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.)
(b) Identify the half-reaction that occurs at the cathode. (Include states-of-matter under the given conditions in your answer.)
That's not the right answer
To identify the cathode and the half-reaction that occurs at the cathode in a silver-beryllium voltaic cell, we need to consider the standard reduction potentials for the half-reactions involving Ag+ and Be2+ ions.
(a) First, let's compare the standard reduction potentials for the silver and beryllium half-reactions:
Ag+(aq) + e- -> Ag(s) E°cell = +0.799 V
(Be2+(aq) + 2e- -> Be(s) E°cell = -1.85 V
The standard reduction potential for the silver half-reaction is positive, indicating that Ag+ ion is being reduced (gaining electrons) to form Ag metal at the cathode. Hence, the cathode in this voltaic cell is where silver is being reduced to form solid silver (Ag(s)).
(b) The half-reaction that occurs at the cathode is the reduction of Ag+ ions to Ag(s):
Ag+(aq) + e- -> Ag(s)
The states-of-matter under the given conditions are:
- Ag+(aq) represents silver ions dissolved in an aqueous solution.
- Ag(s) represents solid silver formed at the cathode.
Therefore, in this silver-beryllium voltaic cell, the cathode is Ag(s), and the half-reaction occurring at the cathode is Ag+(aq) + e- -> Ag(s).