# The following cell is set up: Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu(s) Write down the equation for the reaction which takes place in each half cell. From which electrode do electrons flow away from and into the external circuit? Calculate the e.m.f of this cell.

Jul 11, 2016

See below:

#### Explanation:

List the standard electrode potentials in increasing order:

$\textsf{\textcolor{w h i t e}{\times \times \times \times \times \times \times \times \times \times \times \times \times} {E}^{\circ} \text{ } \left(V\right)}$

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$\textsf{Z {n}_{\left(a q\right)}^{2 +} + 2 e r i g h t \le f t h a r p \infty n s Z {n}_{\left(s\right)} \textcolor{w h i t e}{\times \times \times \times x} - 0.76}$

$\textsf{C {u}_{\left(a q\right)}^{2 +} + 2 e r i g h t \le f t h a r p \infty n s C {u}_{\left(s\right)} \textcolor{w h i t e}{\times \times \times \times x} + 0.34}$

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The 1/2 cell with the most +ve $\textsf{{E}^{\circ}}$ value will take in the electrons.

So we can see that the copper 1/2 cell will be driven left to right and the zinc 1/2 cell right to left as indicated by the arrows.

So the two 1/2 equations are:

$\textsf{Z {n}_{\left(a q\right)} \rightarrow Z {n}_{\left(a q\right)}^{2 +} + 2 e}$

$\textsf{C {u}_{\left(a q\right)}^{2 +} + 2 e \rightarrow C {u}_{\left(s\right)}}$

Adding these gives the cell reaction when it is working:

$\textsf{Z {n}_{\left(s\right)} + C {u}_{\left(a q\right)}^{2 +} \rightarrow Z {n}_{\left(a q\right)}^{2 +} + C {u}_{\left(s\right)}}$

When the zinc atoms lose electrons the zinc ions go into solution and the 2 electrons flow away from this electrode into the external circuit.

When they arrive at the copper electrode, they are picked up by the copper(II) ions to become copper atoms.

The two 1/2 cells are connected by a salt bridge or a membrane. The salt bridge consists of some filter paper soaked in a suitable electrolyte such as saturated potassium nitrate solution.

The function of the salt bridge is to maintain electrical neutrality in each half cell.

As zinc ions go into solution $\textsf{N {O}_{3}^{-}}$ ions flood in to the half cell.

As copper(II) ions leave the solution in the other 1/2 cell, $\textsf{{K}^{+}}$ ions flood in.

To find the emf of the cell, subtract the least +ve $\textsf{{E}^{\circ}}$ value from the most +ve:

$\textsf{{E}_{c e l l}^{\circ} = + 0.34 - \left(- 0.76\right) = + 1.1 V}$

This arrangement is known as "The Daniel Cell".