Question:
The standard Gibbs energy change for Daniell cell reaction is
\[
Zn(s)+Cu^{2+}(aq)\rightarrow Zn^{2+}(aq)+Cu(s)
\]
\[
E^\circ_{\text{cell}}=1.1\ V
\]
Find the standard Gibbs energy change \((\Delta G^\circ)\).
The standard Gibbs energy change for Daniell cell reaction is
\[
Zn(s)+Cu^{2+}(aq)\rightarrow Zn^{2+}(aq)+Cu(s)
\]
\[
E^\circ_{\text{cell}}=1.1\ V
\]
Find the standard Gibbs energy change \((\Delta G^\circ)\).
Show Hint
For an electrochemical cell,
\[
\Delta G^\circ=-nFE^\circ_{\text{cell}}
\]
A positive \(E^\circ_{\text{cell}}\) gives a negative \(\Delta G^\circ\), indicating a spontaneous reaction.
Updated On: Jun 26, 2026
- \(-212.3\ \text{kJ}\)
- \(106.15\ \text{kJ}\)
- \(+212.3\ \text{kJ}\)
- \(100\ \text{kJ}\)
Show Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Step 1: Use the relation between Gibbs energy and cell potential.
The standard Gibbs energy change is related to the standard cell potential by \[ \Delta G^\circ=-nFE^\circ_{\text{cell}} \] where \[ n=\text{number of electrons transferred} \] \[ F=96500\ \text{C mol}^{-1} \] \[ E^\circ_{\text{cell}}=1.1\ V \]
Step 2: Determine the value of \(n\).
The oxidation half-reaction is \[ Zn \rightarrow Zn^{2+}+2e^- \] The reduction half-reaction is \[ Cu^{2+}+2e^- \rightarrow Cu \] Thus, \[ n=2 \]
Step 3: Substitute the values.
\[ \Delta G^\circ = -(2)(96500)(1.1) \] \[ = -212300\ \text{J mol}^{-1} \]
Step 4: Convert into kJ mol\(^{-1}\).
\[ \Delta G^\circ = \frac{-212300}{1000} \] \[ = -212.3\ \text{kJ mol}^{-1} \]
Step 5: Final conclusion.
Therefore, the standard Gibbs energy change is \[ \boxed{-212.3\ \text{kJ mol}^{-1}} \] Hence, the correct option is \[ \boxed{(1)} \]
The standard Gibbs energy change is related to the standard cell potential by \[ \Delta G^\circ=-nFE^\circ_{\text{cell}} \] where \[ n=\text{number of electrons transferred} \] \[ F=96500\ \text{C mol}^{-1} \] \[ E^\circ_{\text{cell}}=1.1\ V \]
Step 2: Determine the value of \(n\).
The oxidation half-reaction is \[ Zn \rightarrow Zn^{2+}+2e^- \] The reduction half-reaction is \[ Cu^{2+}+2e^- \rightarrow Cu \] Thus, \[ n=2 \]
Step 3: Substitute the values.
\[ \Delta G^\circ = -(2)(96500)(1.1) \] \[ = -212300\ \text{J mol}^{-1} \]
Step 4: Convert into kJ mol\(^{-1}\).
\[ \Delta G^\circ = \frac{-212300}{1000} \] \[ = -212.3\ \text{kJ mol}^{-1} \]
Step 5: Final conclusion.
Therefore, the standard Gibbs energy change is \[ \boxed{-212.3\ \text{kJ mol}^{-1}} \] Hence, the correct option is \[ \boxed{(1)} \]
Was this answer helpful?
0
0
Top AP EAPCET Chemistry Questions
- The four quantum numbers of the valence electron of potassium are :
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
- In which of the following the election gain enthalpy of elements is correctly arranged?
- The radii of $ \text{F,}{{\text{F}}^{-}}\text{O} $ and $ {{\text{O}}^{2-}} $ are in the order
- Which of the following has the highest first ionization potential?
View More Questions
Top AP EAPCET Electrochemistry Questions
- Consider the following standard electrode potentials (\(E^\circ\) in volts) in aqueous solution: \includegraphics[]{135.png} Based on this data, which of the following statements is correct?
- The standard reduction potentials of \( 2H^+/H_2 \), \( Cu^{2+}/Cu \), \( Zn^{2+}/Zn \), and \( NO_3^-/HNO_3 \) are 0.0 V, 0.34 V, -0.76 V, and 0.97 V respectively. Identify the correct statements from the following:
I. \( H^+ \) does not oxidize \( Cu \) to \( Cu^{2+} \).
II. \( Zn \) reduces \( Cu^{2+} \) to \( Cu \).
III. \( NO_3^- \) oxidizes \( Cu \) to \( Cu^{2+} \). - At T(K), the electrode potential of A$^{2+}$(X M)$|$A(s) is –0.2285 V. Given $E^\circ_{A^{2+}/A} = -0.14$ V, the value of X in mol L$^{-1}$ is:
- At 298K, the conductivity of KCl solutions of molarity 0.1, 0.01 and 1.0 M are recorded as X, Y and Z, S cm$^{-1}$ respectively. The correct relation between X, Y and Z is
\( \Delta G^\circ \, (\text{in kJ mol}^{-1}) \text{ for the cell reaction is} \)
\( \text{Cu}^{2+}(aq) + \text{Fe}(s) \rightarrow \text{Fe}^{2+}(aq) + \text{Cu}(s) \)
\( \left[\text{Given} \, E^\circ_{\text{Cu}^{2+}/\text{Cu}} = 0.34 \, \text{V}, \, E^\circ_{\text{Fe}^{2+}/\text{Fe}} = -0.44 \, \text{V} \, \text{and} \, F = 96,500 \, \text{C mol}^{-1} \right] \)
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions