Question

If the concentration of fluoride anion and aluminum cation was increased to 5 m, by how much would the measured ecell change?.

Answer 1

If the concentration of fluoride anion and aluminum cation was increased to 5M, then there would be decrease in Ecell by 0.055 volts

The standard reduction potential of Fluoride anion and aluminum cation are -

2Al -----> 2Al3+ + 6e-           $E^o$cell = +1.66

6e- + 3F2 -----> 6F-               $E^o$cell = +2.87

The complete reaction is -

2Al + 3F2 ------> 2Al3+  +  6F-    $E^o$ = +4.53

Using Nernst Equation :-

E = $E^o$ – 0.0592/n*log[Al3+]^2[F-]^6

n = 6 (n = number of transferred electrons)

E = +4.53 - 0.0592/6*log(5)^2(5)^6

E = +4.53 - 0.00987*log(25)(1.56 x 10^4)

E = +4.53 – 0.00987*log(3.9 x 10^5)

E = + 4.53 -0.00987(5.59)

E = + 4.53 - 0.055

E = +4.47

The change in Ecell = 5 - 4.47 = 0.055V

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