14.23 Cathode vs Anode differentiation [ENDORSED]
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14.23 Cathode vs Anode differentiation
I don't understand why in chapter 14 question 23 part B, a value of 0.92 was used as the cathode and 1.09 as the anode. After separating the equation into its half reaction components (with the electrons added to the reactants side), how do we know which half reaction we have to use as anode and which to use as the cathode?
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Re: 14.23 Cathode vs Anode differentiation [ENDORSED]
The anode is the electrode where oxidation takes place and the cathode is the electrode where reduction takes place.
When examining half reactions from a table of standard reduction potentials, the higher the value of Eo, the more likely that half reaction is to be reduced. Thus, comparing the values of Eo for the two half reactions, the one with the "more positive" value will occur at the cathode in a galvanic cell.
However, in problem 14.23 (b), you must write the half reactions as written in the given full redox reaction. Thus, you have not choice in the matter. As written, Br- (aq) must lose electrons (be oxidized) and Hg2+ (aq) must gain electrons (be reduced). Therefore, Hg2+ (aq) is the reduced at the cathode because that is where reduction takes place and Br- (aq) becomes oxidized at the anode. If you were going to use our standard formula to calculate Eocell, then you would look up the reduction forms of each half reaction and your calculation would be (+0.92 V) - (+1.09 V) = -0.17 V. This cell is not spontaneous (not galvanic) because Eocell is negative.
When examining half reactions from a table of standard reduction potentials, the higher the value of Eo, the more likely that half reaction is to be reduced. Thus, comparing the values of Eo for the two half reactions, the one with the "more positive" value will occur at the cathode in a galvanic cell.
However, in problem 14.23 (b), you must write the half reactions as written in the given full redox reaction. Thus, you have not choice in the matter. As written, Br- (aq) must lose electrons (be oxidized) and Hg2+ (aq) must gain electrons (be reduced). Therefore, Hg2+ (aq) is the reduced at the cathode because that is where reduction takes place and Br- (aq) becomes oxidized at the anode. If you were going to use our standard formula to calculate Eocell, then you would look up the reduction forms of each half reaction and your calculation would be (+0.92 V) - (+1.09 V) = -0.17 V. This cell is not spontaneous (not galvanic) because Eocell is negative.
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