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December 20, 2014

December 20, 2014

Posted by **Stephanie** on Wednesday, April 28, 2010 at 12:42pm.

Zn | Zn+2(0.10 M) || Cu+2 (2.50 M)| Cu

The mass of each electrode is 200.0 g. Each half cell contains 1.00 liter of solution.

a) Calculate the cell potential when this battery is first connected.

b) Calculate the cell potential after a current of 10.0 amperes has flowed for 10.0 hours.

c) Calculate the mass of each electrode after 10.0 hours.

d) What is the total life span of this battery, delivering a current of 10.0 amperes, before it goes dead. (Hint: You need to know the limiting reagent. Is it Zn or Cu+2)

- college chemistry -
**DrBob222**, Wednesday, April 28, 2010 at 1:26pmPerhaps I can get you started but that is it.

Since the concns are not standard, first calculate the Eo value for each half cell but BOTH as reductions.

For Zn^+ + 2e = Zn we have

E = Eo-(0.0592/n)*log[(Zn)/(Zn^+2)] =

-.763-(0.0592/2)*log[1/0.1)] = -0.793v. (check my work on this). This will be used as an oxidation reaction; therefore, reverse the equation and change the sign.

Eox = +0.793 v.

For the Cu, as a reduction.

E = Eo-(0.0592/n)*log[(Cu)/(Cu^+2)] =

+0.337 - (0.0592/2)*log(1/2.50) = 0.348v. Check this arithmetic, too.

Ered = 0.348 v.

Ecell = Eox + Ered = 0.793+0.348 = 1.14 v. Check this arithmetic (and chemistry).

The above is part a in DETAIL.

Post your work on the others if you get stuck and someone will help. Some hints:

amperes x seconds = coulombs.

96,485 coulombs will deposit one equivalent of a metal (1 equivalent = 1/2 mole for both Cu and Zn). For (b), you should start by calculating the amount of Zn that goes into solution and how much Cu^+2 comes out of solution, use those new concns to recalculate the half cell voltages just as I did in part (a), then calculate the new cell potential. Part (c) will follow from part (b). Follow the hint for part (d).

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