If hydrogen electrode dipped in 2 solution of pH = 3 and pH = 6 and salt bridge is connected the e.m.f. of resulting cell is :

0.177 V

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0.3 V

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0.052 V

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0.104 V

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Solution

The correct option is A 0.177 V
Given $p H_{1} = 3$
$p H_{2} = 6$
$E_{c e l l} = - 0.059 log \frac{10^{- 6}}{10^{- 3}}$
$= - 0.059 log 1 0^{- 3}$
$= - 0.059 (- 3) = 0.177 V =$ Emf of the cell

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