so much, this is so helpful

**Unknownn1.x**)Sorry how do u know that methanal is being oxidised? And also thank you so much, this is so helpful

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#10

(Original post by

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)

**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 )

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(Original post by

Yikes, not quite sure about this, but this is how i will do it:

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

(let me know if i get it right )

**Yun0**)Yikes, not quite sure about this, but this is how i will do it:

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 )

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**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.

what did I do wrong ?

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#13

(Original post by

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 ?

**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 ?

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#14

**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.

The methanoic acid is further oxidised by excess manganate(VII) in acid.

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#15

(Original post by

There is no mention of using silver ions in the question.

The methanoic acid is further oxidised by excess manganate(VII) in acid.

**charco**)There is no mention of using silver ions in the question.

The methanoic acid is further oxidised by excess manganate(VII) in acid.

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

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(Original post by

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

**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

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**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)

0

reply

**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.

Or is that wrong ?

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(Original post by

It’s 1.51-(-0.03), double negative so it’s plus

**Yun0**)It’s 1.51-(-0.03), double negative so it’s plus

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