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
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a
1.285 V
b
1.193 V
c
1.187 V
d
zero
answer is C.
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Detailed Solution
Cr2O72-+ 14H++ 6e-→2Cr3++7H2O Initial conc10.10 After reaction 0.50.1M1MERP=ERP0-0.066logCr3+2Cr2O72-H+14ERP=1.33-0.066log1(0.5)(0.1)14=1.33-0.066log2×1014=1.187 V
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