For a silver-beryllium voltaic cell containing Ag+(aq) and Be2+(aq) solutions, do the following.
(a) Identify the cathode. (Include states-of-matter under the given conditions in your answer. Type INERT if an inert electrode must be used.)
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(b) Identify the half-reaction that occurs at the cathode. (Include states-of-matter under the given conditions in your answer.)
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To identify the cathode in a voltaic cell, we need to determine which half-reaction occurs at that electrode. In this case, we have a silver-beryllium voltaic cell with Ag+(aq) and Be2+(aq) solutions.
(a) The cathode in this voltaic cell is where reduction occurs. To identify it, we need to compare the reduction potentials of the Ag+ and Be2+ ions. The species with the higher reduction potential will be reduced and therefore will act as the cathode.
We can find the reduction potentials in a table of standard reduction potentials, such as the one found in most chemistry textbooks or online resources. By consulting this table, we find that the reduction potential for Ag+ is positive (Ag+ + e- → Ag) and the reduction potential for Be2+ is negative (Be2+ + 2e- → Be). This indicates that Ag+ has a higher reduction potential compared to Be2+.
Therefore, Ag+ will be reduced and act as the cathode.
(b) Now we need to identify the half-reaction that occurs at the cathode. The half-reaction for the reduction of Ag+ to Ag is:
Ag+ + e- → Ag
Note: The state-of-matter for Ag+ is (aq), indicating it is in aqueous solution.
So, the half-reaction that occurs at the cathode in this voltaic cell is the reduction of Ag+ to Ag.
I fear my answer will be wrong because you mention inert material and under the conditions. You haven't listed any conditions. I will assume 1 M conditions. Anyway, standard reduction potentials are as follows (from Google):
Ag^+(aq) + e ==> Ag(s) Estd red = 0.80 v
Be^2+(aq) + 2e --> Be(s) Estd red = -2.91 v
One of these must be oxidized and one reduced; therefore, make the most negative into an oxidation equation (by reversing it).
Be(s) + 2e --> Be^2+(aq) Eox = 2.91 v
Ag^+(aq) + e ==> Ag(s) Ered = 0.80 v
Reduction in a cell occurs at the cathode. Which electrode gains electrons? That will be the cathode. The equations you need are written also.