Question

Would a reaction occur if a piece of `Sn` metal were placed in a `ZnSO_4` solution?

Answer 1

Yes. There would be a spontaneous redox (oxidation-reduction) reaction between the `Sn` metal and the `Zn^(2+)` ion, given by:

`Sn^(2+) + Zn rarr Sn + Zn^(2+)`

Explanation:

You will have to consult a list of reduction potentials that includes the following two half-reactions:

`Zn^(2+) + 2e^(-) rarr Zn` `E° = -0.76 V`

`Sn^(2+) + 2e^(-) rarr Sn` `E° = -0.14 V`

There is a list here: http://chem.libretexts.org/Core/Analytical_Chemistry/Electrochemistry/Redox_Chemistry/Standard_Reduction_Potential

We are considering that one (and only one) of two possible reactions will occur. Either zinc ion will reduce (the top half-reaction), accompanied by oxidation of tin metal (the reverse of the lower half-reaction), or exactly the reverse will be true - tin ion will reduce (the lower half reaction) accompanied by the oxidation of zinc metal (the reverse of the upper half-reaction)

To determine which is the actual reaction, we note that the following equation must yield a positive result:

E°(reduction) - E°(oxidation) = E°(net)

We get a positive result for the subtraction

`-0.14 - (-0.76) = 0.62 V`

which means that the Sn half-reaction must be the reduction in this system, and the Zn half-reaction furnishes the oxidation.

The actual half-reactions occurring are:

`Sn^(2+) + 2e^(-) rarr Sn`

`Zn rarr Zn^(2+) + 2e^(-)`

To find the overall reaction, add the two halves (after cancelling the electrons):

`Sn^(2+) + Zn rarr Sn + Zn^(2+)`

The sulfate ion is merely a spectator in this situation.