Given the standard electrode potentials. EFe2+/Fe∘=−0.44V and EH+/O2/H2O∘=1.23V

Ecell∘=EH+/O2/H2O∘−EFe2+/Fc∘=1.23−(−0.44)=1.67V

Given the standard electrode potentials. EFe2+/Fe∘=−0.44V and EH+/O2/H2O∘=1.23V

Ecell∘=EH+/O2/H2O∘−EFe2+/Fc∘=1.23−(−0.44)=1.67V

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Given the standard electrode potentials. $E_{{Fe}^{2 +} / Fe}^{\circ} = - 0 .44 V$ and $E_{H^{+} / O_{2} / H_{2} O}^{\circ} = 1 .23 V$

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-0.79 V

-1.67 V

1.67 V

+0.79 V

answer is C.

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$E_{cell}^{\circ} = E_{H^{+} / O_{2} / H_{2} O}^{\circ} - E_{{Fe}^{2 +} / Fc}^{\circ} = 1 .23 - (- 0 .44) = 1 .67 V$

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Given the standard electrode potentials. EFe2+/Fe∘=−0.44V and EH+/O2/H2O∘=1.23V

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Given the standard electrode potentials. $E_{{Fe}^{2 +} / Fe}^{\circ} = - 0 .44 V$ and $E_{H^{+} / O_{2} / H_{2} O}^{\circ} = 1 .23 V$