Redox Reactions And Electrochemistry Question 45
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 $
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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 $ .
BETA
