A) \[-0.10V\]
B) \[+0.18V\]
C) \[-0.54V\]
D) \[0.54V\]
Correct Answer: D
Solution :
\[\overset{+6}{\mathop{C{{r}_{2}}O_{7}^{2-}}}\,+{{I}^{-}}\xrightarrow{{}}{{I}_{2}}+C{{r}^{3+}}\] Here, \[{{I}^{-}}\] is undergoing oxidation while \[C{{r}_{2}}O_{7}^{2-}\] is undergoing reduction, hence \[{{I}_{2}}\] will form anode and \[C{{r}^{3+}}\] will form cathode in the cell. Hence, \[{{E}^{o}}_{cell}{{E}^{o}}_{cathode}-{{E}^{o}}_{anode}\] \[\therefore \] \[0.79=1.33-{{E}^{o}}_{{{I}_{2}}}\] \[\therefore \] \[{{E}^{o}}_{{{I}_{2}}}=1.33-0.79\] \[=+0.54V\]You need to login to perform this action.
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