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Calculate the EMF of the concentration cell at 298 K:Ag/Ag+ (0.005 M)//Ag+ (0.5 M)/Ag. What will be the potential of the above cell. When the concentration of silver ions is changed from 0.005 M to 0.01 M at the same temperature? |
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Answer» Given , Ag/Ag+ (0.005M) // Ag+ (0.5M)/ Ag(s) A+ anode Ag → Ag+ (0.005M)+e A + Cathode Ag+ (0.5M) + e- → Ag(s) over all cell reaction - Ag+ (0.5M) → Ag+ (0.005) as we know concentration cell , the standard electrode potential (E°) equal to zero. using nearest equation -- Ecell = E° - \(\frac{RT}{nF}\) \(ln(\frac{[Ag^+]}{[Ag^+]})\) ∴ Ecell = 0 - \(\frac{RT}{F}\) \(ln\frac{0.005}{0.5}\) Ecell = -0.0591 log 10-2 = 0.0591 x 2 Ecell = 0.1182 v if concentration of Ag+ change from 0.005 M to 0.01 M then Ecell = log \((\frac{0.01}{0.5})\) = -0.0591 log (0.02) = (-0.0591) x (-1.699) Ecell = 0.1 v Hence the potential of cell will be 0.1 v |
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