A) \[-0.10\ V\]
B) \[+0.18\ V\]
C) \[-0.54\ V\]
D) \[0.54\ V\]
Correct Answer: D
Solution :
\[{{I}^{-}}\] get oxidised to \[{{I}_{2}}\]hence will form anode and \[C{{r}_{2}}O_{7}^{2-}\]get reduced to \[C{{r}^{3+}}\]hence will form cathode. \[E_{Cell}^{0}=E_{Cathode}^{0}-E_{Anode}^{0}\]; \[E_{Cell}^{0}=E_{C{{r}_{2}}O_{7}^{-2}}^{{}}-E_{{{I}_{2}}}^{0}\] \[0.79=1.33-E_{{{I}_{2}}}^{0}\]; \[E_{{{I}_{2}}}^{0}=1.33-0.79\]; \[E_{{{I}_{2}}}^{0}=0.54\ V\].You need to login to perform this action.
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