Q. $\mathrm{FeO_4^{2-}}\xrightarrow{+2.0\,V}\mathrm{Fe^{3+}}\xrightarrow{0.8\,V}\mathrm{Fe^{2+}}\xrightarrow{-0.5\,V}\mathrm{Fe^0}$ In the above diagram, the standard electrode potentials are given in volts (over the arrow). The value of $E^\circ_{\mathrm{FeO_4^{2-}/Fe^{2+}}}$ is
A) 2.1 V
B) 1.7 V
C) 1.4 V
D) 1.2 V
Explanation
[Solution by Dr. Abubacker] $\mathrm{FeO_4^{2-}\to Fe^{3+}}$ is a 3-electron step ($\Delta G=-3F(2.0)$) and $\mathrm{Fe^{3+}\to Fe^{2+}}$ a 1-electron step ($-1F(0.8)$). $E^\circ=\frac{3(2.0)+1(0.8)}4=1.7$ V. Answer (2).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.