Kaplan electrical potential: determining half reaction

[451617]

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The relevant reduction potentials are the following:
Au3+ + 3e– → Au E° = 1.50V
Cl2 + 2e– → 2Cl– E° = 1.36V

2 . When the cell described in the passage is operating
spontaneously, what is the overall chemical reaction?

A . 3Cl2(g) + 2Au+3(aq) &#8594; 6Cl–(aq) + 2Au(s) <-- i chose
B . 3Cl2(g) + 2Au(s) &#8594; 6Cl–(aq) + 2Au+3(aq) <---- easy eliminate
C . Cl2(g) + Cl–(aq) &#8594; Au(s) + Au+3(aq) <--- easy eliminate
D . 2Au+3(aq) + 6Cl–(aq) &#8594; 3Cl2 (g) + 2Au(s) <------ answer


the answer states "...you need to look at the half-reactions involved, and decide how to arrange them in such a way so as to get a positive overall cell potential."

i can why you would reverse the direction of 2Cl- from answer above, but if you EVEN OUT the electrons by multiplying Au by 2 and Cl by 3, wouldn't you actually get a positive cell potential if you reverse the Au instead?

do we worry about the cell potential being positive before the electrons are balanced?

thanks so much!

 

[Bomikepa]

The relevant reduction potentials are the following:
Au3+ + 3e– &#8594; Au E° = 1.50V
Cl2 + 2e– &#8594; 2Cl– E° = 1.36V

2 . When the cell described in the passage is operating
spontaneously, what is the overall chemical reaction?

A . 3Cl2(g) + 2Au+3(aq) &#8594; 6Cl–(aq) + 2Au(s) <-- i chose
B . 3Cl2(g) + 2Au(s) &#8594; 6Cl–(aq) + 2Au+3(aq) <---- easy eliminate
C . Cl2(g) + Cl–(aq) &#8594; Au(s) + Au+3(aq) <--- easy eliminate
D . 2Au+3(aq) + 6Cl–(aq) &#8594; 3Cl2 (g) + 2Au(s) <------ answer


the answer states "...you need to look at the half-reactions involved, and decide how to arrange them in such a way so as to get a positive overall cell potential."

i can why you would reverse the direction of 2Cl- from answer above, but if you EVEN OUT the electrons by multiplying Au by 2 and Cl by 3, wouldn't you actually get a positive cell potential if you reverse the Au instead?

do we worry about the cell potential being positive before the electrons are balanced?

thanks so much!

D is the right answer like you stated. Remember that you do not multiply potentials like you do multiply each half reaction to get the electrons balanced.

Set up your Ecell equation:

We know that
Au3+ + 3e– &#8594; Au E° = 1.50V - reducing
Cl2 + 2e– &#8594; 2Cl– E° = 1.36V (oxidizing) -but recall we need to reverse the sign and also rewrite the equation in the other direction when we go to write out the net equation.

Ecell = ox + red = -136 + 1.50 = .14 ( positive e cell) which is what we want bc the reaction is spontaneous

net Eq:

(Au3+ + 3e– &#8594; Au) x 2 = 2Au3+ + 6e– &#8594; 2Au
(2Cl– &#8594; Cl2 + 2e ) x 3 = 6Cl– &#8594; 3Cl2 + 6e


2Au3+ + 6e- + 6Cl– &#8594; 2Au + 3Cl2 + 6e- ( the 6 e- cancel leaving)

2Au3+ + 6Cl &#8594; 2Au + 3Cl2

hope this helps

 

[451617]

thanks for the reply!

i can't think of any off the top of my head, but aren't there questions where you do have to multiply the entire net cell equation?

from my understanding, i believe you DON'T multiply coefficients for entropy and enthalpy questions but DO for cell potential. is it only for half-reactions?

 

[Bomikepa]

thanks for the reply!

i can't think of any off the top of my head, but aren't there questions where you do have to multiply the entire net cell equation?

from my understanding, i believe you DON'T multiply coefficients for entropy and enthalpy questions but DO for cell potential. is it only for half-reactions?

pretty much the only time where you do is for Echem, just like this example ( but you do have to reverse the sign when necessary)

When you calculate all those heat of formations and stuff you have to multiply the associated value with each cell line and/or reverse the value to depending on what you are trying to do.

 

[451617]

pretty much the only time where you dont is for Echem, just like this example ( but you do have to reverse the sign when necessary)

When you calculate all those heat of formations and stuff you have to multiply the associated value with each cell line and/or reverse the value to depending on what you are trying to do.

did you mean "you do" for Echem?

 

[Bomikepa]

did you mean "you do" for Echem?

yes sorry about that my typing sucks. I changed my post to have the correct info lol

 

[451617]

great! thanks so much for the help.

terrific