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I am very much confused: in simple voltaic cell ($\ce{Zn,~Cu}$ in $\ce{H2SO4}$), $\ce{Cu}$ is positively charged and is anode, where in Galvanic cell ($\ce{Zn}$ in $\ce{ZnSO4}$ and $\ce{Cu}$ in $\ce{CuSO4}$), $\ce{Cu}$ is positively charged but is Cathode.so how can i define it, where both are electric or voltaic cell. Please help me

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Cathode is the electrode where the Reduction reaction takes place.

Anode is the electrode where the Oxidation reaction takes place

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Voltaic cell is a synonym for a Galvanic cell. In the mentioned system ($\ce{Zn,~Cu}$ in $\ce{H2SO4}$), $\ce{Cu}$ can be charged both positively and negatively. It depends how you connect it to the power source. In a Galvanic/Voltaic cell (as Babounet mentioned):

$Cathode$ is a negatively charged electrode where $reduction$ take place

and

$Anode$ is a positively charged electrode where $oxidation$ take place

Let's analyse the second cell. Without any additional information, I assume that no external power source is connected. This system is known as Daniell cell: Wikipedia Daniell cell $$Zn \mid Zn^{2+} \parallel Cu^{2+}\mid Cu$$ $Zn - 2e- \rightarrow Zn^{2+}$ = oxidation = anode

$Cu^{2+}+2e- \rightarrow Cu$ = reduction = cathode

The situation is nearly the same for the first cell, but this situation is used and known as Sacrificial Anode. $Zn$ acts in the system as an anode and it is dissolved.

You can have two electrodes that are both positively charged such as $Ag$ and $Cu$. Standard electrode potential of both metals are positive: $\ce{Ag+}$/ $Ag_s$ = +0,80 V and $\ce{Cu2+}$/$Cu_s$= +0,34 V, but if you connect them in a galvanic cell, one will act as cathode ($Au$) and the other as anode ($Cu$).

As far as I can tell (based on provided informations), the copper electrode will be in both cases positively charged and therefore be a cathode. There must be a mistake in the description or in the statement.

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