Cell equation: A + 2B+ → A2+ + 2B A2+ + 2e- → A; E° = +0.34 V and log10 K = 15.6 at 300 K for cell reactions. Find E° for B+ + e- → B. [Given: (2.303 RT/nF)= 0.059 at 300 K]

Question

Cell equation: A + 2B+ → A2+ + 2B A2+ + 2e- → A; E° = +0.34 V and log10 K = 15.6 at 300 K for cell reactions. Find E° for B+ + e- → B. [Given: (2.303 RT/nF)= 0.059 at 300 K] (A) 0.80 V (B) 1.26 V (C) -0.54 V (D) +0.94 V

Tardigrade Admin · Accepted Answer

E°cell=(0.059/2) log K E°B+ / B -E°A2+ / A=(0.059/2) log K E°B+ / B -0.34 V=(0.059/2)×15.6 E°B+ / B=0.46+0.34=0.80 V

Q. Cell equation : $A + 2 B^{+}$ $\to$ $A^{2 +} + 2 B$
$A^{2 +} + 2 e^{-} \to A; E^{\circ} = + 0.34 V$
and $l o g_{10} K = 15.6 a t 300 K$ for cell reactions.
Find $E^{\circ}$ for $B^{+} + e^{-} \to B$ .
[Given : $\frac{2.303 RT}{n F} = 0.059 a t 300 K$ ]

0.80 V

1.26 V

-0.54 V

+0.94 V

$E_{ce ll}^{\circ} = \frac{0.059}{2} l o g K$
$E_{B^{+} / B}^{\circ} - E_{A^{2 +} / A}^{\circ} = \frac{0.059}{2} l o g K$
$E_{B^{+} / B}^{\circ} - 0.34 V = \frac{0.059}{2} \times 15.6$
$E_{B^{+} / B}^{\circ} = 0.46 + 0.34 = 0.80 V$

Q. Cell equation : A+2B+ → A2++2B A2++2e−→A;E∘=+0.34V and log10​K=15.6at300K for cell reactions. Find E∘ for B++e−→B. [Given : nF2.303RT​=0.059at300K]