H2S + Ba2+ → __________ Write the reduction half reaction and the oxidation half reaction.Write the reduction half reaction and the oxidation half reaction. Write the balanced redox equation....

H2S + Ba2+ → __________

Write the reduction half reaction and the oxidation half reaction.

Write the reduction half reaction and the oxidation half reaction. Write the balanced redox equation. Calculate the cell potential, E°. Show your work. Indicate whether the reaction is spontaneous or nonspontaneous.

H2S + Ba2+ → __________

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bandmanjoe | Middle School Teacher | (Level 2) Senior Educator

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When solving redox equations, or oxidation-reduction equations, it is important to know what is being oxidized and what is being reduced.  By remembering the mnemonic "OIL RIG", which means "oxidation is loss and reduction is gain", we can identify which is which.  The H2S is being oxidized because the hydrogen is being separated from the sulfur, resulting in hydrogen ions (H+) being formed.  The barium +2 is being reduced, because it will lose its oxidation number +2.  So the two half reactions are: H2S + 2e- ---> 2H+  + S-2  and Ba+2  + S-2 ---> BaS.  The balanced redox equation is H2S + 2e- + Ba+2 ---> BaS  + H2.  The energy of the cell would involve breaking apart the H2S on the left side, which would be 344kj mol-1  per HS, so that times 2 would equal 688.  The formation of BaS is 400 kj mol-1 plus the formation of H2 is 436 kj mol-1, so to add those together would be 836 kj mol-1.  Subtract the reactants side from the products side and you get a net energy release of 148 kj mol-1.  Since there is a release of energy between the two sides on the products side, we would term this reaction as spontaneous.

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