Refer to the galvanic cell below (the contents of each half-cell are written ben
ID: 953429 • Letter: R
Question
Refer to the galvanic cell below (the contents of each half-cell are written beneath each compartment): he standard reduction potentials are as follows: UO^2+_2 + 4 H + 2 e^- rightarrow U^4- + 2 H_2I^- E = 0.334 V Fe^3+ + e^- rightarrow Fe^2+ E = 0.771 V 20. What is the value of Edegree cell? A) 1.105 V B) -0.437 V C) 0.437 V D) 1.208 V E) 2.29 V 21. When current is allowed to flow, which species is the reducing agent? A) Fe^3+ B) Fe^2 C) UO_2^2- D) U^4+ E) H^+ 22. What is the value of the equilibrium constant at 25degreeC for the net spontaneous cell reaction? A) 239 Times 10^37 B) 1.65 Times 10^-15 C) 606 Times 10^14 D) 7.30 Times 10^40 E) 2.91 Times 10^77 23. What is the value of Q, the reaction quotient, for this cell reaction? A) 1.52 Times 10^11 B) 5.18 Times 10^9 C) 6.60 Times 10^-12 D) 1.93 Times 10^-10 24. In which direction do electrons flow in the external circuit of the cell pictured above? A) left to right B) right to left C) no current Hows; the cell is at equilibrium D) can not be determined25. What is the E_cell expected at 50degreeC for this particular galvanic cell? A) 0.75 V B) 0.72 V C) 1.42 V D) 0.79 V E) 0.00 VExplanation / Answer
accoding to the E0 values Fe 3+ will get reduced and U4+ will get oxidised
Total cell reaction
2Fe 3+ + U4+ + 2H2O------> UO22+ + 2Fe2+ + 4H+
20) E 0 = E0 reduced -E0 oxidised
0.771 - 0.334 = 0.437 v (c)
21) reducing agent oxidizes itself. hence U4+ acts as reducing agent (d)
22) At equiblirium Ecell =0
thus E0cell = 0.0591/ n X log( Kc)
Kc = ([UO22+ ] X [Fe2+]2 X [H+]4)/ ([Fe 3+ ]2 X [U4+])
thus substituting values
logKc = (0.437/0.0591) X 2
Kc = 6.06 X 1014(c)
23) Q = ration of concentraion of products divided by the conc of reactants
from the given velues of reactants and product , Q can be calculated
Q= ([UO22+ ] X [Fe2+]2 X [H+]4/ ([Fe 3+ ]2 X [U4+])
on substituting values
Q = (7.93 X 10-3 X 0.11342 X (1.16X 10-3)4) / ((3.876 X 10-3)2 X (6.37 X 10-3))
= 1.93 X 10-10(d)
24) from th ecell reaction we know that Fe will take electron and U4+ will give electrons
thus electrons flow from right to left (b)
25)
Ecell = E0 cell - (2.303 X 8.314 X 323) /(96500 X 2) log [Qc]
thus substituting vales
Ecell = 0.437 - 0.032 log (1.93 X 10-10)
=0.748 V (a)
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