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 \((\Delta G^\circ)\).

• A. \(-212.3\ \text{kJ}\)
• B. \(106.15\ \text{kJ}\)
• C. \(+212.3\ \text{kJ}\)
• D. \(100\ \text{kJ}\)

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.

Answer 1

The Correct Option is A

Step 1: Recall the relationship between Gibbs energy and cell potential.
The standard Gibbs energy change is related to the standard cell potential by the formula: \[ \Delta G^\circ = -nFE^\circ_{\text{cell}} \] where $n$ is moles of electrons transferred, $F = 96500\ \text{C mol}^{-1}$ is Faraday's constant, and $E^\circ_{\text{cell}}$ is the standard cell potential.
Step 2: Identify $n$ from the half-reactions.
Anode (oxidation): $Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-$. Cathode (reduction): $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$. Two electrons are transferred, so $n = 2$.
Step 3: Write down all known values.
$n = 2$, $F = 96500\ \text{C mol}^{-1}$, $E^\circ_{\text{cell}} = 1.1\ \text{V}$.
Step 4: Substitute into the formula.
\[ \Delta G^\circ = -(2)(96500)(1.1) = -212300\ \text{J mol}^{-1} \]
Step 5: Convert joules to kilojoules.
\[ \Delta G^\circ = \frac{-212300}{1000} = -212.3\ \text{kJ mol}^{-1} \]
Step 6: Interpret the sign.
A negative $\Delta G^\circ$ confirms the Daniell cell reaction is spontaneous under standard conditions, consistent with zinc being a stronger reducing agent than copper.
Step 7: State the final answer.
\[ \boxed{\Delta G^\circ = -212.3\ \text{kJ mol}^{-1}} \]