A) 0.111V
B) 0.330V
C) 1.653V
D) 1.212V
Correct Answer: D
Solution :
[d] 1.212V Since \[E{}^\circ \,of\,F{{e}^{^{2+}}}/Fe\,\,is-\,\,0.441V\]i.e. lower than that of \[F{{e}^{3+}}/Fe\,\,{{24}^{2+}}\,which\,\,is\,\,(+\,0.771V)\]electrode. Therefore, oxidation occurs at \[F{{e}^{2+}}/Fe\] electrode and, reduction occurs at \[F{{e}^{3+}}/F{{e}^{2+}}\] electrode. \[\therefore \,\] \[EMF=E_{cathode\,}^{0}-E_{anode}^{0}\] \[=\,0.771\,V-[-0.441V]\] \[=0.771+0.441\] \[EMF=1.212V\]You need to login to perform this action.
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