Consider a voltaic cell with 2 electrodes (one of them is an Aluminum electrode, and the other is unknown electrode) placed in an electrolytic solution. The half-cell potential value of Aluminum is -1.662 V. If a voltmeter is connected between the two electrodes of this voltaic cell, with the Aluminum electrode is connected to the negative terminal of the voltmeter and the unknown electrode is connected to the positive terminal of the voltmeter, the voltmeter reading is 3.386 V. Find the half cell potential of the unknown electrode.
Consider a voltaic cell with 2 electrodes (one of them is an Aluminum electrode, and the other is unknown electrode) placed in an electrolytic solution. The half-cell potential value of Aluminum is -1.662 V. If a voltmeter is connected between the two electrodes of this voltaic cell, with the Aluminum electrode is connected to the negative terminal of the voltmeter and the unknown electrode is connected to the positive terminal of the voltmeter, the voltmeter reading is 3.386 V. Find the half cell potential of the unknown electrode.
Chemistry: Principles and Practice
3rd Edition
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Chapter18: Electrochemistry
Section: Chapter Questions
Problem 18.40QE: A voltaic cell is based on the reaction Zn(s)+Ni2+(aq)Zn2+(aq)+Ni(s) Voltage measurements show that...
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Consider a voltaic cell with 2 electrodes (one of them is an Aluminum electrode, and the other is unknown electrode) placed in an electrolytic solution. The half-cell potential value of Aluminum is -1.662 V. If a voltmeter is connected between the two electrodes of this voltaic cell, with the Aluminum electrode is connected to the negative terminal of the
voltmeter and the unknown electrode is connected to the positive terminal of the voltmeter, the voltmeter reading is 3.386 V. Find the half cell potential of the unknown electrode.
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