6. Given the electrochemical cell Nio | Ni2+ || Ag+ | Ago at 25.0 oC Half-cell r

Question

6. Given the electrochemical cell Nio | Ni2+ || Ag+ | Ago at 25.0 oC    

Half-cell reaction                                                             Reduction 0 E cell (V)

Ni2+    +     2 e -   ---->    Ni (s)                                             - 0.25

Ag +      +   1e-      -----> Ag (s)                                             +0.80

a. Write a balanced redox reaction for this electrochemical cell.

b. Calculate the cell’s oEcell

c. Determine the electrochemical cell potential (Ecell) if Ni2+ ion is 0.025 M and the Ag+ ion is 0.050 M.

Explanation / Answer

A. The balanced redox reaction will be

Ni (s) + 2Ag+ ---> Ni2+ + 2Ag

B. The E° cell = E cathode - E anode

E° cell = E Ag+|Ag - E Ni+2|Ni

E° cell = 0.80 - (- 0.25)

E° cell= 1.05V

C. E cell = E°cell - (0.059/ n ) log anode / cathode

n = number of electrons =2

E cell = E° cell - (0.059/ 2) log [Ni+2] / [Ag+]2

E cell = 1.05 - 0.0295 log [ 0.025] / [0.050]2

E cell = 1.02 V

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