
The single electrode potential of \[0.1\;M\;\] solution of M+/M ion is:
\[\left( {{E^0} = 2.36\;V} \right)\]
a.) \[2.42\;V\]
b.) $2.30V$
c.) $0.236V$
d.) $0.230V$
Answer
584.4k+ views
Hint: Electrode potential is defined as the potential of a cell consisting of the electrode acting as a cathode and the standard hydrogen electrode acting as an anode. Reduction always takes place at the cathode, and oxidation at the anode. We will use the particular formula for finding the electrode potential of the cell.
Complete step by step answer:
Formula used-
${E_{cell}} = {E^0} - \dfrac{{0.059}}{n}\log \left[ {\dfrac{1}{{{M^ + }}}} \right]$, where,
\[{E_{cell}}\]= electrode potential of a cell,
\[{E^{^{_0}}} = \] standard electrode potential,
$\left[ {\dfrac{1}{{{M^ + }}}} \right]$=reciprocal of the concentration of the ${M^ + }$ions in the reaction,
and $n = $order of the reaction.
Given,
the concentration of the ${M^ + }$ion in the reaction, $\left[ {{M^ + }} \right] = 0.1M$and
standard electrode potential,\[{E^0} = 2.36\;V\].
The reaction according to the question,
${M^ + } + {e^ - } \to M$
So the order of the reaction = $1$
Now using the formula and putting the values in the formula,
${E_{cell}} = 2.36 - \dfrac{{0.059}}{1}\log \left[ {\dfrac{1}{{0.1}}} \right]$
$ \Rightarrow {E_{cell}} = 2.36 - \dfrac{{0.059}}{1}\left[ {\log (1) - \log (0.1)} \right]$
$ \Rightarrow {E_{cell}} = 2.36 - \dfrac{{0.059}}{1}\left[ 1 \right]$
$ \Rightarrow {E_{cell}} = 2.36 - 0.059$
$ \Rightarrow {E_{cell}} = 2.301V$
Hence, the single electrode potential of \[0.1\;M\;\] solution of M+/M ion is ${E_{cell}} = 2.30V$.
So, the correct answer is “Option B”.
Note: To obtain the electrode potential we need the help of the standard electrode potential, so we should know about the standard electrode potential. It is measured with the help of a reference electrode known as the standard hydrogen electrode (abbreviated to SHE. The standard electrode potential, (symbol of the rode potential is ${E^0}$) , is the measure of potential of a reaction that occurs at the electrode when all the substances involved in the reaction are in their standard states. The standard states of the substances are that the concentrations of the solutions are 1M , pressure of the gases are at 1atm and solids and liquids are in pure form with all at 25C.
Complete step by step answer:
Formula used-
${E_{cell}} = {E^0} - \dfrac{{0.059}}{n}\log \left[ {\dfrac{1}{{{M^ + }}}} \right]$, where,
\[{E_{cell}}\]= electrode potential of a cell,
\[{E^{^{_0}}} = \] standard electrode potential,
$\left[ {\dfrac{1}{{{M^ + }}}} \right]$=reciprocal of the concentration of the ${M^ + }$ions in the reaction,
and $n = $order of the reaction.
Given,
the concentration of the ${M^ + }$ion in the reaction, $\left[ {{M^ + }} \right] = 0.1M$and
standard electrode potential,\[{E^0} = 2.36\;V\].
The reaction according to the question,
${M^ + } + {e^ - } \to M$
So the order of the reaction = $1$
Now using the formula and putting the values in the formula,
${E_{cell}} = 2.36 - \dfrac{{0.059}}{1}\log \left[ {\dfrac{1}{{0.1}}} \right]$
$ \Rightarrow {E_{cell}} = 2.36 - \dfrac{{0.059}}{1}\left[ {\log (1) - \log (0.1)} \right]$
$ \Rightarrow {E_{cell}} = 2.36 - \dfrac{{0.059}}{1}\left[ 1 \right]$
$ \Rightarrow {E_{cell}} = 2.36 - 0.059$
$ \Rightarrow {E_{cell}} = 2.301V$
Hence, the single electrode potential of \[0.1\;M\;\] solution of M+/M ion is ${E_{cell}} = 2.30V$.
So, the correct answer is “Option B”.
Note: To obtain the electrode potential we need the help of the standard electrode potential, so we should know about the standard electrode potential. It is measured with the help of a reference electrode known as the standard hydrogen electrode (abbreviated to SHE. The standard electrode potential, (symbol of the rode potential is ${E^0}$) , is the measure of potential of a reaction that occurs at the electrode when all the substances involved in the reaction are in their standard states. The standard states of the substances are that the concentrations of the solutions are 1M , pressure of the gases are at 1atm and solids and liquids are in pure form with all at 25C.
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