-.- asked in 科學化學 · 1 decade ago

誰能幫我解答這題分析化學?!

KMnO4 is used as the titrant in a potentiometric titration of tin(II) to tin(IV), using a S.C.E. reference electrode. The potential of the indicator electrode at the equivalence point would be calculated as:

a) 7 E+ = 5 E°(MnO4-/Mn2+) + 2 E°(Sn4+/Sn2+)

b) 7 E+ = 2 E°(MnO4-/Mn2+) + 5 E°(Sn4+/Sn2+)

c) 7 E+ = 5 E°(MnO4-/Mn2+) + 2 E°(Sn4+/Sn2+) + 0.05916 log [H+]8

答案是C

請幫我詳解...謝謝^^

1 Answer

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  • ?
    Lv 7
    1 decade ago
    Favorite Answer

    半反應為:

    MnO4- + 8H+ + 5e- --> Mn2+ + 4H2O G1=-5FE1=-5FE1o+0.0591log[MnO4-]/[Mn2+]+0.0591log[H+]8 (1)

    Sn4+ + 2e- --> Sn2+ G2=-2FE2=-2FE2o +0.0591log([Sn4+]/[Sn2+] (2)

    全反應=2*(1)-5*(2):

    2MnO4- + 16H+ + 5Sn2+ --> 2Mn2+ + 5Sn4+ + 8H2O G=G1-G2=-7FE=-5FE1+2FE2+0.0591log[H+]8

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