Calculate the emf of the following cell at $298 K$ ?
$F e (s) | F e^{2 +} (0.001 M) | | H^{+} (1 M) | H_{2} (g) (1 b a r), P t (s)$
(Given: $E_{c e l l}^{\circ} = + 0.44 V$ )

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Solution

$F e (s) + 2 H^{+} + (a q) \leftrightarrow F e^{+ 2} (a q) + H_{2} (g)$
Nernst equation
$E_{c e l l} = E_{c e l l}^{0} - \frac{0.0591}{n} l o g k_{c}$
There are two electron are transfering so that n=2
$E_{c e l l}^{0} = E_{r i g h t}^{0} - E_{l e f t}^{0}$
$E_{c e l l}^{0} = 0 - (- 0.44) V$
$E_{c e l l}^{0} = + 0.44 V$
Put the value we get
Concentration of Solid substance is 1 always so that $[F e] = [H_{2}] = 1$

$\Rightarrow E_{c e l l} = 0.44 - \frac{0.0591}{2} l o g \frac{[F e^{2 +}]}{[H^{+}]^{2}}$

$\Rightarrow= 0.44 - \frac{0.0591}{2} l o g (\frac{0.001}{1^{2}})$

$\Rightarrow - E_{c e l l} = 0.44 - \frac{0.0591}{2} l o g (10^{- 3})$

$\Rightarrow E_{c e l l} = 0.44 - \frac{0.0591}{2} (- 3)$

$\Rightarrow E_{c e l l} = 0.44 + 0.089 V$

$\Rightarrow E_{c e l l} = 0.53 V$

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Similar questions

Q. Calculate the emf of the following cell at 298 K:

F e (s) | F e^{2 +} (0.001 M) | | H^{+} (1 M) | H_{2} (g) (1 b a r), P t (s)

(G i v e n E_{c e l l} = + 0.44 V)

Q. Mark the correct Nernst equation for the given cell.

F e_{(s)} | F e^{2 +} (0.001 M) | H^{+} (1 M) | H_{2 (g)} (1 b a r) | P t_{(s)}

Write the Nernst equation and emf of the following cells at 298 K:

(i) $M g (s) | M g^{2 +} (0.001 M) | | C u^{2 +} (0.001 M) | C u (s)$

(ii) $F e (s) | F e^{2 +} (0.001 M) | | H^{+} (1 M) | H_{2} (g) (1 b a r) | P t (s)$

(iii) $S n (s) | S n^{2 +} (0.050 M) | | H^{+} (0.020 M) | H_{2} (g) (1 b a r) | P t (s)$

(iv) $P t (s) | B r_{2} (l) | B r^{-} (0.010 M) | | H^{+} (0.030 M) | H_{2} (g) (1 b a r | P t (s))$

$E_{(\frac{M g^{2 +}}{M g})}^{\circ} = - 2.37 V; E_{(\frac{C u^{2 +}}{C u})}^{\circ} = + 0.34 V; E_{(\frac{F e^{2 +}}{F e})}^{\circ} = - 0.44 V; E_{(\frac{S n^{2 +}}{S n})}^{\circ} = - 0.14 V; E_{(\frac{\frac{1}{B r_{2}}}{B r^{-}})}^{\circ} = + 1.08 V,$