As ECr/cr3+0=−0.72 V and EFe2+/Fe0=−0.42 V2Cr+3Fe2+⟶3Fe+2Cr3+Ecell =Ecell 0−60.0591log(Fe2+)3(Cr3+)2=(−0.42+0.72)−60.0591log(0.01)3(0.1)2=0.30−60.0591log(0.01)3(0.1)2=0.30−60.0591log10−610−2=0.30−60.0591log104Ecell =0.2606 V
JEE Main 2008 — Chemistry Physical Chemistry
Given ECr3+/Cr∘=−0.72 V,EFe2+//Fe∘=−0.42 V. The potential for the cell CrCr3+(0.1M)Fe2+(0.01M)Fe is
Held on 30 Apr 2008 · Verified 6 Jul 2026.
0.26 V
0.399 V
−0.339 V
−0.26 V
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