Question

Calculate the potential for half-cell containing 0.01 M K₂Cr₂O₇(aq), 0.01 M Cr³⁺(aq) and 1.0 × 10⁻⁴ M H⁺(aq). The half cell reaction is

\[\ce{Cr_2O^{2-}_7 + 14 H+_{(aq)} + 6e -> 2Cr^{3+}_{(aq)} + 7H2O_{(l)}}\] 

and the standard electrode potential is given as E^° = 1.33 V.

[Given: log 10 = 1]

Answer 1

Given a half reaction

\[\ce{Cr_2O^{2-}_7 + 14 H+_{(aq)} + 6e -> 2Cr^{3+}_{(aq)} + 7H2O_{(l)}}\] 

E^° or `Cr_2O_7^(2-)//Cr^(+3) = 1.33 V`

`E = E^° - 0.0591/n  log Q`

`= 1.33 - 0.0591/6 log  ([Cr^{+3}][H_3O])/([Cr_2O_7^(-2)][H^+]^14)`

`= 1.33 - 0.0591/6 log  ((0.2)^2)/((0.1)(1 xx 10^(-4))^14)`

`= 1.33 - 0.00985 log  (0.04)/(0.1 xx 1 xx 10^(-56))`

`= 1.33 -0.00985 log  (0.04) /(0.1) xx 10^(56)`

`= 1.33 - 0.00985 log  (0.4) xx 10^(56)`

= 1.33 − 0.00985 log (0.4) + log (10⁵⁶)

= 1.33 − 0.00985 log (0.4) + log (10⁵⁶)

= 1.33 − 0.00985 × (−0.398 + 56)

= 1.33 − 0.00985 × 53.602

= 1.33 − 0.52797

= 0.802 V