The electrode potential of the following half cell at 298 K

$X | X^{2 +} (0.001 M) | | Y^{2 +} (0.01 M) | Y$ is ______ $\times 10^{- 2} V$ (Nearest integer).

Given : $E_{X^{2 +} |X}^{\circ} = - 2.36 V$

$E_{Y^{2 +} |Y}^{\circ} = +0.36 V$

$\frac{2.303 RT}{F} = 0.06 V$ [2023]

Question

The electrode potential of the following half cell at 298 K

$X | X^{2 +} (0.001 M) | | Y^{2 +} (0.01 M) | Y$ is ______ $\times 10^{- 2} V$ (Nearest integer).

Given : $E_{X^{2 +} |X}^{\circ} = - 2.36 V$

$E_{Y^{2 +} |Y}^{\circ} = +0.36 V$

$\frac{2.303 RT}{F} = 0.06 V$ [2023]

Smart Achievers · Accepted Answer

(275)

$E_{cell}^{\circ} = E_{Y^{2 +} | Y}^{\circ} - E_{X^{2 +} | X}^{\circ} = 0.36 - (- 2.36) = 2.72$

$E_{cell} = E_{cell}^{\circ} - \frac{0.06}{2} \log \frac{[X^{2 +}]}{[Y^{2 +}]}$

$= 2.72 - 0.03 log (\frac{11 \times 10^{- 4}}{10^{- 2}})$

$= 2.72 - 0.03 log (11 \times 10^{- 2})$

$= 2.72 - 0.03 [\log 11 - 2]$

$= 2.72 - 0.03 [1 - 2]$

$= 2.72 + 0.03$

$= 2.75 = 275 \times 10^{- 2} V$

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