Calculate the f.e.m. for the following reaction at 25oC
Au3+(aq) (4.4e-3M) + 2Cl-(aq) (1.3e-2M) → Au(s) ( - M) + Cl2(g) ( - M)
Ecell = ? (2 decimals accuracy!)
balanced equation
2Au3+(aq) (4.4e-3M) + 6Cl-(aq) (1.3e-2M) ---> 2Au(s) ( - M) + 3Cl2(g) ( - M)
E0Au3+/Au = 1.52 v
E0Cl2/2cl- = 1.36 v
E0cell = E0cathode - E0anode
= (1.52-1.36)
= 0.16 V
Ecell = E0cell - 0.0591/nlog(1/[Au3+]^2[Cl-]^6)
= 0.16 - (0.0591/6)log(1/((4.4*10^(-3))^2(1.3*10^(-2))^6))
= 0.00211 V
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