Chapter 17: Q37P (page 431)

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 ${RNO}_{2} + 4 e^{-} + 4 H^{+} \to RNHOH + H_{2} O$ . Peak B in the reverse anodic scan could be assigned to $RNHOH \to RNO + 2 H^{+} + 2 e^{-}$ . 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 ×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, “Cyclic Voltammetry,” J. Chem. Ed. 1983, 60, 702.]

Short Answer

Expert verified

The peak C is given in cyclic voltammogram

The reaction for peak $C : RNO + 2 H^{+} + 2 e^{-} \to RNHOH$

Absence of RNO before initial scanning is the reason behind peak C was not seen before the initial scan.

Step by step solution

Introduction

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

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

Peak B: RNHOH $\to RNO + 2 H^{+} + 2 e^{-}$

Peak $C : RNO + 2 H^{+} + 2 e^{-} \to RNHOH$

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