The cell $Ag ∣ Ag X^{+} (c M) ∣∣ Ag X^{+} (0.5 M) ∣ Ag$ has EMF $0.1182 V$ at 298 K with the left electrode as anode. The value of $c$ in millimol per litre is (one decimal; take $2.303 RT / F = 0.0591 V$ ).

Numerical answer type, answer in mM, tolerance 0.1.

Answer: 5 mM

For a concentration cell

E^{\circ} = 0

and

E = \frac{0.0591}{1} lo g \frac{c _{cathode}}{c _{anode}}

lo g \frac{0.5}{c} = \frac{0.1182}{0.0591} = 2

, so

0.5/ c = 100

and

c = 0.005 M = 5.0 mM

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