
In a voltaic cell, oxidation takes place at:
A.Positive electrode
B.Negative electrode
C.Cathode
D.None of the above
Answer
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Hint: To answer this question, recall the concept of electrochemical reaction taking place in a voltaic cell. In oxidation-reduction reactions, electrons are moved from one species to another species. Energy is released if the reaction occurs spontaneously.
Complete step by step answer:
A voltaic cell is also known as the galvanic cell is an electrochemical cell in which electrical energy is obtained from a spontaneous redox reaction that occurs inside the cell. Usually, it contains two different metals that are submerged in electrolytes and connected by a salt bridge. The half-cell is composed of solid metal which is immersed in the electrolytic solution. The full cell is composed of two half-cells.
Generally, a salt bridge connects them both which prevents the ions of the more noble metal from plating out at the other electrode. In a voltaic cell, the anode is a negative electrode and oxidation takes place at the anode while cathode reduction takes place and it is the positive electrode.
Hence, the correct answer is B.
Note:
At the electrodes, electrons are absorbed or released by the atoms and ions. There is gain or loss of electrons which are released and passed into the electrolyte. The uncharged atoms separate from the electrolyte upon exchanging electrons. This process is known as discharging and this concept helps determine the release of ions on different electrodes. In case of electrochemical cells, we can easily remember in which cell oxidation or reduction occurs by: LOAN which means, Left-side Oxidation in Anode Negative.
The increasing order of discharge of few cations is: \[{{\text{K}}^{\text{ + }}}{\text{, C}}{{\text{a}}^{{\text{2 + }}}}{\text{, N}}{{\text{a}}^{\text{ + }}}{\text{, M}}{{\text{g}}^{{\text{2 + }}}}{\text{, A}}{{\text{l}}^{{\text{3 + }}}}{\text{, Z}}{{\text{n}}^{{\text{2 + }}}}{\text{, F}}{{\text{e}}^{{\text{2 + }}}}{\text{, }}{{\text{H}}^{\text{ + }}}{\text{, C}}{{\text{u}}^{{\text{2 + }}}}{\text{, A}}{{\text{g}}^{\text{ + }}}{\text{, A}}{{\text{u}}^{{\text{3 + }}}}\]
Increasing Order of discharge of few anions is: \[{\text{S}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}{\text{, N}}{{\text{O}}_{\text{3}}}^{\text{ - }}{\text{, O}}{{\text{H}}^{\text{ - }}}{\text{, C}}{{\text{l}}^{\text{ - }}}{\text{, B}}{{\text{r}}^{\text{ - }}}{\text{, }}{{\text{I}}^{\text{ - }}}\]
Complete step by step answer:
A voltaic cell is also known as the galvanic cell is an electrochemical cell in which electrical energy is obtained from a spontaneous redox reaction that occurs inside the cell. Usually, it contains two different metals that are submerged in electrolytes and connected by a salt bridge. The half-cell is composed of solid metal which is immersed in the electrolytic solution. The full cell is composed of two half-cells.
Generally, a salt bridge connects them both which prevents the ions of the more noble metal from plating out at the other electrode. In a voltaic cell, the anode is a negative electrode and oxidation takes place at the anode while cathode reduction takes place and it is the positive electrode.
Hence, the correct answer is B.
Note:
At the electrodes, electrons are absorbed or released by the atoms and ions. There is gain or loss of electrons which are released and passed into the electrolyte. The uncharged atoms separate from the electrolyte upon exchanging electrons. This process is known as discharging and this concept helps determine the release of ions on different electrodes. In case of electrochemical cells, we can easily remember in which cell oxidation or reduction occurs by: LOAN which means, Left-side Oxidation in Anode Negative.
The increasing order of discharge of few cations is: \[{{\text{K}}^{\text{ + }}}{\text{, C}}{{\text{a}}^{{\text{2 + }}}}{\text{, N}}{{\text{a}}^{\text{ + }}}{\text{, M}}{{\text{g}}^{{\text{2 + }}}}{\text{, A}}{{\text{l}}^{{\text{3 + }}}}{\text{, Z}}{{\text{n}}^{{\text{2 + }}}}{\text{, F}}{{\text{e}}^{{\text{2 + }}}}{\text{, }}{{\text{H}}^{\text{ + }}}{\text{, C}}{{\text{u}}^{{\text{2 + }}}}{\text{, A}}{{\text{g}}^{\text{ + }}}{\text{, A}}{{\text{u}}^{{\text{3 + }}}}\]
Increasing Order of discharge of few anions is: \[{\text{S}}{{\text{O}}_{\text{4}}}^{{\text{2 - }}}{\text{, N}}{{\text{O}}_{\text{3}}}^{\text{ - }}{\text{, O}}{{\text{H}}^{\text{ - }}}{\text{, C}}{{\text{l}}^{\text{ - }}}{\text{, B}}{{\text{r}}^{\text{ - }}}{\text{, }}{{\text{I}}^{\text{ - }}}\]
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