When ammonia is added to the solution, $p H$ is raised to $11$ . Which half-cell reaction is affected by $p H$ and by how much?

E_{o x d}

will increase by a factor of 0.65 from

{E^{o}}_{o x d}

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E_{o x d}

will decrease by a factor of 0.65 from

{E^{o}}_{o x d}

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E_{r e d}

will increase by a factor of 0.65 from

{E^{o}}_{r e d}

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E_{r e d}

will decrease by a factor of 0.65 from

{E^{o}}_{r e d}

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Solution

The correct option is A $E_{o x d}$ will increase by a factor of 0.65 from ${E^{o}}_{o x d}$
On addition of Ammonia $H^{+}$ concentration will decrease as the following reaction will take place.:
$N H_{3} + H^{+} \to N H_{4}^{+}$
On addition of ammonia $p H$ is raised to $11$ . Therefore, $[H^{+}] = 10^{- 11}$
Oxidation reaction: $H_{2} O \to 2 H^{+} + 2 e^{-}$

$E_{o x} = E_{o x}^{0} - \frac{0.059}{2} l o g \frac{[H^{+}]^{2}}{1}$

$E_{o x} = - 0.05 - 22 \times \frac{0.059}{2} l o g 10$
$= (- 0.05 - 0.65) V$
Therfore, $E_{o x}$ will increase by a factor of $0.65 V$ .

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