Question

Consider a voltaic (galvanic) cell with the following metal electrodes. Identify which metal is the cathode and which is the anode, and calculate the cell potential.

(a)    Pb(II) and Cr(III)

E_cell = ____V

(b)    Au(III) and Sn(II)

E_cell = ____V

(c)    Al and Ni(II)

E_cell = ____V

Answer 1

In any electrolytic /electrochemical (galvanic) cell , the electrode where oxidation takes place s anode and the electrode at which reduction takes place is called cathode.

a) Pb(II) and Cr(III) SRP values of Pb+2/Pb = -0.13V and Cr+3/Cr = -0.74V

Thus chromium is oxidised and lead is reduced. Anode is chromium and cathode is lead.

E cell = -0.13 + 0.74 = 0.61V

b) SRP of Au+3/Au = 1.50V Sn=2/Sn = -0.16v

Thus gold is reduced and tin is oxidised . Thus gold is cathode and tin is anode.

E cell = 1.50 + 0.16 = 1.66v

c) SRP of Al+3/Al = -1.66V and Ni+2/Ni = -0.25 V

Hence Aluminium is oxidised and nickel is reduced. Aluminium is anode and nickel is cathode.

E cell = -0.25 + 1.66 = 1.41 V