For the voltaic cell shown, calculate the standard emf. Pt(s) l F-(aq) l F2(g) l

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Explanation / Answer

For this question, you need to balance the two half-reactions to know how many moles of electrons are being transferred. In cell notation we know that the anode (where oxidation occurs) is on the left, and the cathode (where reduction occurs) is on the right. Mn2+ + 2e- -> Mn (reduction, transfer of two electrons) 2F- -> F2 + 2e- (the two electrons come from this oxidation reaction - so we have balanced the reaction, by adding the 2 stochiometric coefficient) Now we use the formula ?G° = -nFE°cell Where ?G° is the Gibbs free energy for the cell under standard conditions, where n=moles of electrons, F=Faraday's constant (96485 Coloumbs/mol) - that is, every mole of electrons transferred has 96485 Coloumbs So ?G° = - 2 * 96485 C/mol * -4.05 V = 781529 J = 782 kJ (rounded to 3 sig figures)

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