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I am very much confused: in the cell (Zn, Cu in $\ce {H2SO4}$), Cu is positively charged and is anode, where in this cell (Zn in $\ce {ZnSO4}$ and Cu in $\ce {CuSO4}$), Cu is positively charged but is Cathode. So how can I define it, where both are electric or voltaic cell. Please help me. Please see these picture for better understanding.

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enter image description here reference:A Text Book Of Electrical Technology,Volume 1,B.L.Theraja

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  • $\begingroup$ A related question: chemistry.stackexchange.com/questions/26300/… $\endgroup$ – Curt F. Mar 21 '15 at 15:13
  • $\begingroup$ The copper electrode is never the anode (in this cell). It is mislabeled if it is called an anode. The copper ions are reduced at the cathode. The cathode has a negative potential in relation to the anode. $\endgroup$ – LDC3 Mar 21 '15 at 15:51
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Here Cu plate is cathode and Zn plate is anode, it's written wrong in it. In the first case (Zn in $\ce{ZnSO4}$ and Cu in $\ce{CuSO4}$) the electrodes are connected through an external circuit (You see that bulb glowing?). But what this article is trying to say is that in the cell (Zn, Cu in $\ce{H2SO4}$), if you leave the solution as it is without connecting externally, Zn having higher tendency to lose electrons, will lose electrons and these will appear on its plate surface making it negatively charged, the $\ce{Zn^2+}$ thus formed will combine with $\ce{ SO4^2-}$ and form $\ce{ZnSO4}$. The remaining two $\ce{H^+}$ ions will more towards copper plate where they will combine with the electrons on its plate and make it positive charged, releasing $\ce{H2}$. Now if a wire is connected across this setup then the electrons on Zn surface will leave it and go towards positive charged Cu plate.

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