Question #f0862

1 Answer
Dwight
Feb 2, 2018

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^-#

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