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.
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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