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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]
Solution
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
Ecell = `E_(cell)^0 - 0.0591/2 log Q`
Ecell = `E_(cell)^0 - 0.0591/2 log ([Fe^{2+}])/[H^+]^2`
Ecell = `0.44 - (0.0591)/2 log 0.001/(0.01)^2`
Ecell = `0.44 - 0.0591/2 log 10`
Ecell = `0.44 - 0.0591/2 xx 1`
Ecell = 0.41045 V
The e.m.f. of the cell at 298 K is approximately 0.41 V.
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