
How do you calculate the standard cell potential of a cell constructed from \[M{g^{2 + }}/Mg\] and\[N{i^{2 + }}/Ni\]. Which half-cell is the anode and which half-cell is the cathode?
Answer
544.2k+ views
Hint: A reduction potential estimates the propensity of an atom to be reduced by taking up new electrons. The standard reduction potential is the reduction potential of a particle under explicit, standard conditions. Standard reduction potential can be helpful in deciding the directionality of a reaction. The reduction potential of a given species can be viewed as the negative of the oxidation potential.
Complete solution:
The standard reduction potential estimates the strength of the electrode as an oxidizing agent and equivalently the tendency of it to get reduced. Let us consider a platinum-hydrogen electrode under standard conditions and let us assume the reduction potential of it to be zero for reference.
A strong oxidizing agent will have a high reduction potential and a strong reducing agent will have low reduction potential.
A galvanic cell is made up of two half cells with an external circuit and a salt bridge. The half-cell which will have a higher reduction potential will be more likely to be reduced and will experience reduction further, that half-cell behaves as a cathode. And the half-cell which will experience oxidation and will have a lower reduction potential value will behave as an anode.
\[{E^o}(N{i^{2 + }}/Ni) > {E^o}(M{g^{2 + }}/Mg)\]
Therefore we can conclude that, in the galvanic cell, \[N{i^{2 + }}/Ni\]half-cell will be reduced and will behave as a cathode. And \[M{g^{2 + }}/Mg\]will be oxidised and will behave as the anode.
As we know the reduction potentials for cathode and anode (known data) we can easily calculate the cell potential
\[E_{cell}^o = {E^o}(cathode) - {E^o}(anode)\]
\[E_{cell}^o= - 0.257 - ( - 2.372)\]
\[E_{cell}^o = + 2.115\]
Note: Sometimes, the direction of the redox reaction can be dictated by assessing the general qualities of the reductants and oxidants. In circumstances where an electrochemical series is not adequate to absolutely determine the direction of a redox reaction, the standard electrode potential can be used. A negative estimation of cell potential demonstrates a reducing environment, while a positive value shows an oxidizing environment.
Complete solution:
The standard reduction potential estimates the strength of the electrode as an oxidizing agent and equivalently the tendency of it to get reduced. Let us consider a platinum-hydrogen electrode under standard conditions and let us assume the reduction potential of it to be zero for reference.
A strong oxidizing agent will have a high reduction potential and a strong reducing agent will have low reduction potential.
A galvanic cell is made up of two half cells with an external circuit and a salt bridge. The half-cell which will have a higher reduction potential will be more likely to be reduced and will experience reduction further, that half-cell behaves as a cathode. And the half-cell which will experience oxidation and will have a lower reduction potential value will behave as an anode.
\[{E^o}(N{i^{2 + }}/Ni) > {E^o}(M{g^{2 + }}/Mg)\]
Therefore we can conclude that, in the galvanic cell, \[N{i^{2 + }}/Ni\]half-cell will be reduced and will behave as a cathode. And \[M{g^{2 + }}/Mg\]will be oxidised and will behave as the anode.
As we know the reduction potentials for cathode and anode (known data) we can easily calculate the cell potential
\[E_{cell}^o = {E^o}(cathode) - {E^o}(anode)\]
\[E_{cell}^o= - 0.257 - ( - 2.372)\]
\[E_{cell}^o = + 2.115\]
Note: Sometimes, the direction of the redox reaction can be dictated by assessing the general qualities of the reductants and oxidants. In circumstances where an electrochemical series is not adequate to absolutely determine the direction of a redox reaction, the standard electrode potential can be used. A negative estimation of cell potential demonstrates a reducing environment, while a positive value shows an oxidizing environment.
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