At 360 K, the galvanic reaction MnO_4^- (aq)+8 ,H^+(aq)+5 ,Fe^ 2+ (aq) Mn^ 2+ (aq)+4…

CHEMISTRY · Subjective · A Level · Chemistry · Electrode potentials · Electrode potentials

Problem

At 360 K, the galvanic reaction MnO4−(aq)+8 H+(aq)+5 Fe2+(aq)→Mn2+(aq)+4 H2O(l)+5 Fe3+(aq)\mathrm{MnO_4^- (aq)+8\,H^+(aq)+5\,Fe^{2+}(aq)\rightarrow Mn^{2+}(aq)+4\,H_2O(l)+5\,Fe^{3+}(aq)} has a standard cell potential of E∘=0.74 VE^\circ = 0.74\,\text{V}. The reaction involves n=5n=5 electrons. The mixture contains [MnO4−]=5.0 mM[\mathrm{MnO_4^-}]=5.0\,\text{mM}, [Mn2+]=0.20 M[\mathrm{Mn^{2+}}]=0.20\,\text{M}, [Fe2+]=0.10 M[\mathrm{Fe^{2+}}]=0.10\,\text{M}, [Fe3+]=0.50 M[\mathrm{Fe^{3+}}]=0.50\,\text{M}, and [H+]=0.010 M[\mathrm{H^+}]=0.010\,\text{M}. Calculate the cell potential, EE, at 360 K. (Constants: R=8.314 J mol−1K−1R=8.314\,\text{J\,mol}^{-1}\text{K}^{-1}, F=9.65×104 C mol−1F=9.65\times10^{4}\,\text{C\,mol}^{-1}.)

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