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Show all steps please :) A galvanic cell is prepared using the following two half-cells: (i)...

Question:

A galvanic cell is prepared using the following two half-cells: (i) MnO4 (0.273 M), Mn2+ (0.167 M), and H+ (1.0 M), and (ii)

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A galvanic cell is prepared using the following two half-cells: (i) MnO4 (0.273 M), Mn2+ (0.167 M), and H+ (1.0 M), and (ii) Fe2+ (0.247 M) and Fe3+ (0.389 M). The standard reduction potentials for the two half cells are: Mnoa + 8H+ + 5e – Mn2+ + 4H20 E° = 1.507 V Fe3+ + e- Fe2+ E° = 0.770 V a) Calculate the voltage for this galvanic cell. b) Write the balanced equation for the galvanic cell reaction indicating clearly the direction of the spontaneous cell reaction. c) Write the complete cell description. For Mn2+/MnO4 half-cell: name the electrode (anode of cathode), give its polarity (negative or positive) and state the reaction (oxidation or reduction). Note: Both half cells involve redox couples and, as such, use inert Pt electrodes.

Answers

Anode half reactions Conidation at Fee 3+ Fe er Cathode half reactions (Reduction) Mnog +84 + 5e Mn + + 4H2o - - - (11) EmnogE = (Framander unde) - arosa ley (0.2378 x 0.67 X (178 A. E = (Emmen met het z 12) - 6.05 lg (5.9867) M or. E = ((1507-0.77)(Reduction) 6 Mnog +8Ht+ se u mnt + 4 th (This is cathode) It is electropositive. Here reduction reaction occurs

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