Consider the following electrolysis reacuons. Assume T= 298 K. cathode: Ar (aq,
ID: 930091 • Letter: C
Question
Consider the following electrolysis reacuons. Assume T= 298 K. cathode: Ar (aq, 021 M +3e A 3"(aq, 0.21 M) +3e- A1(s) cathode: Aul) Au3+(aq, 0.18 M) +3e anode: Aul Here is a table of standard reduction potentials. a) If current is negligible, what voltage must be applied to drive the net reaction? Number b) If the cell has a resistance of 3.8 and a current of 270 mA, calculate the ohmic potential of the cell. Number c) Calculate the potential that needs to be applied in order drive the reaction considering the ohmic potential calculated in part b and an anodic overpotential of Number scroll down for more quueetExplanation / Answer
(A)
Al3+ (aq, 0.21M) +3e- ------> Al(s)
The potential (E) needed for the above reaction = Eo + (0.0591 / n) log [Al3+ ]
n = no. of electrons involved in the reaction
E = Eo + (0.0591 / 3) log [0.21]
E = Eo + (0.0591 / 3) X (-0.67)
E = -1.66 - 0.0132 Eo = -1.66V from given table
E = - 1.6732V
Similarly E for the below reaction is
Au(s) ---------> Au (aq, 0.81) + 3e-
E = 1.498 + (0.0591 / 3) log [0.18]
E = 1.498 - 0.0146 = 1.48V
Value calculated above is reduction potential. Since Au undergoes oxidatio, Oxidation potential = -1.48
Total potential needed for above process is 3.153V (Sum of above two, Ered(cathode) + Eoxi(Anode)
(b) V = IR
V = 3.8 X 0.27 = 0.945V
270mA = 0.27A
(c) The potential that needs to be applied = 3.153 + ohmic potential + Overvoltage of Cathode + Overvoltage of Anode
The potential that needs to be applied =3.153 + ) 0.945 + 0.40 + 0.29 = 4.788V
(d) The potentials of the two reactions will change and will be given by Nernst equations (as shown above)
for Al
E = - 1.66 + (0.0591 / 3) X log(0.005)
E = - 1.66 - 0.04531 = -1.71V
For Au
E = 1.498 + (0.0591 / 3) log [0.88]
E = 1.498 - 0.0012 = 1.497V
E = Ecath - Eanod = -1.71 - (+1.497) = 3.21V
Total voltage needed = 3.21 + 0.945 + 0.40 + 0.29 = 4.845V