Question

Question #f0862

Answer 1

All the info you request is in the analysis below...

Explanation:

The task here is to create the Mn half-reaction (as it does not seem to appear on any of the standard tables I could find).

The zinc (anode) is quite simple:

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

To build the reduction half reaction, we start with the Mn species:

`MnO_2 rarr Mn(OH)_3`

Next, we balance the hydrogen and oxygen atoms by adding `H_2O` and `OH^-` as needed:

`MnO_2 + 2H_2O rarr Mn(OH)_3 + OH^-`

Finally, add electrons to balance the charge:

`MnO_2 + 2H_2O + e^(-) rarr Mn(OH)_3 + OH^-`

which is a reduction, as expected.

Finally, since we know the net reaction potential (the cell voltage) is 1.50 V, the potential of the reduction must be 0.74 V, because the difference in potential

`E^o (cathode) - E^o (anode) = 1.5 V`

`0.74 - (-0.76) = 1.5`

Finally, to balance the equation, we must balance the # of electrons. This means multiplying the reduction half-reaction by 2, then adding them together:

`Zn + 2MnO_2 + 4H_2O rarr Zn^(2+) + 2Mn(OH)_3 + 2OH^-`