The electrode potential of the following half cell at 298 K X | X 2+(0.001 M ) | Y 2+(0.01 M )| Y is × 10-2 V (Nearest integer). Given: E x 2+ mid x 0=-2.36 V E Y 2+|Y 0=+0.36 V (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 × 10-2 V (Nearest integer). Given: E x 2+ mid x 0=-2.36 V E Y 2+|Y 0=+0.36 V (2.303 RT / F )=0.06 V (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

X + Y 2+ arrow Y + X 2+ E text Cell 0=0.36-(-2.36)=2.72 V E text Cell =2.72-(0.06/2) log (0.001/0.01) =2.72+0.03=2.75 V =275 × 10-2 V

Q. The electrode potential of the following half cell at $298 K$
$X ∣ ∣ X^{2 +} (0.001 M) ∥ Y^{2 +} (0.01 M) ∣ ∣ Y$ is _____ $\times 1 0^{- 2} V$ (Nearest integer).
Given : $E_{x^{2 +} ∣ x}^{0} = - 2.36 V$
$E_{Y^{2 +} ∣ Y}^{0} = + 0.36 V$
$\frac{2.303 RT}{F} = 0.06 V$

$X + Y^{2 +} \to Y + X^{2 +}$
$E_{Cell}^{0} = 0.36 - (- 2.36) = 2.72 V$
$E_{Cell} = 2.72 - \frac{0.06}{2} lo g \frac{0.001}{0.01}$
$= 2.72 + 0.03 = 2.75 V$
$= 275 \times 1 0^{- 2} V$

Q. The electrode potential of the following half cell at 298K X∣∣​X2+(0.001M)∥Y2+(0.01M)∣∣​Y is _____×10−2V (Nearest integer). Given : Ex2+∣x0​=−2.36V EY2+∣Y0​=+0.36V F2.303RT​=0.06V

X+Y2+→Y+X2+ E Cell0​=0.36−(−2.36)=2.72V ECell ​=2.72−20.06​log0.010.001​ =2.72+0.03=2.75V =275×10−2V

Q. The electrode potential of the following half cell at $298 K$
$X ∣ ∣ X^{2 +} (0.001 M) ∥ Y^{2 +} (0.01 M) ∣ ∣ Y$ is _____ $\times 1 0^{- 2} V$ (Nearest integer).
Given : $E_{x^{2 +} ∣ x}^{0} = - 2.36 V$
$E_{Y^{2 +} ∣ Y}^{0} = + 0.36 V$
$\frac{2.303 RT}{F} = 0.06 V$

$X + Y^{2 +} \to Y + X^{2 +}$
$E_{Cell}^{0} = 0.36 - (- 2.36) = 2.72 V$
$E_{Cell} = 2.72 - \frac{0.06}{2} lo g \frac{0.001}{0.01}$
$= 2.72 + 0.03 = 2.75 V$
$= 275 \times 1 0^{- 2} V$