Based on the data given for the half cell M ( x + n ) + , M x + / Pt at 298 K (take 2.303RT/F = 0.059 V, log 3 = 0.48): Case 1 Case 2 % Reduced form 50 10 Potential (V) 0.500 0.514 The value of n is:

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ChemistryElectrochemistryGalvanic cells/Nernst Equation/Concentration CellsHard2 minQB

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Based on the data given for the half cell $M^{(x + n) +}$ , $M^{x +}$ / Pt at 298 K (take 2.303RT/F = 0.059 V, log 3 = 0.48):

	Case 1	Case 2
% Reduced form	50	10
Potential (V)	0.500	0.514

The value of n is:

Answer:

Solution:

E = $E^{\circ}$ - (0.059/n) log([ $M^{x +}$ ]/[ $M^{(x + n) +}$ ]).

At 50% reduced: ratio = 1, log = 0, so $E^{\circ}$ = 0.500 V.

At 10% reduced: ratio = 10/90 = 1/9, log = -2 × 0.48 = -0.96, so E = 0.500 + 0.059×0.96/n.

0.514 - 0.500 = 0.014 = 0.0566/n gives n = 4.04, about 4.

Answer: (C).

Stream:JEESubject:ChemistryTopic:ElectrochemistrySubtopic:Galvanic cells/Nernst Equation/Concentration Cells

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