Question

Given that Eo = -0.34 V for the reduction of Tl+ to Tl, find the voltage developed by a cell consisting of Tl metal dipping in an aqueous solution that is 0.59 M in Tl+, connected by a porous bridge to a 0.53 M aqueous solution of HCl in contact with a Pt electrode over which H2 gas is bubbling at p = 0.81 bar..

Answer 1

Lower the value of standard reduction potential, then the given electrode acts as anode and higher the value of standard reduction potential, then the given electrode acts as cathode.

a. The anode half reaction (oxidation): Tl(s) -------> Tl(+)(aq) + e- ) *2

The cathode half reaction (reduction): 2H+(aq) + 2e- -----> H2(g)

The overall cell reaction : 2H+(aq)   + 2Tl(aq) ---> H2(g) + 2Tl+(aq)

E0cell = Ec - Ea = 0- (-0.34) = 0.34 V

The cell potential of the cell at 25 degree Celsius at diffeerent condition is given by Nernst equation

E = Eo- (0.0591/2)* log([Tl+]^2[H2] /[H+]^2])

    = 0.34 - (0.0591/2)*log((0.59)^2*0.80/(0.53)^2)

The potential for the redox reaction below at 25 degrees celsius = 0.34011 v