Question

Calculate the e.m.f. of the following cell at 298 K:

Fe(s) | Fe²⁺ (0.001 M) | | H⁺ (0.01 M) | H₂(g) (1 bar) | Pt(s)

Given that \[\ce{E^0_{cell}}\] = 0.44 V

[log 2 = 0.3010, log 3 = 0.4771, log 10 = 1]

Answer 1

According to the equation,

\[\ce{Fe(s) + 2H^+(aq) -> Fe^{2+}(aq) + H2(g)}\]

\[\ce{E^0_{cell} = E^0_{cathode} - E^0_{anode}}\]

\[\ce{E^0_{cell}}\] = 0 – (– 0.44) V

\[\ce{E^0_{cell}}\] = + 0.44 V

By applying Nernst Equation

E_cell = `E_(cell)^0 - 0.0591/2 log Q`

E_cell = `E_(cell)^0 - 0.0591/2 log  ([Fe^{2+}])/[H^+]^2`

E_cell = `0.44 - (0.0591)/2 log  0.001/(0.01)^2`

E_cell = `0.44 - 0.0591/2 log 10`

E_cell = `0.44 - 0.0591/2 xx 1`

E_cell = 0.41045 V

The e.m.f. of the cell at 298 K is approximately 0.41 V.