Unformatted text preview: tential will correspond to the reduced species. This provides an overall positive E°. Reverse the appropriate half‐reaction (remembering to flip the sign of E°), multiply them by the appropriate integer to make the electrons lost in one reaction equal those gained in the other (do NOT multiply E° by the integer‐it is an intensive property and does not change), and add the potentials: 2Cr(s)2Cr3+(aq) + 6e‐ E°=+0.74 V E°=‐0.28 V 3Co2+(aq) + 6e‐3Co(s) E°=+0.46 V 23. Reverse the appropriate half‐reaction (remembering to flip the sign of E°), multiply each half‐reaction by the appropriate integer to make the electrons lost in one reaction equal those gained in the other (do NOT multiply E° by the integer‐this does not change), and add the potentials. Verify that the appropriate species cancel out: 3Mn(s)3Mn2+(aq) + 6e‐ E°=+1.18 V E°=+1.00 V 2AuCl4‐(aq) + 6e‐2Au(s) + 8Cl‐(aq) 3Mn(s) + 2AuCl4‐(aq) 3Mn2+(aq) + 2Au(s) + 8Cl‐(aq) E°=+2.18 V 24. Cu(s) + Pd2+(aq)Cu2+(aq) + Pd(s) E°=+0.650 V 2+(aq) + 2e‐Cu(s) Cu E°=+0.337 V Pd2+(aq) + 2e‐Pd(s) E°=+0.987 V 25. Break the total equation into two half‐reactions to solve for the desired reduction potential. Add in the appropriate number of electrons in each to balance charge. Remember, multiplying the half‐reaction by an integer does not affect the potential value! 6I‐(aq) + Cr2O72‐(aq) + 14H+(aq) 3I2(aq) + 2Cr3+(aq) + 7H2O(l) E°=+0.79 V Cr2O72‐(aq) + 14H+(aq) + 6e‐ 2Cr3+(aq) + 7H2O(l) E°=+1.33 V ‐(aq) 3I (aq) + 6e‐ 6I E°=X 2 X+1.33=0.79 X= ‐0.54 V **Remember, we are asked for the STANDARD potential for I2 reduced to I. The value, X, we solved for corresponds to the oxidation of I to I2. Therefore, the answer is of the same magnitude, but opposite sign, of X, namely E°=+0.54 V. 26. Because it is an aqueous solution of a salt instead of solely the ionic compound, this electrolysis is complicated by the presence of water. Therefore, all three products are formed. Thus, the answer is D. Dr. Fus Electrochemistry Problem Set Solutions Chemistry 123 27. For these problems, follow this sequence: current*timecoulombs (charge) coulombs/Faraday’s constant (96,486 C per mole electrons)mol e‐ mol e‐ * __mol metal oxidized or reduced per __mol e‐ mol metal*molecular weight of...
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- Spring '06
- Electrochemistry, Oxidation state, Dr. Fus, Electrochemistry Problem Set Solutions, mol e‐