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Calculate the standard cell potential in (\[V\]) of the cell in which following reaction takes place: \[F{e^{2 + }}(aq) + A{g^ + }(aq) \to F{e^{3 + }}(aq) + Ag(s)\;\]
Given that
 \[E_{A{g^ + }/Ag}^o = xV\]
\[E_{F{e^{2 + }}/Fe}^o{\text{ }}{\text{ }} = yV\]
\[E_{F{e^{3 + }}/Fe}^o = zV\]
A) \[x + 2y - 3z\]
B) \[x - z\]
C) \[x - y\]
D) \[x + y - z\]

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Last updated date: 29th Feb 2024
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Hint: Standard anode potential (\[E^\circ \]) in electrochemistry is considered as the amount of the individual potential of a reversible electrode at standard state with particles at a powerful centralization of \[1mol{\text{ }}d{m^{ - 3}}\]at the pressing factor of\[1{\text{ }}atm\].

Complete step by step answer:
The reason for an electrochemical cell, for example, the galvanic cell, is consistently a redox response which can be separated into two half-responses: oxidation at anode (loss of electron) and decrease at cathode (gain of electron). Power is produced because of electric potential difference between two anodes. This potential difference is made because of the difference between singular potentials of the two metal anodes as for the electrolyte. (Reversible cathode is an electrode that owes its potential to changes of a reversible sort, rather than anodes utilized in electroplating which are obliterated during their utilization.) It is the proportion of decreasing force of any component or compound.
The electrode might be emphatically or adversely accused of regard to the arrangement. A potential difference creates between the electrode and the electrolyte which is known as the anode potential. Point at which the concentrations of the relative multitude of species involved with a half-cell is solidarity then we can say that cathode potential is known as ‘standard electrode potential’. As per IUPAC show, standard decrease potentials are currently called standard electrode potentials. In a galvanic cell, the half-cell wherein oxidation happens is called anode and it has a negative potential as for the arrangement. The other half-cell in which decrease happens is called cathode and it has a positive potential as for the arrangement.
\[F{e^{2 + }}\left( {aq} \right) + A{g^ + }\left( {aq} \right) \to F{e^{3 + }}\left( {aq} \right) + Ag\left( s \right)\]
Cell response,
Anode: \[F{e^{2 + }}\left( {aq} \right) \to F{e^{3 + }}\left( {aq} \right) + {e^ - }\]; \[E_{F{e^{2 + }}/F{e^{3 + }}}^o = mV\]
Cathode: \[A{g^ + }\left( {aq} \right) + {e^ - } \to Ag\left( s \right)\]; \[E_{A{g^ + }/Ag}^o = xV\]
⇒ cell standard potential \[ = \left( {m + x} \right)V\]
Presently, to discover ' \[m\] ',
\[F{e^{2 + }} + 2{e^ - } \to Fe\]; \[E_1^o = yV \Rightarrow \Delta G_1^o = \left( {2Fy} \right)\]
\[F{e^{3 + }} + 3{e^ - } \to Fe\]; \[E_2^o = zV \Rightarrow \Delta G_2^o = \left( {3Fy} \right)\]
\[F{e^{2 + }}\left( {aq} \right) \to F{e^{3 + }}\left( {aq} \right) + {e^ - }\]; \[E_3^o = mV \Rightarrow \Delta G_3^o = - \left( {1Fm} \right)\]
\[\Delta G_3^o = \Delta G_1^0 - \Delta G_2^o = \left( { - 2Fy + 3Fz} \right) = - Fm\]
\[ \Rightarrow m = (2y - 3z)\]
\[ \Rightarrow E_{cell}^o = \left( {x + 2y - 3z} \right)V\]
Hence, the correct option is (A).

Note:
There exists a potential difference between the two anodes and when the switch is in the on position the electrons stream from negative electrode to positive cathode. The heading of the current stream is inverse to that of the electron stream.
\[E^\circ cell{\text{ }} = {\text{ }}E^\circ cathode{\text{ }} - {\text{ }}E^\circ anode\]
where \[E^\circ anode\] is the standard potential at the anode and \[E^\circ cathode\] is the standard potential at the cathode as given in the table of standard electrode potential.
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