For a cell involving one electron E°cell = 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 ]

Question

For a cell involving one electron E°cell = 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 ] (A) 1.0 × 1010 (B) 1.0 × 1030 (C) 1.0 × 102 (D) 1.0 × 105

Tardigrade Admin · Accepted Answer

Ecell = E°cell - (0.059/n) log Q (At equilibrium, Q = Keq and Ecell = 0) 0 = E°cell - (0.059/1) log Keq (from equation (i)) log Keq = (E°cell/0.059) = (0.59/0.059) = 10 Keq = 1010 = 1 × 1010

Q. For a cell involving one electron $E_{ce ll}^{\circ}$ = 0.59 V at 298 K, the equilibrium constant for the cell reaction is : [ Given that $\frac{2.303 RT}{F}$ = 0.059 V at T=298 ]

$1.0 \times 1 0^{10}$

$1.0 \times 1 0^{30}$

$1.0 \times 1 0^{2}$

$1.0 \times 1 0^{5}$

$E_{ce ll} = E_{ce ll}^{\circ} - \frac{0.059}{n} lo g Q$
(At equilibrium, $Q = K_{e q}$ and $E_{ce ll} = 0)$
$0 = E_{ce ll}^{\circ} - \frac{0.059}{1} lo g K_{e q}$ (from equation (i))
$lo g K_{e q} = \frac{E _{ce ll}^{\circ}}{0.059} = \frac{0.59}{0.059} = 10$
$K_{e q} = 1 0^{10} = 1 \times 1 0^{10}$

Q. For a cell involving one electron Ecell∘​ = 0.59 V at 298 K, the equilibrium constant for the cell reaction is : [ Given that F2.303RT​= 0.059 V at T=298 ]

1.0×1010

1.0×1030

1.0×102

1.0×105

Ecell​=Ecell∘​−n0.059​logQ (At equilibrium, Q=Keq​ and Ecell​=0) 0=Ecell∘​−10.059​logKeq​ (from equation (i)) logKeq​=0.059Ecell∘​​=0.0590.59​=10 Keq​=1010=1×1010

Q. For a cell involving one electron $E_{ce ll}^{\circ}$ = 0.59 V at 298 K, the equilibrium constant for the cell reaction is : [ Given that $\frac{2.303 RT}{F}$ = 0.059 V at T=298 ]

$1.0 \times 1 0^{10}$

$1.0 \times 1 0^{30}$

$1.0 \times 1 0^{2}$

$1.0 \times 1 0^{5}$