For a cell involving one electron, at 298 K, the equilibrium constant for the cell reaction is [Given that ] [2019]

Question

For a cell involving one electron, $E_{cell}^{\circ} = 0.59$ $V$ at 298 K, the equilibrium constant for the cell reaction is [Given that $\frac{2.303 R T}{F} = 0.059 V at T = 298 K$ ] [2019]

Smart Achievers · Accepted Answer

(4)

According to Nernst equation,

$E_{cell} = E_{cell}^{\circ} - \frac{0.059}{n} \log Q_{c}$

At equilibrium $E_{cell} = 0,$ $∴ Q_{c} = K_{c}$

$E_{cell}^{\circ} = \frac{0.059}{n} \log K_{c} \Rightarrow 0.59 = \frac{0.059}{1} \log K_{c}$

$K_{c} = antilog 10 \Rightarrow K_{c} = 1 \times 10^{10}$

Solution

Q.
For a cell involving one electron, $E_{cell}^{\circ} = 0.59$ $V$ at 298 K, the equilibrium constant for the cell reaction is [Given that $\frac{2.303 R T}{F} = 0.059 V at T = 298 K$ ] [2019]

1 $1.0 \times 10^{30}$

2 $1.0 \times 10^{2}$

3 $1.0 \times 10^{5}$

4 $1.0 \times 10^{10}$

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Solution

Q. For a cell involving one electron, Ecell∘=0.59 V at 298 K, the equilibrium constant for the cell reaction is [Given that 2.303RTF=0.059 V at T=298 K] [2019]

1 1.0×1030

2 1.0×102

3 1.0×105

4 1.0×1010