Question

a) Write the two half reactions Pb(NO3)2 and Mn(NO3)2. Indicate which is undergoing reduction and which is undergoing oxidation. Determine the net reaction and the net E

Answer 1

Solution :-

Following are the standard reduction potentials for the each element

Pb^2+(aq) + 2e- ---- > Pb(s)   E= -0.126 V

Mn^2+(aq) + 2e- ---- > Mn(s) E=-1.185 V

The two half cell reactions are as follows

Oxidation half reaction

Mn(s) ---- > Mn^2+(aq) + 2e-           anode reaction

Reduction half reaction

Pb^2+(aq) + 2e- ---- > Pb(s)               cathode reaction.

So in this reaction Mn is undergoing oxidation and Pb^2+ is undergoing reduction

Overall reaction

Mn(s) + Pb^2+(aq) ---- > Pb(s) + Mn^2+(aq)

Calculating the E cell

E cell = E cathode – E anode

         = -0.126 V – (-1.185 V)

        = 1.059 V