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# One half-cell in a voltaic cell is constructed from a silver wire dipped in silver nitrate solution of unknown concentration. The other half cell consists of a zinc electrode in a 0.10M solution of $Zn{{(N{{O}_{3}})}_{2}}$.A voltage of 1.48V is measured for this cell. Use this information to calculate the concentration of silver ions in the solution. (Given: ${{E}^{o}}_{Z{{n}^{2+}}/Zn}=-0.763V,{{E}^{o}}_{Ag+/Ag}=0.80V$)

Last updated date: 21st Jun 2024
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Hint: We need to understand the relation between the concentration of the voltaic cells and the standard cell potentials of the cathode and the anode in the cell. We are given the cell potential, so we can find the concentration using the standard equations.

We know that the standard cell potentials of the cells can be used to find the cell potential of a cell in the given concentrations.
We are given the standard cell potentials of the Zn-Ag cells. We can find this information to find the cell potential of the cell combined by the following reactions at anode and cathode –
At Cathode: $A{{g}^{+}}+{{e}^{-}}\to Ag(s),{{E}^{o}}_{Ag+/Ag}=+0.80V$
At Anode: $Zn(s)\to Z{{n}^{2+}}+2{{e}^{-}},{{E}^{o}}_{Z{{n}^{2+}}/Zn}=-0.763V$
The electrochemical cell can be represented as –
$Zn(s)|Z{{n}^{2+}}(0.10M)||A{{g}^{+}}(x)|Ag(s)$
We can find the standard cell potential for this cell as –
\begin{align} & {{E}^{o}}_{cell}={{E}^{o}}_{cathode}-{{E}^{o}}_{anode} \\ & \Rightarrow {{E}^{o}}_{cell}=0.8-(-0.763) \\ & \therefore {{E}^{o}}_{cell}=1.563V \\ \end{align}
We know that the above potential is for a cell consisting of the solution with a unit concentration. The potential of the cell which has concentration other than unit can be found using the logarithmic relation between the concentrations. The formula for this is given as –
${{E}_{cell}}={{E}^{o}}_{cell}-\dfrac{0.0591}{n}\log \dfrac{[Z{{n}^{2+}}]}{{{[A{{g}^{+}}]}^{2}}}$
We know that the ‘n’ is the number of electrons involved for a balanced equation. The balanced equation for the cell reaction can be written as –
$2A{{g}^{+}}+Zn\to 2Ag+Z{{n}^{2+}}$
We can evaluate the cell potential for the reaction as –
\begin{align} & {{E}_{cell}}={{E}^{o}}_{cell}-\dfrac{0.0591}{n}\log \dfrac{[Z{{n}^{2+}}]}{{{[A{{g}^{+}}]}^{2}}} \\ & \Rightarrow 1.48=1.563-\dfrac{0.0591}{2}\log \dfrac{[0.10]}{{{[A{{g}^{2+}}]}^{2}}} \\ & \Rightarrow 2.8088=\log \dfrac{[0.10]}{{{[A{{g}^{2+}}]}^{2}}} \\ & \Rightarrow \dfrac{[0.10]}{{{[A{{g}^{2+}}]}^{2}}}=anti\log (2.8088) \\ & \Rightarrow {{[A{{g}^{2+}}]}^{2}}=\dfrac{[0.10]}{643.9} \\ & \therefore [A{{g}^{2+}}]=0.01246M \\ \end{align}
We get the concentration of the Silver nitrate solution used in the electrochemical cell as –
$[A{{g}^{2+}}]=0.01246M$