Question

Results of an electrochemical experiment are shown in the figure. In each case, a voltage is applied between two electrodes at timeand the absorbance of a solution decreases until the voltage is stepped back to its initial value at time. The upper traces show the average results for 100,300, and 1000 repetitions of the experiment. The measured signal-to-rms noise ratio in the upper trace is 60.0. Predict the signal-to-noise ratios expected for 300 cycles, 100 cycles, and 1 cycle and compare your answers with the observed values in the figure.

Short answer

The signal-to-noise ratios expected for 300 cycles, 100 cycles, and 1 cycle is$32.9;19;1.9$

Step-by-step solution

Definition of electrochemical experiment

The voltage and current of oxidation/reduction reactions are measured and/or controlled in electrochemical corrosion testing.

Manipulation and measurement of these two variables allow for a variety of experiments.

Most experiments include applying a voltage to the working electrode and measuring the current that results.

Determine the square root of each cycle

First, we calculate the square root of each cycle number:

$$\begin{gathered} \sqrt{\mathrm{n}}=\sqrt{1} \\ =1 \\ \sqrt{\mathrm{n}}=\sqrt{100} \\ =10 \\ \sqrt{\mathrm{n}}=\sqrt{300} \\ =17.32 \\ \sqrt{\mathrm{n}}=\sqrt{1000} \\ =31.62 \end{gathered}$$

Determine the S/N ratio

As a result, the expected relative S/N ratio increased in direct proportion to the increase in $\sqrt{n}$:

$\frac{\mathrm{S}}{\mathrm{N}}=1\frac{\mathrm{S}}{\mathrm{N}}=10\frac{\mathrm{S}}{\mathrm{N}}=17.32\frac{\mathrm{S}}{\mathrm{N}}=31.62$

For 1000 cycles, the observed S / N is 60, as seen in the graph. The expected S/N ratio is as follows:

For a total of 1000 cycles:

$\frac{\mathrm{S}}{\mathrm{N}}-60$

The ratio that has been seen is 60.

Determine the ratio of 300 cycle,100 cycle and 1 cycle

For a total of 300 cycles:

$\frac{\mathrm{S}}{\mathrm{N}}-60\times \frac{17.32}{31.62}=32.9$

The measured ratio is 35.9.

For a total of 100 cycles

$\frac{\mathrm{S}}{\mathrm{N}}-60\times \frac{10}{31.62}=19$

20.9% is the measured ratio.

For a single cycle:

$\frac{\mathrm{S}}{\mathrm{N}}-60\times \frac{1}{31.62}=1.9$

1.95 is the observed ratio.