Question

The standard potentials of $C{{u}^{2+}}\left| Cu\text{ and A}{{\text{g}}^{+}}\left| Ag \right. \right.$ electrodes are 0.337 and 0.799 volts. Construct a galvanic cell using these electrodes so that its standard emf is positive. For what concentration of $A{{g}^{+}}$ will the emf of the cell at ${{25}^{\circ }}C$be zero of the concentration of $C{{u}^{2+}}$ is 0.01M?

Answer 1

To construct a galvanic cell for which standard emf is positive, we need to keep the higher reduction potential value at the cathode and the lower one at cathode. To find the concentration of the silver ins, we can use Nernst's equation.

Complete step by step answer:
In the question, the reduction potentials of $C{{u}^{2+}}\left| Cu\text{ and A}{{\text{g}}^{+}}\left| Ag \right. \right.$ electrode is given to us as 0.337 volts and 0.799 volts respectively.
So we can write that $E_{C{{u}^{2+}}\left| Cu \right.}^{\circ }$= 0.337 volt and $E_{\text{A}{{\text{g}}^{+}}\left| Ag \right.}^{\circ }$= 0.799 volt.

We have to construct a galvanic cell using these electrodes so that the value of standard emf is positive.
We know that the standard emf value if given by subtracting the reduction potential value of the cathode by that of the anode. Here, the reduction potential of silver is higher than that of copper. Therefore, if we keep $C{{u}^{2+}}\left| Cu\text{ } \right.$ at the anode, the value of standard emf will be positive.
As we know oxidation takes place at anode and reduction takes place at the cathode so the half-cell reaction is-
At anode, $Cu\to C{{u}^{2+}}+2{{e}^{-}}$and at cathode, $2A{{g}^{+}}+2{{e}^{-}}\to 2Ag$.

Therefore, we can represent the galvanic cell as- $Cu(s)\left| C{{u}^{2+}}\text{(aq}\text{.) }\left\| \text{A}{{\text{g}}^{+}}(aq.)\left| Ag \right. \right.\text{(s) } \right.$.
So the value of standard emf will be, $E_{cell}^{\circ }={{E}_{cathode}}-{{E}_{anode}}=E_{\text{A}{{\text{g}}^{+}}\left| Ag \right.}^{\circ }-E_{C{{u}^{2+}}\left| Cu \right.}^{\circ }=0.799-0.337=0.462V$.
Now, we have to find out the concentration of $A{{g}^{+}}$ such that the emf of the cell becomes zero.
Here, loss/gain of electrons =n = 2.

Using Nernst’s equation, we can write that-
$E_{cell}^{\circ }={{E}_{cell}}-\dfrac{0.059}{2}\log \dfrac{\left[ C{{u}^{2+}} \right]}{{{\left[ A{{g}^{+}} \right]}^{2}}}$
When emf of the cell is zero, the equation becomes-
$E_{cell}^{\circ }=-\dfrac{0.059}{2}\log \dfrac{\left[ C{{u}^{2+}} \right]}{{{\left[ A{{g}^{+}} \right]}^{2}}}$

We know the value of the standard cell and the concentration of$C{{u}^{2+}}$. Therefore putting those values in the above equation, we can find out the concentration of silver ions.
$\begin{aligned} & 0.462=-\dfrac{0.059}{2}\log \frac{\left[ C{{u}^{2+}} \right]}{{{\left[ A{{g}^{+}} \right]}^{2}}} \\\ & or,0=0.462-0.02955\log \frac{\left[ C{{u}^{2+}} \right]}{{{\left[ A{{g}^{+}} \right]}^{2}}} \\\ & or,\dfrac{\left[ C{{u}^{2+}} \right]}{{{\left[ A{{g}^{+}} \right]}^{2}}}=anti\log \left( \dfrac{0.462}{0.02955} \right)=4.3102\times {{10}^{15}} \\\ & or,{{\left[ A{{g}^{+}} \right]}^{2}}=\dfrac{0.01}{4.3102\times {{10}^{15}}}=0.2320\times {{10}^{-17}} \\\ & or,\left[ A{{g}^{+}} \right]=\sqrt{\left( 0.2320\times {{10}^{-17}} \right)}=1.523\times {{10}^{-9}}M \\\ \end{aligned}$
Therefore, the concentration of $A{{g}^{+}}$ required is $1.523\times {{10}^{-9}}M$.

Note: The Nernst’s equation is only valid for a chemical reaction at equilibrium.
In a galvanic cell, the cathode is positive as the ions get reduced by accepting electrons and the anode is negative as the ions are oxidised there by losing electrons. A galvanic cell converts chemical energy into electrical energy by simultaneous oxidation and reduction.