Calculate Ecell at 25°C when [Ag+(aq)] = 0.049 M and [Fe3+(aq)] = 3.1 M.
E cell = E^0 cell - (RT/nF) log([Fe3+]/[Ag+]),
where R = 8.134 J/(K mol); T = 25 + 273 = 298 K; n = 3 (number of electrons transferred); F = 96500 C/mol
E cell = 0.836 - (8.134 * 298)/(3*96500) log(3.1/0.049) = 0.836 - 0.035 = 0.80 V
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