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Reduction Potential

Guest oxaloacetate

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Guest oxaloacetate

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





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?!?!

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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


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