
The standard cell potential of
Zn|\[{\rm{Z}}{{\rm{n}}^{{\rm{2 + }}}}\] (aq)||\[{\rm{C}}{{\rm{u}}^{{\rm{2 + }}}}\] (aq)|Cu cell is 1.10 V. The maximum work done by this cell will be:
(1) 106.15 KJ
(2) 212.30 KJ
(3) 318.45 KJ
(4) 464.60 KJ
Answer
163.2k+ views
Hint: In electrochemical cells, there is a relation between the work done by the electrochemical cell and the Gibbs free energy. The change of Gibbs free energy equates to an electrochemical cell's work done.
Formula used
For used for calculation of work done of electrochemical cells formula is,
\[\Delta {G^0} = - nF{E^0}_{cell}\]
Here, \[\Delta {G^0}\]is Gibbs free energy change, n is the mole of electrons involved and F is Faraday's constant.
Complete Step by Step Solution:
Now, we have to use the above formula to calculate the work done by the electrochemical cell. From the given cell equation, it is clear that the mole number of involved electrons (n) is 2, Faraday's constant is 96500 C and cell potential is 1.10 V.
\[\Delta {G^0} = - nF{E^0}_{cell}\]
\[\Delta {G^0} = 2 \times 96500 \times 1.10\]
\[\Delta {G^0} = 212300\,{\rm{J}}\]
\[\Delta {G^0}{\rm{ = 212}}{\rm{.30}}\,{\rm{KJ}}\]
Therefore, the cell's work done is 212.30 KJ.
Hence, option (2) is right.
Additional Information:nIn the copper-zinc electrochemical cells following reactions take place. The oxidation reaction happens at the cathode of zinc. Zinc lost two numbers of electrons and forms cations and the reaction is,
\[{\rm{Zn}} \to {\rm{Z}}{{\rm{n}}^{2 + }} + 2{e^ - }\]
The reduction reaction happens at the anode of copper. The copper ions take up electrons to form copper metal and get deposited. The chemical reaction is,
\[{\rm{C}}{{\rm{u}}^{{\rm{2 + }}}} + 2{e^ - } \to {\rm{Cu}}\]
Note: There are many useful applications of Cu-Zn cells. They are used in the production of electricity or storage of electricity. They are used in the field of electrical telegraphy and battery development.
Formula used
For used for calculation of work done of electrochemical cells formula is,
\[\Delta {G^0} = - nF{E^0}_{cell}\]
Here, \[\Delta {G^0}\]is Gibbs free energy change, n is the mole of electrons involved and F is Faraday's constant.
Complete Step by Step Solution:
Now, we have to use the above formula to calculate the work done by the electrochemical cell. From the given cell equation, it is clear that the mole number of involved electrons (n) is 2, Faraday's constant is 96500 C and cell potential is 1.10 V.
\[\Delta {G^0} = - nF{E^0}_{cell}\]
\[\Delta {G^0} = 2 \times 96500 \times 1.10\]
\[\Delta {G^0} = 212300\,{\rm{J}}\]
\[\Delta {G^0}{\rm{ = 212}}{\rm{.30}}\,{\rm{KJ}}\]
Therefore, the cell's work done is 212.30 KJ.
Hence, option (2) is right.
Additional Information:nIn the copper-zinc electrochemical cells following reactions take place. The oxidation reaction happens at the cathode of zinc. Zinc lost two numbers of electrons and forms cations and the reaction is,
\[{\rm{Zn}} \to {\rm{Z}}{{\rm{n}}^{2 + }} + 2{e^ - }\]
The reduction reaction happens at the anode of copper. The copper ions take up electrons to form copper metal and get deposited. The chemical reaction is,
\[{\rm{C}}{{\rm{u}}^{{\rm{2 + }}}} + 2{e^ - } \to {\rm{Cu}}\]
Note: There are many useful applications of Cu-Zn cells. They are used in the production of electricity or storage of electricity. They are used in the field of electrical telegraphy and battery development.
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