A direct current is applied to a solution of manganese (II) iodide.
a. Write the balanced equation for the oxidation half reaction.
b. Write the balanced equation for the reduction half reaction.
c. Write the balanced equation for the overall reaction that takes place in the cell.
d. Which reaction takes place at the cathode? Why?
e. Which reaction takes place at the anode? Why?
f. Predict the sign for deltaG^o. Justify your choice.
g. Calculate deltaG^o.
Manganese (II) Iodide
In this Mn2+ will undergo reduction by gain of 2 electrons
I- will undergo oxidation by loss of 2 electrons to give I2
Mn2+ + 2I- ......> Mn (s) + I2(s)
Cathode ...is connected to the negative terminal hence rich in electrons so reduction at cathode
Anode ...is connected to the positive terminal hence poor in electrons so oxidation at cathode
It is a non spontaneous reaction that is the reason current is required to carry out this reaction hence delta G value is positive
Calculate Eo cell = E ocathode + Eo anode
You can pick up values from the standard reduction table
delta G = -nFEoCell
n = 2 (electrons lost or gained )here
F = 96500C