An electrochemical cell is based on the following two half-reactions: Ox: Sn( s

Question

An electrochemical cell is based on the following two half-reactions:
Ox: Sn(s)Sn2+(aq, 1.80 M )+2e
Red:ClO2(g, 0.245 atm )+eClO2(aq, 1.85 M)

Computer the cell potential at 25C.

And then:

A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 C. The initial concentrations of Pb2+ and Cu2+ are 5.00×102 M and 1.50 M , respectively.
A) What is the initial cell potential?
B) What is the cell potential when the concentration of Cu2+ has fallen to 0.220 M ?
C) What are the concentrations of Pb2+ and Cu2+ when the cell potential falls to 0.350 V ?

Any steps would be super helpful! Thank you so much!

Explanation / Answer

Ox: Sn(s)Sn2+(aq, 1.80 M )+2e ; E0 = 0.14 V
Red:ClO2(g, 0.245 atm )+eClO2(aq, 1.85 M) ; E0 = 0.954 V

Overall reaction :- Sn(s) + 2ClO2(g) ------> 2ClO2-(aq) + Sn2+(aq) ; E0cell = 0.954+0.14 = 1.094 V

Now, as per Nernst Equation :-

E = E0 - (0.059/n)*log{[Sn2+]*[ClO2-]2}/[ClO2]2 ; where n = number of electron transfer taking place in the balanced reaction = 2

Thus, E = 1.094 - (0.059/2)*ln{(1.8*1.852)/0.2452} = 0.957 V

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