Calculate emf of the following cell at 25-C-Sn-Sn2-0-001 M-H- 0-01 M-H2 g1 bar-PtsE0Sn2-Sn -0-14 V E0H-H2 -0-00 V

Cell reaction, Sn+2H^+ → Sn^2+ +H_2 ; n =2 E_cell =E^0_cell 0.0591/2 log [Sn^2+]/[H^+]^2 =0 (0.14) 0.0591/2 log 0.001/(0.01)^2 =0.14 0.0591/2log 10 =0.14 ...

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Byju's Answer

Standard XII

Chemistry

Faraday's First Law

Calculate emf...

Question

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 bar)|Pt(s)
$E_{S n^{2 +} / S n}^{0} = - 0.14 V E_{H + / H_{2}}^{0} = 0.00 V$

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Solution

Cell reaction,
$S n + 2 H^{+} \to S n^{2 +} + H_{2}; n = 2$

$E_{c e l l} = E_{c e l l}^{0} - \frac{0.0591}{2} l o g \frac{[S n^{2 +}]}{[H^{+}]^{2}}$

$= 0 - (0.14) - \frac{0.0591}{2} l o g \frac{0.001}{(0.01)^{2}}$

$= 0.14 - \frac{0.0591}{2} l o g 10 = 0.14 - \frac{0.0591}{2}$

$= 0.14 - \frac{0.0591}{2} = 0.14 - 0.02955 = 0.11056 V$

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Calculate emf of the following cell at 25∘C: Sn|Sn2+(0.001M)||H+(0.01M)|H2 (g)(1 bar)|Pt(s) E0Sn2+/Sn=−0.14 V E0H+/H2=0.00 V

Cell reaction, Sn+2H+→Sn2++H2;n=2 Ecell=E0cell−0.05912log[Sn2+][H+]2 =0−(0.14)−0.05912log0.001(0.01)2 =0.14−0.05912log10=0.14−0.05912 =0.14−0.05912=0.14−0.02955=0.11056V

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 bar)|Pt(s)
$E_{S n^{2 +} / S n}^{0} = - 0.14 V E_{H + / H_{2}}^{0} = 0.00 V$