Question

3. For the cell represented by Pt|Fe3+(0.25 M), Fe2+(0.025 M)||Ce4+(0.035 M), Ce3+(0.050 M)|Pt E o (Fe3+/Fe2+ = +0.771V, Eo Ce4+/Ce3+ = + 1.44V

a) Draw the complete cell represented by the above line diagram and label all the components.

b) Write the half-cell reactions and complete redox reaction of the cell.

c) Calculate the standard cell potential.

d) Calculate the cell potential at the non-standard conditions provided above.

e) Was the cell reaction spontaneous?

Answer 1

E' cell = +ve > Spenatenores E cell = -ve Ron Spontaneous Switch ANODE voltmeter Nat NO3- Cathod Pt Salt bridge feat fe3tte: cettteet3 Movement of Cation o Movement of Ponien oxidation half Reaction -fert caq) - test caq, te Reduction half Reaction Cettcaq, te - Ces caq). Ecell = to cell = E cell = cathode - E Anode 1.44V-0.77V 0.67V

fe2+ + Cett (0.025m Co.035m) fe3+ + Cest 0.25m) co-osomy Ray Ecell = Ecel -0.0591 log Qc cell = t cell Ecell = Fielu - 0.0591 lpg [fest] [63] Ecell = Ecell I y I fe2+7 56 4+1 Ecell = 0.67 -0.0591 log . 0.25 × 0.00 0.025 0.035 0.0125 Ecel = 0.67 -0.0591 log 0.000875 Ecell = 0.67 -0.059 1 log 14.285 1.1548 Ecell = 0.67 -0.059 1 X Ecell = 0.67 -0.0682 Ecell = 0.602 v (e. Because Ecel and Ecell = Reaction is spentometers tve so Sxtra fadh Pt[fe3760-25M), 6e?+ (0-625m) $164.035) cete orole Anodic Chamber 11 Cathodic chamber Anode Cathocle oxidation - Reduction Loss of e- / OIL Gain of e- e- RIG -
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