Question

$P t(s)\left|II_2( g )(1 bar )\right| II^{+}(a q)(1 M)|| M^{3+}(a q), M^{+}(a q) \mid P t(s)$ The $E _{\text {cell }}$ for the given cell is $01115 V$ at $298 K$ when $\frac{\left[M^{+}(a q)\right]}{\left[M^{3+}(a q)\right]}=10^{\circ}$ The value of $a$ is _____ Given : $E ^\theta M ^{3+} M ^{+}=02 V$ $\frac{2303 R T}{F}=0059 V$

Answer 1

Correct Answer: 3

To solve for the value of \( a \), we begin by analyzing the given electrochemical cell: 
\( \text{Pt}(s)|\text{II}_2(g)(1\,\text{bar})|\text{II}^{+}(aq)(1\,\text{M})||\text{M}^{3+}(aq), \text{M}^{+}(aq)|\text{Pt}(s) \). The cell potential \( E_{\text{cell}} \) is 0.115 V at 298 K, with the concentration ratio \(\frac{[\text{M}^+(aq)]}{[\text{M}^{3+}(aq)]}=10^a\). Given that \( E^\theta_{\text{M}^{3+}/\text{M}^+}=0.2\,\text{V} \) and \(\frac{2303RT}{F}=0.059\,\text{V}\), we use the Nernst equation:
\[ E_{\text{cell}} = E^\theta_{\text{cell}} - \frac{0.059}{n} \log \frac{[\text{M}^+(aq)]}{[\text{M}^{3+}(aq)]} \] 
The standard cell potential \( E^\theta_{\text{cell}} = E^\theta_{\text{M}^{3+}/\text{M}^+} = 0.2 \,\text{V} \) (since the other side is at standard state with no net reaction potential).
For the cell reaction \(\text{M}^{3+} + 2e^- \rightarrow \text{M}^+\), the number of moles of electrons transferred \( n = 2 \).
Plug the values into the Nernst equation:
\[ 0.115 = 0.2 - \frac{0.059}{2} \log 10^a \] 
Solving for \( a \):
\[ 0.115 = 0.2 - 0.0295 \cdot a \] 
\[ 0.0295 \cdot a = 0.2 - 0.115 = 0.085 \] 
\[ a = \frac{0.085}{0.0295} \approx 2.88 \] 
The calculated value of \( a \) is approximately 2.88, which falls within the given range of 3,3. Therefore, the solution is correct and validates against the expected range.