Question

When an ion-selective electrode for X was immersed in 0.0760 M XCl, the measured potential was 0.0450 V. What is the concentration of X when the potential is 0.0600 V? Assume that the electrode follows the Nernst equation, the temperature is at 25°C, and that the activity coefficient of X is 1.

[X+] = ??M

Answer 1

Voltage of an ion selective electrode can be given using Nerst equation as follows:

E = E^o + (2.303 RT/nF) log (X)

Where ,

E = Potential of electrode ( 0.0450 V)

E^o = Potential of reference electrode ( we have to find out this first)

R = universal gas consatant (8.314 J mol-1 K-1)

T = Absolute temperature (25 + 273 = 298 K )

n = Charge on ionic speceis ( Since species is X+ , it is 1 in this case )

F = Faradys constant ( 96500 Coulomb / mol )

X = Concentration of ionic species ( Since activity coefficient is given as 1 concentration of ion will be 0.076 M )

After plugging all values we get

0.0450 V = E⁰ + 0.0591 log (0.0760 )

E⁰ = 0.0450 V - ( - 0.0661 V) = 0.1111 V

Now since we know the potential of reference electrode, we can calculate the concentration of X when the potential is 0.0600 V

E = E^o + (2.303 RT/nF) log (X)

0.0600 V = 0.1111 V + 0.0591 log A

log X = - 0.0511 / .0591 = - 0.8646

By taking antilog we get the concentration of [X+] = 0.1365 M.