Answer
Verified
378k+ views
Hint: Try to find the concentration of \[Z{{n}^{2+}}\] which is to be used in the reaction Quotient. Electrode potential can be calculated using the Nernst equation.
Complete answer:
Electrode potential refers to the potential difference developed across the electrical double layer in one half cell. In simple words, it refers to the emf of half cell reaction.
The absolute value of the electrode potential cannot be measured as there is no flow of electrons through wire across the electrical double layer.
Let’s find the electrode potential in this question,
Dissociation of \[ZnS{{O}_{4}}\]:
\[ZnS{{O}_{4}}\left( s \right)\to Z{{n}^{2+}}\left( aq \right)+S{{O}_{4}}^{2-}\left( aq \right)\]
As it is given that the \[0.1M\]solution of \[ZnS{{O}_{4}}\] does not dissociate fully, It is dissociating only \[95\%\], therefore the concentration of \[Z{{n}^{2+}}\left( aq \right)\]will be \[0.1M\times \dfrac{95}{100}=0.095M\]
Now let’s write the half cell reaction of which we are supposed to find the electrode potential
\[Z{{n}^{2+}}\left( aq \right)+2{{e}^{-}}\to Zn\left( s \right)\]
The equation to be used is Nernst equation, which is
\[E=\left[ {{E}^{\circ }}-\left( \dfrac{0.0591}{n} \right){{\log }_{10}}Q \right]\]
\[{{E}^{\circ }}=\] Standard emf of the reaction
\[n=\] The n-value for the cell reaction is the number of moles of electrons released at anode or the number of moles of electrons gained at the cathode.
\[Q=\] Reaction quotient. It has the same expression as that of the equilibrium constant but it can be used anytime.
Now, the n-value for the half cell reaction is \[2\]in our case as two electrons are gained at the cathode.
Reaction quotient for the half cell reaction is
\[\dfrac{1}{\left[ Z{{n}^{2+}}(aq) \right]}=\dfrac{1}{0.095}\]
Pure solid \[Zn\left( s \right)\]does not appear in the reaction quotient.
\[{{E}^{\circ }}\]is already given to us which is \[-0.76V\].
Now, putting all these values in the nernst equation, we get
\[\begin{align}
& \rightarrow E=-0.76-\dfrac{0.0591}{2}\times \log \dfrac{1}{0.095} \\
& \rightarrow E=-0.76-\dfrac{0.0591}{2}\times (\log 1000-\log 95) \\
& \rightarrow E=-0.76-\dfrac{0.0591}{2}\times (3-1.9777) \\
& \rightarrow E=-0.76-0.0302 \\
& E=-0.7902V \\
\end{align}\]
Hence, Electrode potential is\[-0.7902V\]
Note:
Electrode potential is of two types:
-Oxidation potential
-Reduction Potential
The greater the oxidation potential, the more is the tendency to be oxidised. A stronger reducing agent has high oxidation potential.
Complete answer:
Electrode potential refers to the potential difference developed across the electrical double layer in one half cell. In simple words, it refers to the emf of half cell reaction.
The absolute value of the electrode potential cannot be measured as there is no flow of electrons through wire across the electrical double layer.
Let’s find the electrode potential in this question,
Dissociation of \[ZnS{{O}_{4}}\]:
\[ZnS{{O}_{4}}\left( s \right)\to Z{{n}^{2+}}\left( aq \right)+S{{O}_{4}}^{2-}\left( aq \right)\]
As it is given that the \[0.1M\]solution of \[ZnS{{O}_{4}}\] does not dissociate fully, It is dissociating only \[95\%\], therefore the concentration of \[Z{{n}^{2+}}\left( aq \right)\]will be \[0.1M\times \dfrac{95}{100}=0.095M\]
Now let’s write the half cell reaction of which we are supposed to find the electrode potential
\[Z{{n}^{2+}}\left( aq \right)+2{{e}^{-}}\to Zn\left( s \right)\]
The equation to be used is Nernst equation, which is
\[E=\left[ {{E}^{\circ }}-\left( \dfrac{0.0591}{n} \right){{\log }_{10}}Q \right]\]
\[{{E}^{\circ }}=\] Standard emf of the reaction
\[n=\] The n-value for the cell reaction is the number of moles of electrons released at anode or the number of moles of electrons gained at the cathode.
\[Q=\] Reaction quotient. It has the same expression as that of the equilibrium constant but it can be used anytime.
Now, the n-value for the half cell reaction is \[2\]in our case as two electrons are gained at the cathode.
Reaction quotient for the half cell reaction is
\[\dfrac{1}{\left[ Z{{n}^{2+}}(aq) \right]}=\dfrac{1}{0.095}\]
Pure solid \[Zn\left( s \right)\]does not appear in the reaction quotient.
\[{{E}^{\circ }}\]is already given to us which is \[-0.76V\].
Now, putting all these values in the nernst equation, we get
\[\begin{align}
& \rightarrow E=-0.76-\dfrac{0.0591}{2}\times \log \dfrac{1}{0.095} \\
& \rightarrow E=-0.76-\dfrac{0.0591}{2}\times (\log 1000-\log 95) \\
& \rightarrow E=-0.76-\dfrac{0.0591}{2}\times (3-1.9777) \\
& \rightarrow E=-0.76-0.0302 \\
& E=-0.7902V \\
\end{align}\]
Hence, Electrode potential is\[-0.7902V\]
Note:
Electrode potential is of two types:
-Oxidation potential
-Reduction Potential
The greater the oxidation potential, the more is the tendency to be oxidised. A stronger reducing agent has high oxidation potential.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Which are the Top 10 Largest Countries of the World?
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Change the following sentences into negative and interrogative class 10 english CBSE
How do you graph the function fx 4x class 9 maths CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE