A galvanic cell consists of a copper cathode immersed in a CuSO 4 solution and a

Question

A galvanic cell consists of a copper cathode immersed in a CuSO4 solution and a cadmium anode immersed in a CdSO4 solution. A salt bridge connects the two half-cells.

(a) Write a balanced equation for the cell reaction.

(b) A current of 1.15 A is observed to flow for a period of 1.51 hours. How much charge passes through the circuit during this time? How many moles of electrons is this charge equivalent to?

(c) Calculate the change in mass of the copper electrode.  g is _______(lost/gained)

(d) Calculate the change in mass of the cadmium electrode.  g is _______(lost/gained0

___________(aq)(s)(l)(g) + _________ (aq)(s)(l)(g)    ___________ (aq)(s)(l)(g) + ___________(aq)(s)(l)(g)

Explanation / Answer

(a)

Balanced cell reaction :

Cd (s) + CuSO4 (aq.) ------------> CdSO4 (aq.) + Cu (s)

(b)

Charge = Current * time in s

Q = 1.15 * 1.51 * 60 * 60

Q = 6251.4 C

We know that,

96485 C = 1mol of electrons

Then,

6251.4 C = 1 * 6251.4 / 96485 = 0.0648 mol of electrons.

(c)

According to first law of electrolysis,

W = M * Q / ( Z * F )

W = 63.5 * 6251.4 / ( 2 * 96485 )

W = mass gained by copper electode = 2.06 g.

(d)

W = 112.4 * 6251.4 / ( 2 * 96485 )

W = Mass Lost by cadmium electrode = 3.64 g.

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