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A galvanic cell is powered by the following redox reaction: 2IO₃⁻ (aq) + 12H⁺ (aq) + 10Cu⁺ (aq) → I₂ (s) + 6H₂O (I) + 10Cu²⁺ (aq) Answer the following questions about this cell. 1. Write a balanced equation for the half-reaction that takes place at the cathode. 2.Write a balanced equation for the half-reaction that takes place at the anode. 3.Calculate the cell voltage under standard conditions. Round your answer to 2 decimal places.

Answer

Part 1: At the cathode, Cu2+ ions undergo reduction and form Cu+ ions; the corresponding half-reaction is 10Cu2+ (aq) + 2e- → 10Cu+ (aq). Part 2: At the anode, IO3- ions experience oxidation and produce I2, with the corresponding half-reaction being 2IO3- (aq) → I2 (s) + 6e-. Part 3: The cell voltage indicates the difference in electrical potential between two electrodes. Under standard conditions, the cell voltage equals the difference between the standard reduction potentials of the half-reactions, which can be calculated as follows: Ecell = E°cathode - E°anode, where Ecell equals -0.20 V, E°cathode equals 0.34 V, and E°anode equals +0.54 V. Hence, Ecell = 0.34 V - (+0.54 V) = -0.20 V. Rounded off to 2 decimal places, Ecell equals -0.20 V. For more information on cell voltage, please visit brainly.com/question/30226316 #SPJ1.