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The cyclic voltammogram of the antibiotic chloramphenicol (abbreviated) is shown here. The first cathodic scan goes from 0 to -1.0 V. The first cathodic wave, , is from the reaction ${\mathbf{RNO}}_{\mathbf{2}}\mathbf{+}\mathbf{4}{\mathbf{e}}^{\mathbf{-}}\mathbf{+}\mathbf{4}{\mathbf{H}}^{\mathbf{+}}\mathbf{鈫�}\mathbf{RNHOH}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$. Peak B in the reverse anodic scan could be assigned to $\mathbf{RNHOH}\mathbf{鈫�}\mathbf{RNO}\mathbf{+}\mathbf{2}{\mathbf{H}}^{\mathbf{+}}\mathbf{+}\mathbf{2}{\mathbf{e}}^{\mathbf{-}}$. In the second cathodic scan from +0.9 to -0.4 V, the new peak C appears. Write a reaction for peak C and explain why peak C was not seen in the initial scan.

Cyclic voltammogram of 3.7 脳10^-4 chloramphenicol in 0.1 M acetate buffer, pH 4.62. The voltage of the carbon paste working electrode was scanned at a rate of 350 mV/s. [Data from P. T. Kissinger and W. R. Heineman, 鈥淐yclic Voltammetry,鈥� J. Chem. Ed. 1983, 60, 702.]

Step by step solution

01

Introduction

Voltammetry is one type of electrochemical reactions which expresses the relationship between current and voltage. Voltammogram is the graph representing current vs voltage .

02

Determine the reaction between peak current and voltage for the given reaction.

Peak B: RNHOH$鈫�\mathrm{RNO}+2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}$

Peak $\mathrm{C}:\mathrm{RNO}+2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}鈫�\mathrm{RNHOH}$

During the first scan, Nitrate gets converted into hydrazine by reduction. In the anodic scan, it undergoes oxidation to give nitric oxide. Peak appears in the anodic scan. Peak C does not appear in the initial scan because nitric oxide forms only in anodic scan. Absence of RNO before initial scanning is the reason behind peak C was not seen before the initial scan

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