Question

Calculate the potential for half cell containing 0.10M K2Cr2O7(aq), 0.20M Cr^3+ and 1.0 × 10^-4M H^1+ . The half reaction is -
Cr2O7^2- + 14H^1+ + 6e^- ---------- 2Cr^3+ + 7H2OCl
E not = 1.33V

Answer 1

0.79 V

Explanation:

Given: The half reaction is -Cr2O7^2- + 14H^1+ + 6e^- ---------> 2Cr^3+ + 7H2OCl

E not = 1.33V

Find: Calculate the potential for half cell containing 0.10M K2Cr2O7(aq), 0.20M Cr^3+ and 1.0 × 10^-4M H^1+ .

Solution:

Cr2O7 -2 +14 H++ 6e ----------------> 2Cr+3 +7H20

Ecell = E°cell - (0.0591 / 6) log { [Cr+3] 2 / [H+] 14 [Cr2O7-2]}

 =1.33 - (0.0591 / 6) log {[0.20] 2 / [(1*10^-4) 14 (0.10) ] }

 =1.33 - 0.54

 = 0.79 V

Answer 2

Answer:

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