Question 14.19


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Robert Estrada
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Joined: Fri Sep 29, 2017 7:05 am

Question 14.19

Postby Robert Estrada » Thu Feb 15, 2018 8:15 pm

Can someone please explain question 19:
a student was given a standard cu(s)ICu^2+(aq) half cell and another half cell containing an unknown metal M immersed in 1.00M M(NO3)2(aq); When the copper was connected as the anode at 25 degrees Celsius, the cell potential was found to be -0.689V. What is the reduction potential for the unknown M^2/M couple?

Tanaisha Italia 1B
Posts: 55
Joined: Fri Sep 29, 2017 7:04 am

Re: Question 14.19

Postby Tanaisha Italia 1B » Fri Feb 16, 2018 12:18 am

You would use the equation E˚cell = E˚cathode - E˚anode. The E˚cell is the value that you were given for the cell potential. The E˚anode is the potential of Cu is this case, because it is being oxidized. You would substitute these values and then solve for E˚cathode, which is the E˚of the metal.

Tanaisha Italia 1B
Posts: 55
Joined: Fri Sep 29, 2017 7:04 am

Re: Question 14.19

Postby Tanaisha Italia 1B » Fri Feb 16, 2018 12:20 am

You can find the E˚anode value for Cu in Table 14.1.

torieoishi1A
Posts: 32
Joined: Sat Jul 22, 2017 3:00 am

Re: Question 14.19

Postby torieoishi1A » Fri Feb 16, 2018 11:59 pm

Looking at the cell:
Cu/Cu2+ electrode is the anode (oxidation is occurring)
M2+/M electrode is the cathode
Then use the formula: E=E(cathode)-E (anode)
-.689V=E(cathode)-(+.34V)
E(cathode)=-0.349V

Robert Estrada
Posts: 50
Joined: Fri Sep 29, 2017 7:05 am

Re: Question 14.19

Postby Robert Estrada » Tue Feb 20, 2018 8:43 pm

Thanks Guys!


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