Consider the following half cell reaction Cr2O72-(aq) + 6e- + 14H+(aq) 2Cr3+(aq) + 7H2O( ) The reaction was conducted with the ratio of [ Cr3+]2[ Cr2O72-] = 10-6. The pH value at which the EMF of the half cell will become zero is . (nearest integer value) [Given: standard half cell reduction potential E Cr2O72-,H+/Cr3+ = 1.33 V, 2.303RTF = 0.059 V.]
Answer
10
Solution
Nernst equation for the reduction half cell (n = 6):
E = E - 0.0596 [ Cr3+]2[ Cr2O72-][H+]14
E = E - 0.0596 ( 10-6 - 14 [H+] )
E = E - 0.0596 (-6 + 14 pH )
For E = 0:
1.33 = 0.0596 (14 pH - 6 )
14 pH - 6 = 1.33 60.059 = 135.25
pH = 141.2514 = 10.09 10
Answer: 10