A) - 0.10 V
B) + 0.18 V
C) - 0.54 V
D) 0.54 V
Correct Answer: D
Solution :
Key Idea: (i) Decide cathode and anode with the help of given cell reaction. (ii) Find electrode potential by using formula \[E_{cell}^{o}=E_{C}^{o}-E_{A}^{o}\] \[C{{r}_{2}}{{O}_{7}}^{2-}+{{1}^{-}}\to {{I}_{2}}+C{{r}^{3+}}\] In this reaction \[{{I}^{-}}\] is oxidised to \[{{I}_{2}},\,\therefore {{I}_{2}}\] is anode. \[C{{r}_{2}}{{O}_{7}}^{2-}\] is reduced to \[C{{r}^{3+}}\therefore Cr\] forms cathode. Given, \[{{E}^{o}}_{cell}=0.79\,V\,\,{{E}^{o}}_{cathode}=1.33\,V\] \[{{E}^{o}}_{cell}={{E}^{o}}_{cathode}-{{E}^{o}}_{anode\,({{I}_{2}})}\] \[0.79=1.33-{{E}^{o}}_{anode\,({{I}_{2}})}\] or \[{{E}^{o}}_{anode}=1.33-0.79=0.54V\]
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