Answer
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Hint: This question can be solved with the help of the direct formula for EMF, i.e. $EMF = E_{cell}^0 - \dfrac{{0.059}}{{{n_{cell}}}}\log \dfrac{{[P]}}{{[R]}}$ where $E_{cell}^0$ is the standard electrode potential of the cell, ${n_{cell}}$ is the number of transfer electrons in the cell, $[P]$ is the concentration of the product side and $[R]$ is the concentration at the reactant side.
Complete step-by-step answer:
Given that $E_{cell}^0$ is 0.268V, the number of transfer electrons in a decinormal electrode ${n_{cell}}$ is 2.
We know that 1 decinormal = ${10^{ - 1}}$ normal
Equation for decinormal electrode is
$H{g_2}C{l_2} + 2{e^ - } \to 2Hg + 2C{l^ - }$
We know that
Using the formula for EMF,
$EMF = E_{cell}^0 - \dfrac{{0.059}}{{{n_{cell}}}}\log \dfrac{{[P]}}{{[R]}}$
Now substituting the values we get,
$ \Rightarrow EMF = 0.268 - \dfrac{{0.059}}{2}\log {[{10^{ - 1}}]^2}$
$ \Rightarrow EMF = 0.327V$
Therefore, option (B) is the correct answer.
Note- The formula for electrode potential should be memorized. The balanced chemical equation for the electrodes should be kept in mind to know the number of electrons required in the formula.Calomel electrode consists of mercury at the bottom over which a paste of mercury-mercurous chloride is placed over which a paste of potassium chloride is placed. The potential of the calomel electrode is dependent on the concentration of the potassium chloride solution. A saturated potassium chloride solution means the electrode will be called as saturated calomel electrode and if the concentration of the potassium chloride solution is 1 N, the electrode is known as normal calomel electrode (NCE) and if the concentration of potassium chloride solution is 0.1N, then the electrode is referred to as decinormal calomel electrode (DNCE). The balanced electrode reaction when the electrode acts as cathode is:
$H{g_2}C{l_2} + 2{e^ - } \to 2Hg + 2C{l^ - }$
Complete step-by-step answer:
Given that $E_{cell}^0$ is 0.268V, the number of transfer electrons in a decinormal electrode ${n_{cell}}$ is 2.
We know that 1 decinormal = ${10^{ - 1}}$ normal
Equation for decinormal electrode is
$H{g_2}C{l_2} + 2{e^ - } \to 2Hg + 2C{l^ - }$
We know that
Using the formula for EMF,
$EMF = E_{cell}^0 - \dfrac{{0.059}}{{{n_{cell}}}}\log \dfrac{{[P]}}{{[R]}}$
Now substituting the values we get,
$ \Rightarrow EMF = 0.268 - \dfrac{{0.059}}{2}\log {[{10^{ - 1}}]^2}$
$ \Rightarrow EMF = 0.327V$
Therefore, option (B) is the correct answer.
Note- The formula for electrode potential should be memorized. The balanced chemical equation for the electrodes should be kept in mind to know the number of electrons required in the formula.Calomel electrode consists of mercury at the bottom over which a paste of mercury-mercurous chloride is placed over which a paste of potassium chloride is placed. The potential of the calomel electrode is dependent on the concentration of the potassium chloride solution. A saturated potassium chloride solution means the electrode will be called as saturated calomel electrode and if the concentration of the potassium chloride solution is 1 N, the electrode is known as normal calomel electrode (NCE) and if the concentration of potassium chloride solution is 0.1N, then the electrode is referred to as decinormal calomel electrode (DNCE). The balanced electrode reaction when the electrode acts as cathode is:
$H{g_2}C{l_2} + 2{e^ - } \to 2Hg + 2C{l^ - }$
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