Guest oxaloacetate Posted April 30, 2003 Report Share Posted April 30, 2003 I am yet again working on another potential mcat question and am not sure where I am going wrong with this question.... A student attempts to id the metals involved in several oxidation reduction reactions of the form: Y2+ (aq) + Z (s) ----- Y (s) + Z 2+ (aq) The student has been told that all half-reactions occurring in the cells are listed in Table 1. Table 1: Half Reaction Ba (aq) + 2 e ----- Ba (s) -2.9V Mg (aq) + 2e ----- Mg (s) -2.4V Be (aq) + 2e ------ Be (s) -1.9V Zn (aq) + 2e ----- Zn (s) -0.8V Cd (aq) + 2e ------ Cd (s) -0.4V Experiment 1: The voltmeter is connected to the electrodes of one cell and registers a potential difference of 0.5 V QUESTION: If it is known that Z(s) for experiment 1 is solid magnesium, the student should determine that the identity of Y (s) is: a. Barium b. Beryllium c. Zinc d. Cadmium In order to determine what the id of Y was I used the forumula E cell = E cathode - E anode. Using this... 0.5 V = x - 2.4V x = 2.9 V This answer led me to believe that barium is the correct answer. However, it is not. Beryllium is the right answer. What am I doing wrong?!?! Link to comment Share on other sites More sharing options...
Guest SoHo Posted April 30, 2003 Report Share Posted April 30, 2003 You're going from solid Mg to aq. therefore the sign of the reduction potential as given should be changed from negative to positive (-2.4 to 2.4). So you're equation should be: 2.4+x = 0.5 x=-1.9 Link to comment Share on other sites More sharing options...
Guest Talon01 Posted May 1, 2003 Report Share Posted May 1, 2003 The MCAT was LAST saturday :b Link to comment Share on other sites More sharing options...
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