Question

Calculate the electrode potentials of the following half cells at 298 K.

Co_(s) | Co²⁺ (0.01 M)

Given: \[\ce{E^{\circ}_{Ag^+/Ag}}\] = +0.80 V, \[\ce{E^{\circ}_{{Co^{2+}/Co}}}\] = −0.28 V

Answer 1

Given: The half cell is Co_(s) | Co²⁺ (0.01 M)

Temperature = 298 K

Standard electrode potential for Co_(s) | Co²⁺ is

\[\ce{E^{\circ}_{{Co^{2+}/Co}}}\] = −0.28 V

The nernst equation is E = `E^circ - 0.0591/n log Q`

The half reaction is \[\ce{Co{^{2+}_{(aq)}} + 2e- -> Co_{(s)}}\]

The number of electrons transferred, n, is 2.

Q = `1/([text(Co)^(2+)]` for the reduction reaction

Q = `1/0.01`

Substitute the values into the Nernst equation

E = `E^circ - 0.0591/n log Q`

E = `-0.28 - (0.0591/2) log (1/0.01)`

= `-0.28 - 0.0591/2 log(100)`

= `-0.028 - 0.0591/2 xx 2`

= −0.28 − 0.0591

E = −0.3391 V