Question

a) Calculate the expected slope of potential E, vs. pH for an ideal electrode to four significant figures (use ln 10, not 2.303)

b) Show that the units of the slope are V/pH-unit.

c) Estimate the real slope value expected from your E vs. pH calibration.

We are given the equations:

E = constant₂ + 2.303 (RT/zF)pH_external

E = constant₂ + (0.05916 V / 1)pH_external

E = constant + 0.05916 ∆pH

E = constant + ß 0.05916 ∆pH

For an ideal electrode ß = 1. For real electrodes, ß is typically between 0.98 and 1.00.

Answer 1

a) The given equation: E = constant + ß 0.05916 ∆pH, Here, ß = 1 (since ideal electrode)

If you compare the above equation with the straight line equation, i.e. y = mx + c, where m = slope of the plot

the expected slope of the potential E, vs. pH for an ideal electrode = 0.0592 V/pH-unit

b) The unit of the slope: E = constant + 0.05916 ∆pH

Here, the unit of potential (E) is volt, i.e. V

Let's the unit of pH is pH-unit

The constant must have the units same as that of potential, i.e. V (since it's a y-intercept)

Hence the unit of slope = unit of E/unit of pH

= V/pH-unit

c) The real slope value expected from your E vs. pH calibration = 2.303RT/F (where R = 8.314 J mol^-1 K^-1, T = 298 K and F = 96500 C)

= 0.0591 V/pH-unit