Question

A galvanic cell consists of a cadmium cathode immersed in a Cdso solution and a manganese anode immersed in a MnSo4 solution. A salt bridge connects the two half- cells a) Write a balanced equation for the cell reaction. (b) A current of 1.01 A is observed to flow for a period of 1.67 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 cadmium electrode. g is d Calculate the change in mass of the manganese electrode 1S

Answer 1

Mn(s) ---> Mn2+(aq) +2e- at anode oxidation E0=-1.185 V
Cd2+ +2e- ---> Cd(s) -at cathode reduction E0=-0.4 v
net cell reaction
Mn(s) +Cd2+(aq) ----> Mn2+(aq) + Cd(s)
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Charge Q = Current xtime= 1.01 Ax 1.67*60*60 S
= 6072.12 C
1 mole electron =1 F= 96485 C
SO 6072.12 C=1 mole/96485 x 6072.12 = 0.0629 mols of electrons
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From balanced equation 1 mole Mn(s) release 2mole electron
0.0629 mole e- makes 0.03145 mol Mn2+ goes in to the solution at anode
0.0629 mole e- makes 0.03145 mol Cd at cathode
convert this mols to g using molar mass
Mass of Cd = mols of Cdx molar mass of Cd ='112.4 x0.03145 =3.54 g
Mass of Mn = Mols of Mnx molar mass of Mn = 55 x0.03145 =1.73 g

Thus for Cd 3.535 G is increased
For Mn 1.73 g is decreased
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Thank you