emf of the cell reaction 3Sn4+ + 2Cr--> 3Sn2+ + 2Cr3+ is 0.89 V.Calculate delta G.{F=96500 C+mol and VC=J}

Dear Student,

G°=-nFE°Here, E°=0.89 VAt cathode: Sn4+ + 2e- Sn2+  ..(i)At anode: Cr  Cr3+ + 3e-   ...(ii)Multiplying (i) by 3 and (ii) by 2 and adding both, we get 3Sn4+ + 2Cr 3Sn2+ +2 Cr3+ Here, n=6So, G°=-6×96500×0.89 = -515.3 kJ

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!I too need its Ans
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 n = 6 here now use the formula -nFe*
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