
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: \[{{E}^{o}}_{Z{{n}^{2+}}/Zn}=-0.763V,{{E}^{o}}_{Ag+/Ag}=0.80V\])
Answer
467.4k+ views
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.
Complete answer:
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 –
\[[A{{g}^{2+}}]=0.01246M\]
This is the required answer.
Note:
The electrochemical cell potentials are dependent on the molar concentration of the solutions used as electrolytes in both the cells. The standard cell potential can be used to determine the cell potential of the cells with varying concentrations as we see here.
Complete answer:
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 –
\[[A{{g}^{2+}}]=0.01246M\]
This is the required answer.
Note:
The electrochemical cell potentials are dependent on the molar concentration of the solutions used as electrolytes in both the cells. The standard cell potential can be used to determine the cell potential of the cells with varying concentrations as we see here.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Trending doubts
Who is Mukesh What is his dream Why does it look like class 12 english CBSE

Who was RajKumar Shukla Why did he come to Lucknow class 12 english CBSE

The word Maasai is derived from the word Maa Maasai class 12 social science CBSE

What is the Full Form of PVC, PET, HDPE, LDPE, PP and PS ?

Why is the cell called the structural and functional class 12 biology CBSE

Which country did Danny Casey play for class 12 english CBSE
