
By how much will be the potential of a copper electrode change if the solution of $CuS{O_4}$ which is immersed is diluted to 10 times at $298K$ ?
A. Increases by $30mV$
B. Decreases by $30mV$
C. Increases by $60mV$
D. Decreases by $60mV$
Answer
559.2k+ views
Hint: Dilution is the process of decreasing the concentration of a solute in a solution, usually simply by mixing with more solvent like adding more water to a solution. To dilute a solution means to add more solvent without the addition of more solute. The resulting solution is thoroughly mixed so as to ensure that all parts of the solution are identical.
Complete step by step answer:
The electrode potential has its origin in the potential difference developed at the interface between the electrode and the electrolyte. It is common, for instance, to speak of the electrode potential of the \[{M^ + }/M\] redox couple.
The change in electrode potential is given by the following expression:
${E_2} - {E_1} = \dfrac{{2.303RT}}{{nF}}\log \dfrac{{[{C_2}]}}{{[{C_1}]}}$ …. (i)
Where, ${E_2} - {E_1} = $ change in electrode potential
$R = $ universal gas constant
$T = $ temperature = $298K$
$n = $ n-factor = number of electrons transferred in the solution
The reaction that takes place in the electrolyte is as follows:
$Cu \to C{u^{2 + }} + 2{e^ - }$
Let the initial concentration be 1 = $[{C_1}]$
On dilution, the concentration becomes = 0.1 = $[{C_2}]$
Substituting the values, we have equation (i) as:
${E_2} - {E_1} = \dfrac{{0.06}}{2}\log \dfrac{{[0.1]}}{{[1]}} = - 30mV$
Thus, the potential of a copper electrode decreases by $30mV$ if the solution of $CuS{O_4}$ which is immersed is diluted to 10 times at $298K$.
The correct option is B. Decreases by $30mV$.
Note:
In electrochemistry, electrode potential is the electromotive force of a galvanic cell built from a standard reference electrode and another electrode to be characterized. By convention, the reference electrode is the standard hydrogen electrode (SHE). It is defined to have a potential of zero volts.
Complete step by step answer:
The electrode potential has its origin in the potential difference developed at the interface between the electrode and the electrolyte. It is common, for instance, to speak of the electrode potential of the \[{M^ + }/M\] redox couple.
The change in electrode potential is given by the following expression:
${E_2} - {E_1} = \dfrac{{2.303RT}}{{nF}}\log \dfrac{{[{C_2}]}}{{[{C_1}]}}$ …. (i)
Where, ${E_2} - {E_1} = $ change in electrode potential
$R = $ universal gas constant
$T = $ temperature = $298K$
$n = $ n-factor = number of electrons transferred in the solution
The reaction that takes place in the electrolyte is as follows:
$Cu \to C{u^{2 + }} + 2{e^ - }$
Let the initial concentration be 1 = $[{C_1}]$
On dilution, the concentration becomes = 0.1 = $[{C_2}]$
Substituting the values, we have equation (i) as:
${E_2} - {E_1} = \dfrac{{0.06}}{2}\log \dfrac{{[0.1]}}{{[1]}} = - 30mV$
Thus, the potential of a copper electrode decreases by $30mV$ if the solution of $CuS{O_4}$ which is immersed is diluted to 10 times at $298K$.
The correct option is B. Decreases by $30mV$.
Note:
In electrochemistry, electrode potential is the electromotive force of a galvanic cell built from a standard reference electrode and another electrode to be characterized. By convention, the reference electrode is the standard hydrogen electrode (SHE). It is defined to have a potential of zero volts.
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