6M.1

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Xuan Kuang 2L
Posts: 31
Joined: Wed Nov 14, 2018 12:23 am

6M.1

Postby Xuan Kuang 2L » Thu Feb 28, 2019 10:49 pm

A student was given a standard Cu(s)|Cu2+ (aq) half-cell and another half-cell containing an unknown metal M in 1.00 M M(NO3)2(aq) and formed the cell M(s)|M+(aq)||Cu2+ (aq)|Cu(s). The cell potential was found to be -0.689 V. What is the value of E(M2+/M)?

From the cell diagram, I though Cu/Cu2+ was the cathode because it was on the right side, so when I calculate Ecell, I set it up like Ecell=Ecathode-Eanode, so -0.689V= 0.34V-Eanode, and Eanode= 1.029V, but the correct answer was -0.349V. Could someone explain why this is?

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Re: 6M.1

Postby Chem_Mod » Fri Mar 01, 2019 6:10 pm

I believe this is a typo. Copper should be the anode. If you follow that notation, you should get the right answer. Come to office hours or peer learning sessions if you still have trouble with this.


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