Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5

Q.

What would be the reduction potential of an electrode at 298 K, which originally contained 1M K2Cr2O7 solution in acidic buffer soluton of pH = 1.0 and which was treated with 50% of the Sn necessary to reduce all Cr2O72- to Cr3+. Assume pH of solution remains constant. 

 Given :ECr2O72-/Cr3+,H+0=1.33 V,log2=0.32.303RTF=0.06

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away
An Intiative by Sri Chaitanya

a

zero

b

1.187 V

c

1.193 V

d

1.285 V

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

                                                            Cr2O72-+      14H++  6e-2Cr3++7H2O   Initial conc10.10 After reaction 0.50.1M1M

ERP=ERP0-0.066logCr3+2Cr2O72-H+14

ERP=1.33-0.066log1(0.5)(0.1)14

=1.33-0.066log2×1014=1.187 V

 

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon