Calculate emf of the following cell at 25- C -Sn- Sn -2-0-001 M H -0-01 M - H - 2 g 1 bar - Pt s -E -circ- Sn 2- Sn -0-14 V -E -circ-lfrac H - H 2-0-00 V right

According to Nernst equation, E_cell=E^∘_cell 0.0592/nlog [P]/[R]E^∘_cell=E^∘_Oxi(anode) E^∘_Oxi(cathode) E^∘_Sn | Sn^2+ E^∘_H_2 | 2H^2+ = 0.14 0.0 = 0.1 ...

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Byju's Answer

Standard XII

Chemistry

Secondary Cells

Calculate emf...

Question

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

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

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Solution

According to Nernst equation,

$E_{c e l l} = E_{c e l l}^{\circ} - \frac{0.0592}{n} l o g \frac{[P]}{[R]}$ $E_{c e l l}^{\circ} = E_{O x i (a n o d e)}^{\circ} - E_{O x i (c a t h o d e)}^{\circ}$

$- E_{S n | S n^{2 +}}^{\circ} - E_{H_{2} | 2 H^{2 +}}^{\circ}$

= -0.14 - 0.0

= -0.14

Reactions :

Anode (oxidation) : $S n (s) \to S n^{2 +} (a q) + 2 e^{-}$

Cathode (reduction) : $2 H^{+} (a q) + 2 e^{-} \to H_{2} (g)$

Net reaction : $S n (s) + 2 H^{+} (a q) \to S n^{2 +} (a q) + H_{2} (g)$

$∴$ n = 2 (number of electrons transferred)

$[P] = [S n_{2}^{+}]$ = 0.001 M

$[R] = [H^{+}]$ = 0.01 M

So, putting the above values in the formula,

$E_{c e l l} = - 0.14 - \frac{0.0592}{2} l o g \frac{[0.001]}{[0.01]^{2}}$

$E_{c e l l} = - 0.1696 V$

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

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

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