14.19

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Tina Nguyen_1L
Posts: 19
Joined: Fri Sep 25, 2015 3:00 am

14.19

Postby Tina Nguyen_1L » Sat Feb 06, 2016 2:19 pm

Problem 14.19 says "A student was given a standard Cu(s)|Cu^2+(aq) half-cell and another half-cell containing an unknown metal M immersed in 1.00 m M(NO3)2 (aq). When the copper was connected as the anode at 25 C, the cell potential was found to be -0.689 V. What is the reduction potential for the unknown M^2+|M couple?"

The solutions manual's answer is -0.689 V=E(cathode)-(0.34 V)
Why do you not reverse the sign of the anode (copper at 0.34 V) when calculating this?

Noah Reid 4C
Posts: 23
Joined: Fri Sep 25, 2015 3:00 am

Re: 14.19

Postby Noah Reid 4C » Sat Feb 06, 2016 3:41 pm

There are two ways to calculate the cell potential. One way is by subtracting the cathode half reduction reaction from the anode half reduction reaction. The other way is to reverse the half reaction for whichever reaction will be oxidized and adding the two half reactions together. The book used the reductions values for each half reaction and subtracted the anode from the cathode. If we changed the sign of Cu to -0.34V we would have the E of oxidation. We would then be adding it to the cathode, changing the equation. We would be solving for E(Total) = E(reduction of cathode) + E(oxidation of anode), rather than the original method which is E(Total) = E (Reduction of cathode) - E( reduction of anode). Either way is acceptable.

-0.689V = E(cathode) - (0.34V) = -0.349V

-0.689V = E(cathode) + (-0.34V) = -0.349V

Adriana_4F
Posts: 74
Joined: Fri Sep 28, 2018 12:29 am

Re: 14.19

Postby Adriana_4F » Wed Feb 19, 2020 9:47 pm

Since the copper is being reduced doesn't it shouldn't 0.34 be the cathode potential?


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