Calculate the E Cell
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# Calculate the E Cell

[From: ] [author: ] [Date: 11-05-18] [Hit: ]
As the E0cell is coming out to be negative, in such a cell anode and cathode are incorrectly placed,......
I need help figuring this question out...

Calculate the E Cell

Hg^2+ + Cu^2+

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Well, the information provided in this question is rather incomplete. So, i'll make the Cell as per my choice and show you how to solve.
Let the cell be
Hg+(c1M)/Hg+2(c2M) || Cu+2(c3M)/Cu+(c4M)

Anodic Reaction: Hg+ ---------> Hg+2 + e
Cathodic Reaction: Cu+2 + e ----> Cu+
=====================
Hg+ + Cu+2 -----> Hg+2 + Cu+
======================
Ecell = E0Cell - 0.0591log [Hg+2][Cu+] / [Hg+][Cu+2]
= E0Cell - 0.0591log (c2 x c4) / (c1 x c3)
Now, E0cell = E0cathode - E0Anode
= (+0.15) - (+0.92) = -0.77Volt. As the E0cell is coming out to be negative, in such a cell anode and cathode are incorrectly placed, so we must replace anode by the cathode and vice versa
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