Given standard electrode potential Fe2++2e−→Fe;E0=−0.440VFe3++3e−→Fe;E0=−0.036V. Calculate the electrode potential of cell.
A
+0.772V
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B
−0.772V
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C
+0.569V
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D
−0.569V
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Solution
The correct option is D+0.772V ΔGo=−nFEo For the reaction, Fe2++2e−→Fe ΔGo=−2×F×(−0.440V)=0.880F ....(I) For the reaction, Fe3++3e−→Fe ΔGo=−3×F×(−0.036)=0.108F ....(II) On subtracting equation (I) from equation (II) Fe3++e−→Fe2+ ΔGo=0.108F−0.880F=−0.772F Eo=−ΔGonF=−−0.772F1×F