Question

Question 4 5 pts Given the following reaction, calculate the electrical potential after adding 10ml of cesium. Assume the analyte initially is present at a concentration of 0.0100 M and that a 25.0-ml sample is taken for analysis. The titrant, which is the underlined species, has a concentration of 0.0100 M. V2+(aq) + Ce4#lag) → V3+(aq) + Ce3+(ag) O 0.245 V 0 -0.265 V O 0.265 V 0 -0.245 V

Answer 1

Our Analyte solution have 25 ml of 0.0100 M V²⁺,

so initial  [V²⁺] = 0.0100 mole/L * 25 ml / 1000 ml = 2.5*10^-4 moles

We added 10 ml of Ce³⁺ ,  [ Ce³⁺] = 0.0100 M , so  [ Ce³⁺] = 1.0*10^-4 moles

After addition of 10 ml of Ce³⁺ , remaining  [V²⁺] = (2.5*10^-4 -1.0*10^-4 ) = 1.5*10^-4 moles

and now we have  [V³⁺] =  1.0*10^-4 moles

Now total volume is 35 ml , so,  [V²⁺] =0.00429 M and  [V³⁺] = 0.00286 M

Since, we have not reached at equivalence point, It's easier to use the V half-reaction because we know how much was originally present and how much remains addition titrant, simpler to solve.

our reaction is ,  V²⁺_(aq) + Ce³⁺$\rightarrow$ V³⁺_(aq) + Ce⁴⁺

We have,  V³⁺_(aq) + e^– V²⁺_(aq)   E⁰= –0.255 V

so, E⁰ = +0.255 V

Using, Nernst equation,

E = E⁰ - (RT/nF) ln ([V³⁺] /[V²⁺] ) =  +0.255 V - 0.0591*log([0.00286] /[0.00429] )

E =+0.255 V + 0.01 V = +0.265 V