Question

Problem 1

Part 1:

(a) Calculate 2.303RT in joules per mole and calories per mole at 25◦C.

(b) Show that 2.303RT/F is equal to 0.0591 V equiv/mol at 25◦C.

Part 2:

(a) Calculate the electrode potential at 25◦C for the Peterson–Groover Ag/AgCl reference electrode when it is used in natural seawater, for which the Cl– concentration is 0.6 M.

(b) How does this electrode potential compare with that for the saturated calomel electrode?

Part 3:

Dissolved lead can enter the public drinking supply through the use of lead piping systems and lead solder. The present standard for allowable dissolved Pb2+ ions is 15 μg/L. Calculate the reduction potential for lead metal in contact with this concentration of dissolved Pb2+ ions:

Pb2+(aq) + 2e− → Pb(s)

Answer 1

(a) value of R (Gaseous Constant)

R=8.314J/K mol

Given 2.303 RT Now put R=8.314 and T=298K

=2.303*8.314J/K mol *298K = 5705.84 J/lol

2) We know 1Cal=4.18 J

R in calorie

=2.303*1.987 cal/K mol *298K

= 1363.66 cal/mol

2) To show 2.303 RT/F = 0.0591 V/mol

Now put the values of R , T AND F, 1 Faraday =96500 columb

So 2.303*8.314J/K mol*298K/96500 columb

=0.0591 J/mol columb

0.0591 volt/mol because workdone/charge = volt