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Please help The following electrochemical potentials were measured for reactions

ID: 961341 • Letter: P

Question

Please help The following electrochemical potentials were measured for reactions between the following set of metals: Cs, K, Na, Rb. Rb(s) + Na^+(aq) rightarrow Rb^+(aq) + Na(s) E^0 = 0.17 V Na(s) + Cs^+(aq) rightarrow Na^+(aq) + Cs(s) E^0 = -0.22 V Rb(s) + K^+(aq) rightarrow Rb^+(aq) + K(s) E^0 = 0.05 V K(s) + Cs^+(aq) rightarrow K^+(aq) + Cs(s) E^0 = -0.10 V If you arbitrarily assume that the reduction potential of K^+ is 0.0 V. K^+(aq) + e^- rightarrow K(s) E^0 = 0.0 V calculate the reduction potentials for the remaining three metals relative to the reduction potentail for K.

Explanation / Answer

ANSWER

In reaction 3rd Rb is oxidised byK

Rb + K+ -----> Rb+ + K

Eo = Eocathode - Eoanode

K acts cathode and Rb as anode, hence we can wright

0.05 = Eo (K) - Eo(Rb)

Electrode potential of K (Eo (K)) = 0.00V

0.05 = 0.00V - Eo(Rb)

Eo(Rb) = - 0.05V

2. Rb + Na+ -------> Rb+ + Na

Eo =  Eo (Na) - Eo(Rb)

0.17V = Eo (Na) -(-0.05)

Eo (Na) = 0.17 - 0.05 = 0.12V

3. Na + Cs+ -----> Na+ + Cs

Eo =  Eo (Cs) - Eo(Na)

- 0.22V = Eo (Cs) - 0.12

Eo (Cs) = - 0.10V