Question

The electrode potential of the following half cell at $298\, K$ $X \left| X ^{2+}(0.001 M ) \| Y ^{2+}(0.01 M )\right| Y$ is _____$\times 10^{-2} V$ (Nearest integer)
 Given: $E _{ x ^{2 *} \mid x }^0=-2.36\, V$ 
$ E _{ Y ^{0-1 Y }}^0=+0.36\, V$ 
$\frac{2303 RT }{ F }=0.06 \,V$

Answer 1

Correct Answer: 275

To find the electrode potential of the given cell, we use the Nernst equation:

\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n}\log\left( \frac{[\text{Products}]}{[\text{Reactants}]} \right) \]

First, compute the standard cell potential \(E^\circ_{\text{cell}}\):
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.36\, \text{V} - (-2.36\, \text{V}) = 2.72\, \text{V} \]

For the cell: \( \text{X} | \text{X}^{2+} | \text{Y}^{2+} | \text{Y} \), the concentration term in the Nernst equation becomes:

\[ \frac{[\text{Y}^{2+}]}{[\text{X}^{2+}]} = \frac{0.01}{0.001} = 10 \]

Assuming \( n = 2 \) as both \(\text{X}^{2+}\) and \(\text{Y}^{2+}\) involve a two-electron exchange, apply the Nernst equation:

\[ E_{\text{cell}} = 2.72\, \text{V} - \frac{0.059}{2} \log(10) = 2.72\, \text{V} - 0.059 \times 0.5 = 2.72\, \text{V} - 0.0295\, \text{V} \approx 2.6905\, \text{V} \]

Converting \(E_{\text{cell}}\) to the desired form: number \(\times 10^{-2}\), we have:

\[ E_{\text{cell}} = 269.05 \times 10^{-2} \text{V} \]

Rounding to the nearest integer, the result is:
269. The computed value fits the range 275,275, validating our calculation.