Answer to Question #115378 in General Chemistry for Kiara

Question

If the battery is permitted to discharge for a period of two hours at a constant current of 0.50 A, what will be the concentration of the manganese ion in the anode electrolyte solution at the end of the discharge?
Express the molarity to two decimal places of Mn2+ solution.

Reaction:
Mn(s)+Cu2+(aq) arrow Mn2+(aq)+Cu(s)

Half reaction: E°(V)
Mn2+(aq)+2e- arrow Mn(s). -1.18
Cu2+(aq)+2e- arrow Cu(s). 0.34

Answer 1

Under the conditions of the problem, the initial concentration of manganese ions is missing. Add them please. From the given conditions we can find the number of moles of manganese passed into the solution:

n(Mn²⁺) = It/nF = 0.0186 mol

where I = 0.50 A, t = 2 hours = 7200 s, n = 2 (number of electrons in half-reaction), F = 96500 C/mol. Knowing the initial concentration of manganese ions, you can calculate the final.

Or (more correct solution of the problem) knowing the change in the electrode potential of the cathode to calculate the concentration of magnesium ions from the following formula:

E = E^o_Mn2+/Mn + (0.059/n) * lgC(Mn²⁺), where E^o_Mn2+/Mn = -1.18 V and n = 2 (number of electrons in half-reaction)

So, C(Mn²⁺) = 10^{(E + 1.18)/0.0295}