Answer:
2Rb(s) + Sr^+(aq) → 2Rb^+ (aq) + Sr(s)
Explanation:
Rubidium has a more negative reduction potential (-2.98 V) compared to strontium (-2.89 V).
Hence, in a redox reaction involving rubidium and strontium, rubidium will be oxidized while strontium is reduced.
The balanced redox reaction equation is obtained from;
Oxidation half equation;
2Rb(s) ---->2Rb^+(aq) + 2e
Reduction half equation;
Sr^2+(aq) + 2e ----> Sr(s)
Overall reaction equation;
2Rb(s) + Sr^+(aq) → 2Rb^+ (aq) + Sr(s)
