Question

In: Chemistry

Consider the cell described below at 261 K: Fe | Fe2+ (0.903 M) || Cd2+ (0.991...

Consider the cell described below at 261 K:

Fe | Fe2+ (0.903 M) || Cd2+ (0.991 M) | Cd

Given EoCd2+→Cd = -0.403 V, EoFe2+→Fe = -0.441 V. Calculate the cell potential after the reaction has operated long enough for the Fe2+ to have changed by 0.373 mol/L

Solutions

Expert Solution

Fe --> Fe+2 + 2e- ... Eo = - 0.0.441 volts
Cd+2 + 2 e- --> Cu ... Eo = - 0.403 volts
cell potential ... Eo = -0.403 (-0.441) = 0.038V
                       Fe + Cd+2 --> Fe+2 + Cd
Initial                          0.991   0.903
Change                  0.373           0.373

Fina                       0.618          1.276
-------

nernst Equation
E = Eo - (0.0592/n) log Q

E = 0.038 volts - (0.0592/2) log [Fe+2] / [Cd+2]

E = 0.038 volts - (0.0296) log [1.276] / [0.618]

E = 0.038 volts - (0.0296) log 2.064
E = 0.038 volts - (0.0296) (0.314)

E = 0.038 volts - 0.00929
your answer is: E = 0.0287 volts

queen_honey_blossom answered 16 hours ago
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