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In which of t...

Question

In which of the following cell (S) : $E_{c e l l} = E_{c e l l}^{o}$ ?

C u (s) ∣ ∣ {C u}^{2 +} (0.01 M) ∣ ∣ ∣ ∣ {A g}^{+} (0.1 M) ∣ ∣ A g (s)

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P t, H_{2} (g) 1 a t m | p H = 1 | ∣ ∣ {Z n}^{2 +} (0.01 M) ∣ ∣ Z n (s)

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P t, H_{2} (g) 1 a t m | p H = 1 | ∣ ∣ {Z n}^{2 +} (1 M) ∣ ∣ Z n (s)

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P t, H_{2} (g) 1 a t m ∣ ∣ H^{+} = 0.1 M ∣ ∣ ∣ ∣ {Z n}^{2 +} (0.01 M) ∣ ∣ Z n (s)

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Solution

The correct options are
A $C u (s) ∣ ∣ {C u}^{2 +} (0.01 M) ∣ ∣ ∣ ∣ {A g}^{+} (0.1 M) ∣ ∣ A g (s)$
B $P t, H_{2} (g) 1 a t m | p H = 1 | ∣ ∣ {Z n}^{2 +} (0.01 M) ∣ ∣ Z n (s)$
D $P t, H_{2} (g) 1 a t m ∣ ∣ H^{+} = 0.1 M ∣ ∣ ∣ ∣ {Z n}^{2 +} (0.01 M) ∣ ∣ Z n (s)$
$E_{c e l l} = E_{c e l l}^{o} - \frac{0.0591}{n} l o g Q$

if Q =1 $\Rightarrow E_{c e l l} = E_{c e l l}^{o} - 0$

$E_{c e l l} = E_{c e l l}^{o}$

$(A) Q = \frac{[C u^{2 +}]}{[A g^{+}]^{2}} = \frac{0.01}{{0.1}^{2}} = 1 (B) Q = \frac{[H^{+}]^{2}}{[Z n^{2 +}]} = \frac{{0.1}^{2}}{0.01} = 1 (C) Q = \frac{[H^{+}]^{2}}{[Z n^{2 +}]} = \frac{{0.1}^{2}}{1} = 0.01 (D) Q = \frac{[H^{+}]^{2}}{[Z n^{2 +}]} = \frac{{0.1}^{2}}{0.01} = 1$

Options A, B and D are correct.

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