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A hydrogen electrode placed in a buffer solution of sodium cyanide and hydrogen cyanide in the ratio x:y and y:x has the electrode potential values “a” and “b” volts respectively at 25⁰C. If a – b = 35.52 mV. Then y:x is

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answer is B.

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Detailed Solution

$H^{+} + e^{-} \to \frac{1}{2} H_{2}$

$a = E_{1} = 0.0591 \log {[H^{+}]}_{1}$

$b = E_{2} = 0.0591 \log {[H^{+}]}_{2}$

$a - b = \frac{35.52}{1000} = 0.0592 \log {[\frac{y}{x}]}^{2}$

$\frac{y}{x} = Antilog 0.30 = 2$

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