Write the balanced redox reaction: a) MnO4 – (aq) + Fe2+ (aq) → Mn2+ (aq) + Fe3+ (aq)

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Here is Reaction with Balance 10FeC2O4 + 6KMnO4 + 24H2SO4 ----> 3K2SO4+ 6MnSO4 +  5Fe2(SO4)3 + 24H2O + 20CO2 So we see that  6 moles of KMnO4 is required to oxidize 10 moles of  FeC2O4  Then, 1 mole of FeC2O4 would be...

We have Fe3+=x and Fe2+=.93-x then using charge balance.. 3*x+2*(0.93-x)=2*1 ( 1 for oxygen) x=.14 % of Fe3+=(.14/.93)*100=15% % of Fe2+=(.79/.93)*100=85%

Oxidizing agent = MnO4 – and Reducing agent = Fe2+  Oxidizing agent = Cr2O7 2– and Reducing agent = Fe2+ . 

Fe2+ => Fe3+ + 1e  So, 1F = 1 x 96500 C = 96500 C

Cl2 < MnO4 _ < Ce4+ 

Mn+2 has half-filled electronic configuration.

(a) MnO4 - + Fe2+ + H+ → Mn+2 + Fe+3 (b) Cr2O7 2- + Sn2+ + H+→ Cr+3 + Sn+4

Electronic configuration of Mn2+ is [Ar]18 3d5. Electronic configuration of Fe2+ is [Ar]18 3d6. It is known that half-filled and fully-filled orbitals are more stable. Therefore, Mn in (+2) state has a...