Calculate emf of the following cell at $25^{o} C$ :
$S n ∣ ∣ {S n}^{2 +} (0.001 M) ∣ ∣ ∣ ∣ H^{+} (0.01 M) ∣ ∣ H_{2} (g) (1 b a r) | P t (s)$
$E_{({S n}^{2 +} / S n)}^{0} = - 0.14 V E_{(H^{+} / H_{2})}^{0} = 0.00 V$

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Solution

$S n ⟶ {S n}^{2 +} + {2 e}^{-} 2 H^{+} + {2 e}^{-} ⟶ H_{2} E_{c e l l}^{o} = 0.00 + 0.14 = 0.14 V E_{c e l l} = 0.14 - \frac{0.0591}{2} l o g \frac{(0.001)}{{(0.01)}^{2}} = 0.14 - \frac{0.0591}{2} \times l o g 10 = 0.110 V$

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Similar questions

Q. Calculate emf of the following cell at

25^{\circ} C

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Q. Calculate the emf of the following cell at

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Q. Calculate emf of the following cell at

25^{\circ} C

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(g)(1 bar)|Pt(s)

E_{S n^{2 +} / S n}^{0} = - 0.14 V E_{H + / H_{2}}^{0} = 0.00 V

Calculate emf of the following cell at $25^{\circ} C$ :

$S n | S n^{2 +} (0.001 M) | | H^{+} (0.01 M) | H_{2} (g) (1 b a r) | P t (s) E_{S n 2 + | S n}^{\circ} = - 0.14 V E_{\frac{H +}{H 2}}^{\circ} = 0.00 V$

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Calculate emf of the following cell at 25oC:Sn∣∣Sn2+(0.001 M)∣∣∣∣H+(0.01 M)∣∣H2(g)(1 bar)|Pt(s)E0(Sn2+/Sn)=−0.14VE0(H+/H2)=0.00V

Sn⟶Sn2++2e−2H++2e−⟶H2Eocell=0.00+0.14=0.14VEcell=0.14−0.05912log(0.001)(0.01)2=0.14−0.05912×log10=0.110V

Calculate emf of the following cell at $25^{o} C$ :
$S n ∣ ∣ {S n}^{2 +} (0.001 M) ∣ ∣ ∣ ∣ H^{+} (0.01 M) ∣ ∣ H_{2} (g) (1 b a r) | P t (s)$
$E_{({S n}^{2 +} / S n)}^{0} = - 0.14 V E_{(H^{+} / H_{2})}^{0} = 0.00 V$

$S n ⟶ {S n}^{2 +} + {2 e}^{-} 2 H^{+} + {2 e}^{-} ⟶ H_{2} E_{c e l l}^{o} = 0.00 + 0.14 = 0.14 V E_{c e l l} = 0.14 - \frac{0.0591}{2} l o g \frac{(0.001)}{{(0.01)}^{2}} = 0.14 - \frac{0.0591}{2} \times l o g 10 = 0.110 V$