Question

The \( E^\Theta \) of \( M^{2+}|M \) is 0.3 V. At what concentration of \( Cu^{2+} \) (in mol \( L^{-1} \)), the \( E_{\text{cell}} \) value becomes zero?
\( \left(\frac{2.303RT}{F} = 0.06\right) \) (Conc. of \( M^{2+} = 0.1M \)).

• A. \( 10^{-9} \)
• B. \( 10^{-8} \)
• C. \( 10^{-11} \)
• D. \( 10^{-10} \)

Hint:

Use the Nernst equation to determine equilibrium concentrations in electrochemical cells: \[ E_{\text{cell}} = E^\Theta - \frac{0.06}{n} \log Q \]

Answer 1

The Correct Option is C

Step 1: Apply the Nernst Equation. \[ E_{\text{cell}} = E^\Theta - \frac{0.06}{2} \log \frac{[M^{2+}]}{[Cu^{2+}]} \] Since \( E_{\text{cell}} = 0 \), we set up the equation: \[ 0 = 0.3 - \frac{0.06}{2} \log \frac{0.1}{[Cu^{2+}]} \] Step 2: Solve for \( Cu^{2+} \). \[ \frac{0.06}{2} \log \frac{0.1}{[Cu^{2+}]} = 0.3 \] \[ \log \frac{0.1}{[Cu^{2+}]} = \frac{0.3}{0.03} = 10 \] \[ \frac{0.1}{[Cu^{2+}]} = 10^{10} \] \[ [Cu^{2+}] = 10^{-11} \text{ M} \]