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Q.

For a cell involving one electron, $E °_{cell} = 0.59 V at 298 K,$ the equilibrium constant for the cell reaction is
$[Given that \underline{2.303 RT} = 0.059 V atT = 298 K]$

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a

$1.0 \times 10^{30}$

b

$1.0 \times 10^{5}$

c

$1.0 \times 10^{10}$

d

$1.0 \times 10^{2}$

answer is D.

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Detailed Solution

$According to Nernst equation, E_{c e l l} = E_{c e l l} - \frac{0.059}{n} \log Q_{c}$

${At equilibrium E}_{c e l l} = 0, ∴ Q_{c} = K_{c}$

$E_{c e l l} = \frac{0.059}{n} \log K_{c} \Rightarrow 0.59 = \frac{0.059}{1} \log K_{c}$

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

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For a cell involving one electron, E°cell =0.59 V at 298 K, the equilibrium constant for the cell reaction is [Given that 2.303RT¯=0.059 V atT=298 K ]