Try it on paper first. A hint, the key idea and the worked solution are here when you need them.
The standard potentials of two couples are and . Each 2+ ion can reduce hydrogen ions: . Use . Find the standard emf for samarium and for ytterbium, and the Gibbs energy change for each. One electron is transferred.
The 3+ ion of a lanthanoid is with . Unpaired electrons: up to 7, then . Spin-only moment: . An empty, half-full or full 4f subshell (, , ) makes an ion unusually stable. That explains , , and . For a cell, with .
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Step 1. The emf for samarium. The hydrogen ion is reduced, so it is the cathode. .
Step 2. The emf for ytterbium. .
Step 3. Gibbs energy. For samarium, . For ytterbium, .
Answer: and , with and .
Check. Both values are negative, so each 2+ ion is able to reduce acid. The samarium reaction has the larger drive.
Watch out. The emf says how far a reaction is able to go. It does not say how fast. In practice, these ions keep for a short time in water without air, because the reaction with hydrogen ions can be slow.
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