# Anode and Cathode

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

Cathode is the electrode where the Reduction reaction takes place.

Anode is the electrode where the Oxidation reaction takes place

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: $$Zn \mid Zn^{2+} \parallel Cu^{2+}\mid Cu$$ $Zn - 2e- \rightarrow Zn^{2+}$ = 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.