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 : $E_{X^{2 +} ∣ X}^{0} = - 2.36 V$
$E_{Y^{2 +} Y}^{0} = + 0.36 V$
$\frac{2.303 RT}{F} = 0.06 V$

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 : $E_{X^{2 +} ∣ X}^{0} = - 2.36 V$

$E_{Y^{2 +} Y}^{0} = + 0.36 V$

$\frac{2.303 RT}{F} = 0.06 V$

TestHub · Accepted Answer

The net cell reaction can be written as follows,

$X + Y^{2 +} \to Y + X^{2 +}$

$E_{cell}^{0} = E_{cathode}^{0} - E_{anode}^{0}$ (Both are SRP values)

$E_{Cell}^{0} = 0.36 - (- 2.36) = 2.72 V$

Now, the Nernst equation is,

$E_{cell} = E_{cell}^{0} - \frac{0.06}{n} \log \frac{[X^{2 +}]}{[Y^{2 +}]}$

$n =$ The number of electrons involved in the net reaction.

$E_{Cell} = 2.72 - \frac{0.06}{2} \log \frac{0.001}{0.01}$

$= 2.72 + 0.03 = 2.75 V$

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

Chemistry - Electrochemistry Question with Solution | TestHub