E cell value for 1 is (.851) and the second one is (-.071) Answer 1 and 2 please

Question

E cell value for 1 is (.851) and the second one is (-.071)

Answer 1 and 2 please for B. Thanks

B. Effect of Concentration on Cell Potential -Nernst Equation Eca, after addiing 6 M NH, to the copper cell Calculate the concentration o that free copper (11) ion that is in equilibrium with the complexed copper (II) ion, CuNH in the solution. Does the calculated valuc make sense and what does it imply? 2t ce Eveli, after adding 1.0 M HCI only to the silver cell Calculate the concentration of the free silver () ion that is in Does the calculated value make sense. Justify your answer equilibrium with the precipitated silver () ion, AgCi 124

Explanation / Answer

B) 2. We know that at equilibrium

E = -. 059/n log Q

Where n=number of free electrons and Q =reaction quotient.

So for Ag+ +Cl- - - - > AgCl

Ksp for AgCl is 1.8 *10-11

Q=AgCl /[ Ag+] [ Cl-]

Q= Ksp / Ag+ * 1M

n=1 and E=-0. 71

Putting values in the equation we get  

-0.71=-.059/1*log(1.8*10-11)/Ag +

Ag+=2.85*10-10 M

Similarly use the value of Ksp in the 1st question.

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