5 Calculating standard cell potential

Using electrode potentials The most useful application of electrode potentials is to show the direction of spontaneous change for redox reactions The easiest way to use electrode potentials is as follows: For any two half equations The more negative half cell will always oxidise (go backwards) Mg2+ (aq) + 2e-
Mg(s) E= -2.37V Cu2+ (aq) + 2e-
Cu (s) E = +0.34V The more positive half cell will always reduce (go forwards) The reaction would be Mg + Cu2+
Cu + Mg 2+ If we want to work out the Ecell that corresponds to this spontaneous change then use Ecell = Ered – Eox A spontaneous change will always have a positive Ecell Zn2+(aq) + 2e-
Zn(s) E= – 0.76V Fe2+(aq) + 2e-
Fe(s) E= -0.44V The more positive electrode will reduce and go from left to right Fe2+ (aq) +2e-
Fe(s) Electrons arrive at this electrode and are absorbed (gained) To get the full equation of the reaction add the two half reactions together, cancelling out the electrons. Zn + Fe2+
Fe + Zn2+ The most negative electrode will oxidise and go from right to left The half equation is therefore Zn(s)
Zn2+ (aq) +2eElectrons are given off (lost) and travel to positive electrode Using series of standard electrode potentials Li+ + e-
Li -3.03V Mn2+ + 2e-
Mn -1.19V 2H+ + 2e-
H2 0V Ag+ + e-
Ag +0.8V F2 + 2e-
2F- +2.87 As more +ve increasing tendency for species on left to reduce, and act as oxidising agents As more -ve increasing tendency for species on right to oxidise, and act as reducing agents oxidation reduction Most strong reducing agents found here Most strong oxidising agents found here The most powerful reducing agents will be found at the most negative end of the series on the right (ie the one with the lower oxidation number) The most powerful oxidising agents will be found at the most positive end of the series on the left (ie the one with the higher oxidation number) If we want to work out the Ecell from two standard electrode potentials then use Ecell = Ered – Eox N Goalby chemrevise.org 8 O2 (g) + 4H+ (aq) + 4e– → 2H2O(I) Eo+1.23V F2 (g) + 2e– → 2F– (aq) Eo +2.87V Use electrode data to explain why fluorine reacts with water. Write an equation for the reaction that occurs. Cl2 (aq) + 2e– → 2Cl– (aq) Eo+1.36V 2HOCl(aq) + 2H+ (aq) + 2e– → Cl2 (aq) + 2H2O(I) Eo+1.64V H2O2 (aq) + 2H+ (aq) + 2e– → 2H2O(I) Eo +1.77V O2 (g) + 4H+ (aq) + 4e– → 2H2O(I) Eo +1.23V Use data from the table to explain why chlorine should undergo a redox reaction with water. Write an equation for this reaction. Use the half-equations to explain in terms of oxidation states what happens to hydrogen peroxide when it is reduced. H2O2 (aq) + 2H+ (aq) + 2e– → 2H2O(I) Eo+1.77V O2 (g) + 2H+ (aq) + 2e– → H2O2 (aq) Eo +0.68V Example 1 First apply idea that more positive Eo will reduce (go forward) and more negative Eo will oxidise (go backwards) reduce oxidise Explanation to write As Eo F2 /F- > Eo O2 /H2O, F2 will oxidise H2O to O2 Equation 2F2 (g) + 2H2O(I) → 4F– (aq) + O2 (g) + 4H+ (aq) Can also work out Ecell and quote it as part of your answer Ecell = Ered – Eox = 2.87-1.23 =1.64V Remember to cancel out electrons in full equation Example 2 First select relevant half equations by considering the Eo values and applying the idea that more positive Eo will reduce (go forward) and more negative Eo will oxidise (go backwards) Cl2 (aq) + 2e– → 2Cl– (aq) Eo+1.36V O2 (g) + 4H+ (aq) + 4e– → 2H2O(I) Eo +1.23V reduce oxidise Explanation to write As Eo Cl2 /Cl- > Eo O2 /H2O, Cl2 will oxidise H2O to O2 Equation 2Cl2 (g) + 2H2O(I) → 4Cl– (aq) + O2 (g) + 4H+ (aq) Fe3+ (aq) + e– → Fe2+ (aq) Eo +0.77V 2H+ (aq) + 2e– → H2 (g) Eo 0.00V Fe2+ (aq) + 2e– → Fe(s) Eo–0.44V Suggest what reactions occur, if any, when hydrogen gas is bubbled into a solution containing a mixture of iron(II) and iron(III) ions. Explain your answer. First select relevant half equations by considering the Eo values and applying the idea that more positive Eo will reduce (go forward) and more negative Eo will oxidise (go backwards) Fe3+ (aq) + e– → Fe2+ (aq) Eo +0.77V 2H+ (aq) + 2e– → H2 (g) Eo 0.00V oxidise reduce Explanation to write Fe3+ will be reduced to Fe2+ by H2 oxidising to H+ because Eo Fe3+ /Fe2+ > Eo H+ /H2 Equation 2Fe3+ (aq) + H2 (g) → 2Fe2+ (aq) + 2H+ (aq) Example 3 reduce oxidise Explanation to write As Eo H2O2 /H2O > Eo O2 /H2O2 , H2O2 disproportionates from -1 oxidation state to 0 in O2 and -2 in H2O 2H2O2 (aq) → 2H2O(I) + O2