For Cr 2 O 7 2 - + 14 H + + 6 e - yields ⟶ 2 Cr 3 + + 7 H 2 O , E 0 = 1 . 33 V at Cr 2 O 7 2 - = 4 . 5 millimole, Cr 3 + = 1 . 5 millimole and E = 1 . 067 V , then calculate the pH of the solution.

Step-by-step solution

According to the Nernst's equation: E   =   E ° - RT nF lnQ So, by given value of concentrations the above formula can be modified as E   =   E ° - 0 . 059 nF log Cr 3 + 2 Cr 2 O 7 2 - H + 14 1 . 067   =   1 . 33 - 0 . 06 6 × 96500 log 15 × 10 - 3 2 4 . 5 × 10 - 3 H + 14 H +   =   10 - 2   M Hence, the concentration of H + ion will be pH   = - log H + pH = - log 10 - 2 = - ( - 2 ) × log 10 10 pH   =   2