Questions

The standard electrode potential for the following reaction is +1.33 V. The potential at pH = 2.0 cannot be
${Cr}_{2} O_{7}^{2-} (aq,1M) + {14 H}^{+} (aq) + {6e}^{-} \to {2Cr}^{3+} (aq, 1M) + {7H}_{2} O (l)$

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+1.820 V

+1.990 V

+1.608 V

+1.0542 V

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detailed solution

Correct option is A

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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