Calculate emf of the following cell at 25-C- Fe-Fe2-0-001M-H-0-01M-H2g1 barPtsE-Fe2-Fe-0-44 VF-H-H2-0-00V

For the given cell representation, the cell reaction will be #160; Fe(s)+2H^+(aq)→ Fe^2+(aq)+H_2(g) The standard emf of the cell will be #160; E^∘_c ...

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Byju's Answer

Standard XII

Chemistry

EMF

Calculate emf...

Question

Calculate emf of the following cell at $25^{\circ} C$ .
$F e | F e^{2 +} (0.001 M) | H^{+} (0.01 M) | H_{2} (g) (1 b a r) P t (s) E^{\circ} (F e^{2 +} | F e) = - 0.44 V F^{\circ} (H^{+} / H_{2}) = 0.00 V$

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Solution

For the given cell representation, the cell reaction will be
$F e (s) + 2 H^{+} (a q) \to F e^{2 +} (a q) + H_{2} (g)$
The standard emf of the cell will be
$E_{c e l l}^{\circ} = E_{H_{-}^{+} / H_{2}}^{\circ} - E_{F e^{2 +} / F e}^{\circ}$
$E_{c e l l}^{\circ} = 0 - (- 0.44) = 0.44 V$
The Nernst equation for the cell reaction at $25^{\circ} C$ will be
$E_{c e l l} = E_{c e l l}^{\circ} \frac{0.0891}{n} l o g \frac{[F e^{2 +}]}{[H^{+}]^{2}}$
$= 0.44 - \frac{0.0591}{2} l o g \frac{0.001}{(0.01)^{2}}$
$= 0.44 - \frac{0.0591}{2} l o g 10$
$= 0.44 - \frac{0.0591}{2}$
$E_{c e l l} = 0.4105 V \approx 0.41 V$

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Calculate emf of the following cell at 25∘C.Fe|Fe2+(0.001M)|H+(0.01M)|H2(g)(1bar)Pt(s)E∘(Fe2+|Fe)=−0.44VF∘(H+/H2)=0.00V

Calculate emf of the following cell at $25^{\circ} C$ .
$F e | F e^{2 +} (0.001 M) | H^{+} (0.01 M) | H_{2} (g) (1 b a r) P t (s) E^{\circ} (F e^{2 +} | F e) = - 0.44 V F^{\circ} (H^{+} / H_{2}) = 0.00 V$