Ok so i have looked at the question form every angle but I\'m not able to break

Question

Ok so i have looked at the question form every angle but I'm not able to break it down. Any help would be greatly appreciated. Okay here goes:

Consider the two half reactions: MnO4- + 8H+ +5e = Mn2+ + 4H2O E01/2 = +1.51v; Sn4+ + 2e = Sn2+ E01/2 = 0.154v. If these reactions are used to make a cell write the spontaneous cell reaction. Which half-reaction would go anodically? Which would be at the cathode? If we use 0.02M permanganate as a titrant to titrate 25 ml of 0.05M Sn2+ how many ml of titrant are needed to reach the equivalence point? What will be the half-cell voltage after adding 24.5 ml of MnO4-. After adding 25.5 ml of titrant? The combined solutions are buffered to pH2

Explanation / Answer

(a) 2MnO4- + 16H+ + 5Sn2+= 2Mn2+ + 5Sn4+ + 8H2O

2 moles of permanganate requires 5 moles of Sn2+

ml of Sn2+ = 5 *(0.02*25)/2*0.05 ml

  ml of Sn2+ = 25 ml

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