Question

In: Chemistry

Consider the cell described below at 275 K: Sn | Sn2+ (0.845 M) || Pb2+ (0.937...

Consider the cell described below at 275 K: Sn | Sn2+ (0.845 M) || Pb2+ (0.937 M) | Pb Given EoPb2+→Pb = -0.131 V, EoSn2+→Sn = -0.143 V. Calculate the cell potential after the reaction has operated long enough for the Sn2+ to have changed by 0.359 mol/L.

Expert Solution

Sn(s) --------> Sn2+(aq) + 2e Anode E° = -0.143V

Pb2+(aq) + 2e --------> Pb(s) Cathode E° = -0.131V

Sn(s) + Pb2+(aq) --------Sn2+(aq) + Pb(s)

E°cell = E°cathode - E° Anode

= -0.131V - ( - 0.143V)

= 0.012V

Nernst equation is

Ecell = E°cell - ( 2.303RT/nF)logQ

where,

no of electron transfer , n =2

Temperature T = 275K

F = 96485C/mol

R = 8.314 J/K mol

Q = [Sn2+]/[ Pb2+ ]

If [ Sn2+ ] Changed by 0.359M

[ Sn2+ ] = 1.204M

[ Pb2+ ] = 0.578M

Q = 1.204M/0.578M= 2.083

Therefore,

Ecell = 0.012V - ((2.303 × 8.314(J/Kmol)) ×275K)/2×96485C/mol) log 2.083

Ecell = 0.012V - (0.0273V) log 2.083

= 0.012V - 0.0087V

= 0.0033V

queen_honey_blossom answered 2 years ago

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