
#1
Nov1007, 09:37 AM

P: 303

I needed to find the E°cell (always positive for a galvanic cell), based on the following (unbalanced) reaction:
IO3(aq) + Fe2+(aq) > Fe3+(aq) + I2(s) So E°cell = Ecathode  Eanode I made the half reactions Fe3+ (aq) > Fe2+(aq) EĜ = +0.77 V (i found this reaction in my table) IO3(aq) > I2(s) balanced the 2nd one to 12 H + 2IO3 + 5e > I2 + 6H2O EĜ = I cannot find this value in that table I FOUND IT ONLINE ON A WEBSITE TO BE EĜ = +1.19V, HOW DID THEY GET THIS VALUE! Anyone know? 



#2
Nov1107, 09:15 AM

P: 1,504

E° = +1.19  0.77 = +0.42V. About the reaction: (1) 2IO3(aq) + 12 H+ + 10e > I2(s) + 6H2O You can find its E° if you know the E° of other similar reactions from which you can get that one, for example: (2) IO3(aq) + 6H+ +6e > I + 3H2O;_______>E°(2) = +1.08V (3) I2(s) + 2e > 2I ;____________________>E°(3) = +0.53V Reaction (1) is given by: 2*Reaction(2)  Reaction(3) and so: Delta G°(1) = 2*Delta G°(2)  Delta G°(3) Now you use the equation: Delta G° = nF*Delta E° where n is the number of electrons of the reaction, so you have: 10F*Delta E°(1) = 2*[6F*Delta E°(2)]  [2F*Delta E°(3)] 10*Delta E°(1) = 12*Delta E°(2)  2*Delta E°(3) Delta E°(1) = (1/10)[12*Delta E°(2)  2*Delta E°(3)] = = (1/5)[6*Delta E°(2)  Delta E°(3)] = = (1/5)[6*1.08  0.53] = +1.19V. 



#3
Nov1507, 10:37 AM

P: 303

wowi would never have that of that...:S
how is the delta G of the first equation equal to the delta G if the second and third with those operations. Like how do we know to relate delta G? 



#4
Nov1507, 01:06 PM

P: 1,504

How to Find E°cell Values2*Reaction(2) = 2IO3(aq) + 12H+ +12e > 2I + 6H2O  Reaction(3) = 2I > I2(s) + 2e Now you sum them: 2IO3(aq) + 12H+ +12e + 2I > 2I + 6H2O + I2(s) + 2e you simplify canceling 2I and 2e on both members: 2IO3(aq) + 12H+ +10e > 6H2O + I2(s). About summing ΔG: if you have (1) A > B you know that ΔG(1) = G(B)  G(A) (2) C > D you know that ΔG(2) = G(D)  G(C) (3) A + C > B + D you know that ΔG(3) = G(B + D)  G(A + C) But G(B + D)  G(A + C) = G(B) + G(D)  G(A)  G(C) = = G(B)  G(A) + G(D)  G(C) = ΔG(1) + ΔG(2) 


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