Question

Use the appropriate standard reduction potentials below to determine the equilibrium constant at 356 K for the following reaction under acidic conditions. 4H+(aq) + MnO2(s) + 2Fe2+(aq) ? Mn2+(aq) + 2Fe3+(aq) + 2H2O(l) Standard reduction potentials: MnO2(s) + 4H+(aq) + 2e- ? Mn2+(aq) + 2H2O(l) E° = 1.23 V Fe3+(aq) + e- ? Fe2+(aq) E° = 0.770 V

          Use the appropriate standard reduction potentials below to determine the equilibrium constant at 356 K for the following reaction under acidic conditions.

4H+(aq) + MnO2(s) + 2Fe2+(aq) ? Mn2+(aq) + 2Fe3+(aq) + 2H2O(l)

Standard reduction potentials:
MnO2(s) + 4H+(aq) + 2e- ? Mn2+(aq) + 2H2O(l) E° = 1.23 V
Fe3+(aq) + e- ? Fe2+(aq) E° = 0.770 V

Use the appropriate standard reduction potentials below to determine the equilibrium constant at 356 K for the following reaction under acidic conditions.

4H+(aq) + MnO2(s) + 2Fe2+(aq) ? Mn2+(aq) + 2Fe3+(aq) + 2H2O(l)

Standard reduction potentials:
MnO2(s) + 4H+(aq) + 2e- ? Mn2+(aq) + 2H2O(l) E° = 1.23 V
Fe3+(aq) + e- ? Fe2+(aq) E° = 0.770 V

Added by Erika C.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert David Collins

04/23/2023

Step 1

23 V 2) Fe3+(aq) + e- -> Fe2+(aq) E° = 0.770 V Now, we need to balance the electrons in both half-reactions. Since the first half-reaction involves 2 electrons and the second half-reaction involves 1 electron, we need to multiply the second half-reaction by 2 to Show more…

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Use the appropriate standard reduction potentials below to determine the equilibrium constant at 356 K for the following reaction under acidic conditions. 4H+(aq) + MnO2(s) + 2Fe2+(aq) ⟶ Mn2+(aq) + 2Fe3+(aq) + 2H2O(l) Standard reduction potentials: MnO2(s) + 4H+(aq) + 2e- ⟶ Mn2+(aq) + 2H2O(l) E° = 1.23 V Fe3+(aq) + e- ⟶ Fe2+(aq) E° = 0.770 V