Question

Consider the cell:
$Cd(s)|C{d^{2 + }}(1.0M)||C{u^{2 + }}(0.1M)|Cu(s)$
If we wish to make a cell with a more positive voltage using the same substance, we should:
This question has multiple correct options:
A. Increase both $[C{d^{2 + }}]$ and $[C{u^{2 + }}]$ to 2.0 M
B. Decrease the $[C{d^{2 + }}]$ to 0.1 M
C. Increase $[C{u^{2 + }}]$ to 1.0M
D. Decrease both the $[C{d^{2 + }}]$ and $[C{u^{2 + }}]$ to 0.01M

Answer 1

In this reaction Cadmium is oxidized to cadmium ion by losing two electrons and copper ion is reduced to copper by gaining two electrons. The concentration of cadmium ion and copper ion is changed to increase the cell potential.

Complete step by step answer:
Given,
Concentration of $C{d^{2 + }}$ is 1.0 M.
Concentration of $C{u^{2 + }}$ is 0.1M.
The given cell is shown below.$Cd(s)|C{d^{2 + }}(1.0M)||C{u^{2 + }}(0.1M)|Cu(s)$
The reaction for the given cell is shown below.
$Cd(s) + C{u^{2 + }}(aq) \to C{d^{2 + }}(aq) + Cu(s)$
In this reaction, a redox reaction is taking place. The redox reaction is defined as the reaction where transfer of electrons take place during the reaction and oxidation and reduction occurs simultaneously.
The oxidation is defined as loss of electrons or gain of hydrogen.
The oxidation half reaction is shown below.
$Cd(s) \to C{d^{2 + }}(aq) + 2e$
In this reaction solid cadmium is oxidized by releasing two electrons to cadmium ions.
The reduction is defined as gain of electrons or loss of hydrogen.
The reduction half reaction is shown below.
$C{u^{2 + }}(aq) + 2e \to Cu(s)$
In this reaction, copper ion is reduced by gaining two electrons to solid copper.
The formula for calculating the ${E_{cell}}$ is shown below.
${E_{cell}} = E_{cell}^0 - \dfrac{{0.0592}}{n}\operatorname{logQ}$
Where,
${E_{cell}}$is cell potential.
$E_{cell}^0$ is standard state cell potential
n is the number of electrons exchanged
Q is the reaction quotient
The reaction quotient for the given reaction is shown below.
$Q = \dfrac{{[C{d^{2 + }}][Cu]}}{{[Cd][C{u^{2 + }}]}}$
The cadmium and copper are excluded as they are in solid form.
The number of exchanged electrons is 2.
Substitute the values in the equation.
$\Rightarrow {E_{cell}} = E_{cell}^0 - \dfrac{{0.0592}}{2}\log \dfrac{{C{d^{2 + }}}}{{C{u^{2 + }}}}$
The ${E_{cell}}$ is increased when the concentration of $[C{d^{2 + }}]$ is decreased to 0.1 M and concentration of $[C{u^{2 + }}]$is increased to 1.0 M.
Therefore, the correct option is B and C.

Note:
The given formula of ${E_{cell}}$ is for non-standard conditions where the concentration is not 1 molar or pressure is not 1 atm. The ${E_{cell}}$ equation can also be written as shown below.
${E_{cell}} = E_{cell}^0 - \dfrac{{0.0257}}{n}\operatorname{lnQ}$