Consider the following electrochemical cell notation: M(s) / M2+ (aq)// H+ (aq)/ H2(g). The value of the electrode potential is positive when?
electrons flow from the metal electrode, M(s) to hydrogen electrode, H2(g)
electrons fow from hydrogen electrode, H2(g) to metal electrode, M(s)
the flow of current is high
there is equilibrium between the flow of electrons from the hydrogen electrode, H2(g) to metal electrode, M(s)
Explanation
No explanation available
Video Explanation
No video available
Post your Contribution
Discussions (4)
For current to be produced in electrochemical cells, the anode must be able to reduce the cathode and the cathode must be able to oxidize the anode (remember that oxidation occurs at the anode while reduction occurs at the cathode). In this case where the metal is a better reducing agent than hydrogen, the metal becomes the anode while hydrogen becomes the cathode (It should also be noted that unlike Electrolytic cells, in electrochemicals cells, the cathode is the positive electrode while the anode is the negative electrode). By convention Hydrogen is arbitrarily given an electrode potential of 0.0V, whereas metals which are higher than hydrogen in the electrochemical series have electrode potential values which are higher than 0.0V. Having made the metal our anode and hydrogen our cathode:
emf= electrode potential (cathode) - electrode potential (anode)
= +ve emf.
Therefore, with this kind of setup current/electricity will be generated
The provided answer is wrong
The correct answer is B
Too many wrong answers on the app, this app has lost its credibility for me
What If a student fails some questions due to the numerous wrong answers provided on the app?
when posting anwers that are contradictory to what we read in the textbooks at least we should be given explanations to clarify our confusions
please do something about that
The correct answer is B, you should review the question please. Negative electrode potential means the metal electrode is donating electron(s) to another electrode.
