The standard e.m.f of a cell, involving one electron change is found to be 0.591 V at 25°C. The equilibrium constant of the reaction is (F = 96,500 C mol-1: R = 8.314 JK-1 mol-1)

Question

The standard e.m.f of a cell, involving one electron change is found to be 0.591 V at 25°C. The equilibrium constant of the reaction is (F = 96,500 C mol-1: R = 8.314 JK-1 mol-1) (A) 1.0×101 (B) 1.0×1030 (C) 1.0×1010 (D) 1.0×105

Tardigrade Admin · Accepted Answer

∴ log Keq = (nE°/0.0591) = (1×0.591/0.0591) â Keq = 1010

Q. The standard e.m.f of a cell, involving one electron change is found to be $0.591 V$ at $2 5^{\circ} C$ . The equilibrium constant of the reaction is $(F = 96, 500 C m o l^{- 1} : R = 8.314 J K^{- 1} m o l^{- 1})$

$1.0 \times 1 0^{1}$

$1.0 \times 1 0^{30}$

$1.0 \times 1 0^{10}$

$1.0 \times 1 0^{5}$

$∴ l o g K_{e q} = \frac{n E ^{\circ}}{0.0591} = \frac{1 \times 0.591}{0.0591}$
$\Rightarrow K_{e q} = 1 0^{10}$

Q. The standard e.m.f of a cell, involving one electron change is found to be 0.591V at 25∘C. The equilibrium constant of the reaction is (F=96,500Cmol−1:R=8.314JK−1mol−1)

1.0×101

1.0×1030

1.0×1010

1.0×105