For a reaction, A + B^ 2 + ⟶ B + A^ 2 + , at 25 ^ 0 CE^ 0 = 0....

Given that,

For a reaction,&

$A + B^{2 +} ⟶ B + A^{2 +}, E ° = 0. 2955V$

$T = {25}^{°}C, ⇒ T = 298Kn = 2$

Using Nernst equation,

$E ° = \dfrac{2. 303RT}{nF}{logk}_{eq}$

$E ° = \dfrac{0. 0591}{n}{logk}_{eq} \left(at{25}^{o}C\right)$

$⇒ 0. 2955 = \dfrac{0. 0591}{2}$ log{k}_{eq}$</p><p>${k}_{eq} = {10}^{10}$</p><p>Therefore, the value of${K}_{eq}$is${10}^{10}$.</p>$