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following available on the polarization curve of the active- surface TAEFL alloy M-metal. If Q1/ The information is the corrosion current density without protection and with protection of the alloy to be 104 and 10 Amp per square centimeter, respectively. Which protection method (cathodic or anodic) is more cost- effective in the case of alloys? = OV in = 10“ A/ cm² -6 E H* IH2 Н(4) Epp = -0.09 V ipp = 10-² A/ cm² i, = 10-$ A/ cm² B = 0.1V / decad 2 PP lp

following available on the polarization curve of the active- surface TAEFL alloy M-metal. If Q1/ The information is the corrosion current density without protection and with protection of the alloy to be 104 and 10 Amp per square centimeter, respectively. Which protection method (cathodic or anodic) is more cost- effective in the case of alloys? = OV in = 10“ A/ cm² -6 E H* IH2 Н(4) Epp = -0.09 V ipp = 10-² A/ cm² i, = 10-$ A/ cm² B = 0.1V / decad 2 PP lp

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Q1/ The following information is available on the polarization curve of the active- surface TAEFL alloy M-metal. If the corrosion current density without protection and with protection of the alloy to be 104 and 10 Amp per square centimeter, respectively. Which protection method (cathodic or anodic) is more cost- effective in the case of alloys? E H* IH, = 0V OV iH = 10 A/ cm? %3D = -0.09V ipp = 10- A/ cm² i, = 10- A/ cm? B=0.1V / decad E pp PP ст
Transcribed Image Text:Q1/ The following information is available on the polarization curve of the active- surface TAEFL alloy M-metal. If the corrosion current density without protection and with protection of the alloy to be 104 and 10 Amp per square centimeter, respectively. Which protection method (cathodic or anodic) is more cost- effective in the case of alloys? E H* IH, = 0V OV iH = 10 A/ cm? %3D = -0.09V ipp = 10- A/ cm² i, = 10- A/ cm? B=0.1V / decad E pp PP ст
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