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>>Electrochemistry
>>Nernst Equation
>> $pH of the solution in the anode compartm$

Question

$p H$ of the solution in the anode compartment of the following cell at $2 5^{o} C$ is $x$ when $E_{c e l l} - E_{c e l l}^{o} = 0.0591 V^{'} P t (H_{2}) ∣ p H = x ∣ ∣ N i^{2 +} (1 M) ∣ N i, x$ is $1 a t m$

A

$4$

B

$3$

C

$2$

D

$1$

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Solution

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Correct option is A)

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Similar questions

A reaction is spontaneous if $E_{c e l l} = + v e$ .

For $E_{c e l l} = + v e, △ G$ is always $- v e$ .

For the above cell.

For the above cell:

The standard Gibbs free energy change $(△ G^{\circ}$ in $k J m o l^{- 1})$ , in a Daniel cell $(E_{c e l l}^{\circ} = 1.1 V)$ , when $2$ moles of $Z n (s)$ is oxidized at $298 K$ , is closest to:

$E_{F e^{+ +} ∣ F e}^{0} < E_{N i^{+ +} ∣ N i}^{0}$ Fe electrode act as cathode and Ni electrode act as anode

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