Question

The cell potential for the following cell notation is approximately \[ M(s)\mid M^{3+}(aq,0.01M)\parallel N^{2+}(aq,0.1M)\mid N(s) \] Given \[ E^\circ_{M^{3+}/M}=0.6\ V \quad \text{and} \quad E^\circ_{N^{2+}/N}=0.1\ V \]

• A. \(0.51\ V\)
• B. \(1.5\ V\)
• C. \(2.0\ V\)
• D. \(2.5\ V\)

Hint:

For electrochemical cells, use \[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n}\log Q \] at \(298K\). Always balance the electron transfer first to find the correct value of \(n\).

Answer 1

The Correct Option is A

Step 1: Identify anode and cathode from the cell notation.
The given cell is \[ M(s)\mid M^{3+}(aq,0.01M)\parallel N^{2+}(aq,0.1M)\mid N(s) \] In cell notation, the left electrode acts as anode and the right electrode acts as cathode.
Thus, \[ M(s)\rightarrow M^{3+}+3e^- \] and \[ N^{2+}+2e^-\rightarrow N(s) \]

Step 2: Calculate standard cell potential.
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}}-E^\circ_{\text{anode}} \] Here, \[ E^\circ_{\text{cathode}}=E^\circ_{M^{3+}/M}=0.6\ V \] and \[ E^\circ_{\text{anode}}=E^\circ_{N^{2+}/N}=0.1\ V \] Therefore, \[ E^\circ_{\text{cell}}=0.6-0.1 \] \[ E^\circ_{\text{cell}}=0.5\ V \]

Step 3: Balance the electron transfer.
For oxidation: \[ M\rightarrow M^{3+}+3e^- \] For reduction: \[ N^{2+}+2e^-\rightarrow N \] The least common multiple of electrons is \[ 6 \] Thus, \[ n=6 \] Overall reaction is \[ 2M+3N^{2+}\rightarrow 2M^{3+}+3N \]

Step 4: Write the reaction quotient.
\[ Q=\frac{[M^{3+}]^2}{[N^{2+}]^3} \] Substituting values, \[ Q=\frac{(0.01)^2}{(0.1)^3} \] \[ Q=\frac{10^{-4}}{10^{-3}} \] \[ Q=10^{-1} \]

Step 5: Apply Nernst equation.
At \(298K\), \[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n}\log Q \] \[ E_{\text{cell}} = 0.5 - \frac{0.0591}{6}\log(10^{-1}) \] Since, \[ \log(10^{-1})=-1 \] \[ E_{\text{cell}} = 0.5 + \frac{0.0591}{6} \] \[ E_{\text{cell}} = 0.5+0.00985 \] \[ E_{\text{cell}} \approx0.51\ V \]

Step 6: Final conclusion.
Therefore, the cell potential is approximately \[ \boxed{0.51\ V} \] Hence, option (1) is correct.