Electrochemistry Homework Set Worked Out Solutions

Cyesbr2isfavorablyreducedandthustendstooxidizeother

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Unformatted text preview: rably, while Ni2+ is not. The reverse of the reduction reaction is an oxidation reaction, and the sign of E° flips. Thus the oxidation of Ni (s) is favorable. 2+ is not favorably reduced and thus would not do (A) NO‐Ni so to oxidize Sn2+. The oxidation of Sn2+ is likewise unfavorable. (B) NO‐ although Sn4+ is favorably reduced and thus tends to oxidize other species, Br‐is not favorably oxidized to give Br2. The E° values for Sn4+ and Br2 are equal, thus neither has the upper hand. (C) YES‐Br2 is favorably reduced and thus tends to oxidize other species, and Ni (s) is favorably oxidized. (D) NO‐Ni2+ is not favorably reduced. Also, Br2 is reduced, not oxidized, to give Br‐. 14. Flip half‐reactions as needed so that the two combined half‐reactions give the full chemical equation. **Make sure to switch the sign of E°** Multiply the half‐reactions to give the right stoichiometric coefficients, but do not change the value of E°! Flip the first half‐reaction and multiply by 2: 2Ce3+ 2e‐ + 2Ce4+ E°=‐1.61 V Do not change the second half‐reaction: ‐ 2Br‐ E°=+1.06 V Br2 + 2e Add the two half‐reactions (the electrons should cancel on either side) and add the two E° values to obtain the overall E° for the reaction: 2Ce3+ 2e‐ + 2Ce4+ E°=‐1.61 V ‐ 2Br‐ E°=+1.06 V Br2 + 2e 3+2Br‐+2Ce4+ E°=‐0.55 V Br2+2Ce 15. Use the equation: E=E°­(0.0592/n)*logQ (valid at 298 K, n=number of electrons transferred in the reaction) For Cu2+(aq) + Fe (s) Cu (s) + Fe2+(aq), Q=[Fe2+]/[Cu2+]=0.40/0.040=10 Plug into the equation: E=0.78‐(0.0592/2)*log(10)=0.75 V Dr. Fus Electrochemistry Problem Set Solutions Chemistry 123 16. To balance a redox reaction in acidic aqueous solution: 1. Divide the equation into 2 half‐reactions (one oxidation, one reduction) 2. Balance each half‐reaction: ‐first, balance elements other than H and O. ‐next, balance O atoms, adding H2O as needed. ‐balance H atoms, adding H+ as needed. ‐balance charge, adding e‐ as needed. 3. Multiply half‐reactions by integers if needed in order to make the number of electrons lost in one half‐reaction equal to the number of electrons gained in the other. 4. Add the half‐reactions, cancelling species on both sides of the combined equation if possible....
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This homework help was uploaded on 03/28/2014 for the course CHEM 123 taught by Professor Larosa during the Spring '06 term at Ohio State.

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