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How are overall equations of the galvanic cell written? In a zinc-copper cell, the overall equation is

$$ \ce{Cu^2+(aq) + Zn(s) -> Zn^2+(aq) + Cu(s)} $$

So in order to come up with this equation, do you take the net ionic equation that occurs when $ \ce{Zn} $ is placed into $ \ce{CuSO4} $ solution (the electrolyte for copper half cell)? In other words, do you calculate the overall equation of the galvanic cell by coming up with the net ionic equation of the single displacement reaction that occurs between the anode electrode and the cathode electrolyte?

In this example, it would be

$$ \ce{Zn(s) + Cu^2+(aq) + SO_4^2- (aq) -> Zn^2+(aq) + Cu(s) + SO4^2- (aq)} $$

where $ \ce{Zn(s)} $ would be the anode electrode and $ \ce{CuSO_4} $ would be the cathode electrolyte.

And the $ \ce{SO_4^{2-}} $ will cancel out since it is a spectator ion to give the overall equation of the galvanic cell.

Am I doing this right?

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Since $ \ce{Zn} $ never touches any $ \ce{Cu^2+} $ ions in a galvanic cell - the salt bridge/porous plate is preventing this - it would not be helpful to take this chemical equation as a starting point. As the aim of such cells is to separate the reduction and oxidation process, this is where the equations should originate.


Direction of the Reaction

To start with, one should find out which of the 2 electrode materials is more noble, hence will be the oxidizing agent aka cathode material, and will get itself reduced during the process through receiving electrons from the anode over the external circuit. A higher standard electrode potential ($E^\circ$) correlates with higher nobility. Information on different $E^\circ$ values can be found in galvanic series (Wikipedia: Standard electrode potential (data page)). A lookup of $\ce{Cu}$ and $\ce{Zn}$ gives:

$$ \begin{array}{rclc} \hline \text{Oxidant} & & \text{Reductant} & E^\circ/\mathrm{V}\\ \hline \ce{Zn^2+(aq) + 2 e− & <=> & Zn(s)} & −0.7618\\ \ce{Cu^2+(aq) + 2 e− & <=> & Cu(s)} & +0.337\\ \hline \end{array} $$

Since $\ce{Cu}$ has the higher $E^\circ$ (is more noble), $\ce{Cu}$ will act as cathode, and $\ce{Zn}$ as anode, to provide the reactions spontaneity.


Setting up the Reaction

Looking at the electrochemical process, first the $ \ce{Zn} $ from the zink electrode gets oxidized according to:

$$ \text{Ox - Anode half equation:}\qquad\ce{Zn(s) -> Zn^2+(aq) + 2 e-} $$

After passing through the external circuit the electrons get to reduce the $ \ce{Cu^2+} $ ions according to:

$$ \text{Red - Cathode half equation:}\qquad\ce{Cu^2+(aq) + 2 e- -> Cu(s)} $$

Which shows, that the anode is going to lose mass through corrosion, while the cathode is going to gain mass through the formation of solid copper on its surface.

After having these two half equation we can sum them up and cancel out the $ \ce{2 e-} $ respectively to yield:

$$ \text{Redox - Total equation:}\qquad\ce{Cu^2+(aq) + Zn(s) -> Cu(s) + Zn^2+(aq)} $$

While these processes happen, the salt bridge/porous plate is allowing the balancing of charges in both half cells.


Note: Of course someone could begin with the reduction, rather than with the oxidation, but it would anyway end in the same total equation, and it's harder then to imagine the electrons wandering through the external circuit.


Edit: As requested in the comments: The reactions with spectator ions:

$$\begin{align} \text{Ox - Anode:}&& \ce{Zn(s) + SO4^{2-}(aq) &-> Zn^2+(aq) + SO4^{2-}(aq) + 2 e-}\\ \text{Red - Cathode:}&& \ce{Cu^2+(aq) + SO4^{2-}(aq) + 2 e- &-> Cu(s) + SO4^{2-}(aq)}\\ \text{Redox - Cell:}&& \ce{Cu^2+(aq) + Zn(s) + 2SO4^{2-}(aq) &-> Cu(s) + Zn^2+(aq) + 2SO4^{2-}(aq)} \end{align}$$

Edit 2: Added Direction of the Reaction

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  • $\begingroup$ but how do you know zinc is getting organized. What is the equation in half cell (i.e. are there spectator ions that cancel out and just leave Zn(s)⟶Zn2+(aq)+2e−. How do they come up with these half cell reactions. $\endgroup$ – user510 Feb 24 '16 at 2:39
  • $\begingroup$ Zinc is oxidized, since it is less noble than copper. This information, including the half equations can be found in the galvanic series (Sry for the german wiki, but in the english version there is no table). Of course you could write the anions in both half equations, I omitted them, since they are spectator ions and would cancel out anyways. $\endgroup$ – lcnittl Feb 24 '16 at 7:54
  • $\begingroup$ could you please show them with it, I am a little confused. $\endgroup$ – user510 Feb 24 '16 at 22:39
  • $\begingroup$ Show what with it? $\endgroup$ – lcnittl Feb 24 '16 at 23:32
  • $\begingroup$ the reaction with spectator ions $\endgroup$ – user510 Feb 25 '16 at 2:29

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