Redox Reactions And Electrochemistry Ques 223

Question: $ Cr_2O_7^{2-}+{I^{-}}\to I_2+C{r^{3+}} $ $ E^0_{cell}=0.79\ V $ $ E_{Cr_2O_7^{2-}}^{0}=1.33\ V,\ E^0_{I_2} $ is [BVP 2004]

Options:

A) $ -0.10\ V $

B) $ +0.18\ V $

C) $ -0.54\ V $

D) $ 0.54\ V $

Show Answer

Answer:

Correct Answer: D

Solution:

$ {I^{-}} $ get oxidised to $ I_2 $ hence will form anode and $ Cr_2O_7^{2-} $ get reduced to $ C{r^{3+}} $ hence will form cathode.

$ E_{cell}^{0}=E_{Cathode}^{0}-E_{Anode}^{0} $ ;

$ E_{Cell^{0}}=E_{Cr_2O_7^{-2}}^{{}}-E_{I_2^{0}} $

$ 0.79=1.33-E_{I_2^{0}} $ ; $ E_{I_2^{0}}=1.33-0.79 $ ; $ E_{I_2^{0}}=0.54\ V $ .



जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक