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What is the standard potential of the $\text{T}{{\text{l}}^{3+}}\text{/ Tl}$ electrode?
$\text{T}{{\text{l}}^{3+}}\ \text{+ 2}{{\text{e}}^{-}}\text{ }\to \text{ T}{{\text{l}}^{+}}\text{ ; E}{}^\circ \text{ = 1}\text{.26 volt}$
$\text{T}{{\text{l}}^{+}}\ \text{+ }{{\text{e}}^{-}}\text{ }\to \text{ Tl ; E}{}^\circ \text{ = -0}\text{.336 volt}$
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Hint: The measurement of electrode potential under certain conditions is known as standard electrode potential. The formula for the standard Gibbs free energy is $\text{nFE}{}^\circ $ where n is the moles of the electron, F is the Faraday's constant and $\text{E}{}^\circ $is the standard electrode potential.
Complete answer:
-As we know that Gibbs free energy tells us about the feasibility of the cell reaction.
-The calculation of Gibbs free energy under the standard condition is known as Standard Gibb's free energy $\Delta \text{G}{}^\circ $.
-To calculate the standard electrode potential, we know that $\Delta \text{G}{}^\circ $for both the equations will be:
$\Delta \text{G}{}^\circ \text{ = nFE}{}^\circ \text{ = 2 }\times \text{ 1}\text{.26 F }.....\text{ (1)}$
-Here, the value of n is 2 because a total of 2 electrons participate in the reaction.
-The $\Delta \text{G}{}^\circ $ for second reaction will be:
$\Delta \text{G}{}^\circ \text{ = nFE}{}^\circ \text{ = 1 }\times \text{ -0}\text{.336 F }.....\text{(2)}$
-Here, the value of n is 1 because only one electron participates in the reaction.
-Now, by adding equation 1 and 2 we will get,
\[\Delta \text{G}_{3}^{{}^\circ }\text{ = }\Delta \text{G}_{1}^{{}^\circ }\text{ + }\Delta \text{G}_{2}^{{}^\circ }\]
\[\text{nFE}{}^\circ \text{ = 2}\text{.52F + (-0}\text{.336)F}\]
\[\text{3FE}{}^\circ \text{ = 2}\text{.18F}\]
-Here the F will cancel out because it is present on both sides and it is the same for both i.e. 96,485 C/mol and the value of n is 3 because the total of 3 electrons participates in the reaction.
The reaction after the addition of first and second becomes:
$\text{T}{{\text{l}}^{3+}}\text{ + 3}{{\text{e}}^{-}}\text{ }\to \text{ Tl}$
-So, the value of Standard electrode potential will be:
\[\text{E}{}^\circ \text{ = }\frac{\text{2}\text{.18}}{3}\]
\[\text{E}{}^\circ \text{ = 0}\text{.718 volt}\]
Therefore, the value of Standard electrode potential is 0.718 volt.
Note: If the value of $\Delta \text{G}{}^\circ $ is negative, then the cell reaction is spontaneous which means the reaction is possible in the forward direction i.e. from reactant to the product in the electrochemical cell. Whereas If the value of $\Delta \text{G}{}^\circ $ is positive, then the cell reaction is spontaneous which means the reaction is possible but in the reverse direction i.e. from product to the reactant.
Complete answer:
-As we know that Gibbs free energy tells us about the feasibility of the cell reaction.
-The calculation of Gibbs free energy under the standard condition is known as Standard Gibb's free energy $\Delta \text{G}{}^\circ $.
-To calculate the standard electrode potential, we know that $\Delta \text{G}{}^\circ $for both the equations will be:
$\Delta \text{G}{}^\circ \text{ = nFE}{}^\circ \text{ = 2 }\times \text{ 1}\text{.26 F }.....\text{ (1)}$
-Here, the value of n is 2 because a total of 2 electrons participate in the reaction.
-The $\Delta \text{G}{}^\circ $ for second reaction will be:
$\Delta \text{G}{}^\circ \text{ = nFE}{}^\circ \text{ = 1 }\times \text{ -0}\text{.336 F }.....\text{(2)}$
-Here, the value of n is 1 because only one electron participates in the reaction.
-Now, by adding equation 1 and 2 we will get,
\[\Delta \text{G}_{3}^{{}^\circ }\text{ = }\Delta \text{G}_{1}^{{}^\circ }\text{ + }\Delta \text{G}_{2}^{{}^\circ }\]
\[\text{nFE}{}^\circ \text{ = 2}\text{.52F + (-0}\text{.336)F}\]
\[\text{3FE}{}^\circ \text{ = 2}\text{.18F}\]
-Here the F will cancel out because it is present on both sides and it is the same for both i.e. 96,485 C/mol and the value of n is 3 because the total of 3 electrons participates in the reaction.
The reaction after the addition of first and second becomes:
$\text{T}{{\text{l}}^{3+}}\text{ + 3}{{\text{e}}^{-}}\text{ }\to \text{ Tl}$
-So, the value of Standard electrode potential will be:
\[\text{E}{}^\circ \text{ = }\frac{\text{2}\text{.18}}{3}\]
\[\text{E}{}^\circ \text{ = 0}\text{.718 volt}\]
Therefore, the value of Standard electrode potential is 0.718 volt.
Note: If the value of $\Delta \text{G}{}^\circ $ is negative, then the cell reaction is spontaneous which means the reaction is possible in the forward direction i.e. from reactant to the product in the electrochemical cell. Whereas If the value of $\Delta \text{G}{}^\circ $ is positive, then the cell reaction is spontaneous which means the reaction is possible but in the reverse direction i.e. from product to the reactant.
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