Question

Calculate the electrode potentials of the following half cell at 25°C.

Pt, H₂ (5 atm) | HCl (0.5 M)

Given: \[\ce{E^{\circ}_{\frac{1}{2}Cl_2/Cl^-}}\] = +1.36 V and \[\ce{E{^{\circ}_{H^+/\frac{1}{2}H_2}}}\] = 0.00 V.

Answer 1

Given: Pt, H₂ (5 atm) | HCl (0.5 M),

\[\ce{E^{\circ}_{\frac{1}{2}Cl_2/Cl^-}}\] = +1.36 V,

\[\ce{E{^{\circ}_{H^+/\frac{1}{2}H_2}}}\] = 0.00 V,

log 20 = 1.3010,

Temperature = 25°C,

n = 2,

For the hydrogen half-cell:

`E = E^circ - 0.0591/n log (5/(0.5)^2)`

`E = 0.00 - 0.0591/2 log (5/0.25)`

E = −0.02955 log 20

∴ E = −0.02955 × 1.3010

E = −0.03845 V