pls help me electrode potential exam question
MnO4– + 8H+ + 5e– ---> Mn2+ + 4H2O +1.51
Cl2(g) + 2e– ----> 2Cl–(aq) +1.36
Cr2O72– + 14H+ + 6e– ---> 2Cr3+ + 7H2O +1.33
Use information from the table in part (a) to determine the minimum volume, in cm3, of 0.500 mol dm–3 sulfuric acid that is required for a titre of 25.0 cm3 of 0.0200 mol dm–3 potassium manganate(VII) solution. Show your working
the markschme says work moles of H+ =25x0.02 x8/1000 = 4x10-3
then work moles h2so4 = 2x10-3 , why do u half the moles H+, I don't know what the overall equation is, the reduction of MnO4- is one of the half equations, but I don't know what the other half equation is
Cl2(g) + 2e– ----> 2Cl–(aq) +1.36
Cr2O72– + 14H+ + 6e– ---> 2Cr3+ + 7H2O +1.33
Use information from the table in part (a) to determine the minimum volume, in cm3, of 0.500 mol dm–3 sulfuric acid that is required for a titre of 25.0 cm3 of 0.0200 mol dm–3 potassium manganate(VII) solution. Show your working
the markschme says work moles of H+ =25x0.02 x8/1000 = 4x10-3
then work moles h2so4 = 2x10-3 , why do u half the moles H+, I don't know what the overall equation is, the reduction of MnO4- is one of the half equations, but I don't know what the other half equation is
(edited 5 years ago)
Original post by usernamenew
MnO4– + 8H+ + 5e– ---> Mn2+ + 4H2O +1.51
Cl2(g) + 2e– ----> 2Cl–(aq) +1.36
Cr2O72– + 14H+ + 6e– ---> 2Cr3+ + 7H2O +1.33
Use information from the table in part (a) to determine the minimum volume, in cm3, of 0.500 mol dm–3 sulfuric acid that is required for a titre of 25.0 cm3 of 0.0200 mol dm–3 potassium manganate(VII) solution. Show your working
the markschme says work moles of H+ =25x0.02 x8/1000 = 4x10-3
then work moles h2so4 = 2x10-3 , why do u half the moles H+, I don't know what the overall equation is, the reduction of MnO4- is one of the half equations, but I don't know what the other half equation is
Cl2(g) + 2e– ----> 2Cl–(aq) +1.36
Cr2O72– + 14H+ + 6e– ---> 2Cr3+ + 7H2O +1.33
Use information from the table in part (a) to determine the minimum volume, in cm3, of 0.500 mol dm–3 sulfuric acid that is required for a titre of 25.0 cm3 of 0.0200 mol dm–3 potassium manganate(VII) solution. Show your working
the markschme says work moles of H+ =25x0.02 x8/1000 = 4x10-3
then work moles h2so4 = 2x10-3 , why do u half the moles H+, I don't know what the overall equation is, the reduction of MnO4- is one of the half equations, but I don't know what the other half equation is
You only need the first half-equation, which shows you the ratio of manganatge(VII) ions to hydrogen ions is 1:8
Hence you need 8 times the concentration of hydrogen ions.
Each sulfuric acid molecule provides two hydrogen ions, so you divide the moles of hydrogen ions by 2 to get moles of sulfuric acid.
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