/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 The corrosion rate is to be dete... [FREE SOLUTION] | 91影视

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Chapter 17: Problem 18

The corrosion rate is to be determined for some divalent metal M in a solution containing hydrogen ions. The following corrosion data are known about the metal and solution: \begin{tabular}{rr} \hline \multicolumn{1}{c}{ For Metal \(M\)} & For Hydrogen \\ \hline\(V_{\left(M M^{2}+\right)}=-0.47 \mathrm{~V}\) & $V_{\left(\mathrm{H}^{+} / H_{2}\right)}=0 \mathrm{~V}$ \\ \(i_{0}=5 \times 10^{-10} \mathrm{~A} / \mathrm{cm}^{2}\) & $i_{0}=2 \times 0^{-9} \mathrm{~A} / \mathrm{cm}^{2}$ \\ \(\beta=+0.15\) & \(\beta=-0.12\) \\ \hline \end{tabular} (a) Assuming that activation polarization controls both oxidation and reduction reactions, determine the rate of corrosion of metal $\mathrm{M}\left(\mathrm{in} \mathrm{mol} / \mathrm{cm}^{2} \cdot \mathrm{s}\right)$ (b) Compute the corrosion potential for this reaction.

Short Answer

Expert verified
After solving the equation for E (corrosion potential) and finding E_corr, we can now determine the corrosion current density (i_corr): i_corr = i_ox = i_red Using the Tafel equation for either the oxidation or reduction reaction at E_corr: i_corr = i鈧�_ox (exp((E_corr-E鈧�_ox)伪_oxF/RT) - exp(-(E_corr-E鈧�_ox)伪_oxF/RT)) or i_corr = i鈧�_red (exp((E_corr-E鈧�_red)伪_redF/RT) - exp(-(E_corr-E鈧�_red)伪_redF/RT)) Because we've found E_corr, we can now calculate i_corr. #tag_title#Step 2: Convert Corrosion Current Density to Corrosion Rate#tag_content# To convert the corrosion current density (i_corr) to the rate of corrosion in mol/cm虏路s, we need to divide i_corr by the number of electrons transferred in the reaction times Faraday's constant (F): Corrosion rate = i_corr / (2F) With the calculated i_corr, we can find the corrosion rate. #tag_title#Step 3: Compute the Corrosion Potential#tag_content# The corrosion potential was already calculated in Step 1 when solving the Tafel equation for E (corrosion potential). This value, E_corr, represents the corrosion potential of the metal. In conclusion, we've calculated the corrosion potential (E_corr) and used it to find the corrosion current density (i_corr). By converting the i_corr to a corrosion rate in mol/cm虏路s, we've successfully determined the corrosion rate and the corrosion potential for this reaction.

Step by step solution

01

Calculate the Corrosion Current Density

Let's apply the Tafel equation to both oxidation and reduction reactions: For Metal M (Oxidation): M -> M虏鈦� + 2e鈦� i_ox = i鈧�_ox (exp((E-E鈧�_ox)伪_oxF/RT) - exp(-(E-E鈧�_ox)伪_oxF/RT)) For Hydrogen (Reduction): 2H鈦� + 2e鈦� -> H鈧� i_red = i鈧�_red (exp((E-E鈧�_red)伪_redF/RT) - exp(-(E-E鈧�_red)伪_redF/RT)) At the corrosion potential (E_corr), the oxidation and reduction current densities are equal: i_ox = i_red i鈧�_ox (exp((E-E鈧�_ox)伪_oxF/RT) - exp(-(E-E鈧�_ox)伪_oxF/RT)) = i鈧�_red (exp((E-E鈧�_red)伪_redF/RT) - exp(-(E-E鈧�_red)伪_redF/RT)) Now, let's plug in the given values: 5*10^{-10} (exp((E-(-0.47))0.15F/RT) - exp(-(E-(-0.47))0.15F/RT)) = 2*10^{-9} (exp((E-0)(-0.12)F/RT) - exp(-(E-0)(-0.12)F/RT)) We can solve this equation for E (corrosion potential), assuming a room temperature of 298K, knowing that R = 8.314 J K鈦宦� mol鈦宦�, and F = 96,485 C/mol.

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Most popular questions from this chapter

According to Table \(17.3,\) the oxide coating that forms on silver should be nonprotective, and yet Ag does not oxidize appreciably at room temperature and in air. How do you explain this apparent discrepancy?

Briefly explain why cold-worked metals are more susceptible to corrosion than noncoldworked metals.

An electrochemical cell is composed of pure copper and pure cadmium electrodes immersed in solutions of their respective divalent ions. For a \(6.5 \times 10^{-2} M\) concentration of \(\mathrm{Cd}^{2+},\) the cadmium electrode is oxidized yielding a cell potential of \(0.775 \mathrm{V}\). Calculate the concentration of \(\mathrm{Cu}^{2+}\) ions if the temperature is \(25^{\circ} \mathrm{C}\)

An \(\mathrm{Fe} / \mathrm{Fe}^{2+}\) concentration cell is constructed in which both electrodes are pure iron. The \(\mathrm{Fe}^{2+}\) concentration for one cell half is \(0.5 M,\) for the other \(, 2 \times 10^{-2} M .\) Is a voltage generated between the two cell halves? If so, what is its magnitude and which electrode will be oxidized? If no voltage is produced, explain this result.

For a concentration cell, briefly explain why corrosion occurs at that region having the lower concentration.
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