Q.

The standard electrode potential for the following reaction is +1.33 V. The  potential at pH = 2.0 cannot beCr2O72-aq,1M+14 H+aq+6e- → 2Cr3+aq,  1M+7H2Ol

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a

+1.820 V

b

+1.990 V

c

+1.608 V

d

+1.0542 V

answer is A.

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Detailed Solution

Given, Standard electrode potential E°= + 1.33V Electrode potential, E=?,PH=2E=Eo−0.059/n  logOPRP⇒ECr2O72-/Cr3+=ECr2O72-/Cr3+o−0.05916×logCr3+2Cr2O72-×1H+14⇒ECr2O72-/Cr3+=1.33−0.05916log1(0.01)14 ⇒1.0542 V
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