Question

For a cell involving one electron E⁻cₑ = 0.59 V at 298 K, the equilibrium constant for the cell reaction is .
[Given that (2.303 RT)/F = 0.059 V at T = 298 K ]
(1) 1.0 × 10²
(2) 1.0 × 10⁵
(3) 1.0 × 10¹⁰
(4) 1.0 × 10³⁰

Answer 1

The equilibrium constant of the reaction is $1.0\times 10^{10}$

Step by step explanation:

Equilibrium constant of the reaction can be calculated by the Nernst equation.

Nernst equation:

$\bold{E_{cell}^{o}=\frac{0.0592}{n}\,logQ}$......................(1)

From the given,

n = 1(cell involving one electron)

At equilibrium,

$E_{cell}=0$

$Q=K_{eq}$

Substitute the all values in equation (1)

$0=E_{cell}^{o}-\frac{0.059}{1}logK_{eq}$

$logK_{eq}=\frac{E_{cell}^{o}}{0.059}$

$\Rightarrow =\frac{0.59}{0.059}=10\,\,\,=10^{10}=1\times 10^{10}$

Therefore,  The equilibrium constant of the reaction is $1.0\times 10^{10}$

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