Question

A galvanic cell consists of a manganese anode immersed in a MnSO4 solution and a copper cathode immersed in a CuSo4 solution. A salt bridge connects the two half-cells (a) Write a balanced equation for the cell reaction. (b) A current of 1.34 A is observed to flow for a period of 1.99 hours. How much charge passes through the circuit during this time? How many moles of electrons is this charge equivalent to? mol (c) Calculate the change in mass of the manganese electrode. g is (d) Calculate the change in mass of the copper electrode 1S

Answer 1

a)

Mn(s)   + Cu2+(aq) ------> Mn2+(aq) + Cu(s)

b)

current = charge / time

charge = current * time

= 1.34 * 1.99 *60*60

charge = 9599.76 C

1 mole of electron contain 96485 C charge

no of moles of electron = 9599.76 / 96485 = 0.0995 mol

c)

moles of Mn oxidized = 0.0995 / 2 = 0.0497 mol

mass of Mn oxidized = 0.0497 * 54.938 = 2.73 g

so

change in mass of Mn electrode = 2.73 g is decrease

d)

moles of Cu Reduced = 0.0995 / 2 = 0.0497 mol

mass of Cu Reduced =   0.0497 * 63.546 = 3.158 g

change in mass of Copper electrode = 3.158 g is increase