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I'm pretty sure you do the reaction in the anode and then the cathode. Make sure to separate different states with a single line and those with the same states with a comma. A double line goes where the salt bridge is. Do not also forget to add platinum where there is no solid on the anode or cathode side.
In a galvanic cell, Ecell is always greater than zero. When you are given two half equations, determine in which reaction the reactant is being oxidized and in which reaction the reactant is being reduced. In problem 14.13 part c in the first half reaction, Cl's oxidation number goes from 0 to 1-. Therefore, we know Cl is being oxidized (it gained electrons). In the second half reaction, H's oxidation number goes from 0 to 1+. Therefore, we know H is being reduced (it lost electrons). Oxidation happens at the anode, so we know the first half reaction will go on the left side of the cell diagram. Reduction happens at the cathode, so we know the second half reaction will go on the right of the cell diagram.
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