FeO^2- 4 → +2.0 V Fe^3+ 0.8V → Fe^2+ →-0.5V Fe^0 In the above diagram, the standard electrode potentials are given in volts (over the arrow).

Question

\(FeO^{2-} _ 4 \xrightarrow{+2.0V} Fe^{3+} \xrightarrow{0.8V} Fe^{2+} \xrightarrow{0.5V} Fe^0\)

In the above diagram, the standard electrode potentials are given in volts (over the arrow). The value of \(E^\ominus _{FeO^{2-} _4/ Fe^{2+}}\) is

(1) 1.7 V 

(2) 1.2 V 

(3) 2.1 V 

(4) 1.4 V

Answer 1

Correct option is: (1) 1.7 V 

\(\Delta G^\circ _4 = \Delta G_1 ^{\circ} + \Delta G^{\circ} _2\)

\(\Rightarrow -n_4 FE^{\circ}_4 = -n_1FE^{\circ}_1 - n_2FE^\circ_2\)

\(\Rightarrow +4 E^\circ_4 = 3 \times 2 + (1 \times 0.8)\)

\(\Rightarrow E^\circ_4= \frac{6.8}{4}V\)

\(\Rightarrow E^\circ_4 = 1.7V\)