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

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Solution

$E_{c e l l} = ? E_{c e l l} = E_{c e l l}^{o} - \frac{0.0591}{2} l o g ⎛ ⎝ \frac{P_{H_{2}} [{F e}^{2 +}]}{{[H^{+}]}^{2}} ⎞ ⎠ = 0.44 - \frac{0.0591}{2} \times l o g \frac{1 \times 0.001}{1} E_{c e l l} = 0.35 V$

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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)

(Given:

E_{c e l l}^{\circ} = + 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) $P t (s) | B r_{2} (l) | B r^{-} (0.010 M) | | H^{+} (0.030 M) | H_{2} (g) (1 b a r | P t (s))$

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

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

(i) Mg(s) | Mg²⁺(0.001M) || Cu²⁺(0.0001 M) | Cu(s)

(ii) Fe(s) | Fe²⁺(0.001M) || H⁺(1M)|H₂(g)(1bar) | Pt(s)

(iii) Sn(s) | Sn²⁺(0.050 M) || H⁺(0.020 M) | H₂(g) (1 bar) | Pt(s)

(iv) Pt(s) | Br₂(l) | Br⁻(0.010 M) || H⁺(0.030 M) | H₂(g) (1 bar) | Pt(s).

Q. 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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Calculate the emf of the following cell at 298 K:Fe(s)|Fe2+(0.001M)||H+(1M)|H2(g)(1bar),Pt(s)(Given Ecell=+0.44V)

Ecell=?Ecell=Eocell−0.05912log⎛⎝PH2[Fe2+][H+]2⎞⎠=0.44−0.05912×log1×0.0011Ecell=0.35V

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

$E_{c e l l} = ? E_{c e l l} = E_{c e l l}^{o} - \frac{0.0591}{2} l o g ⎛ ⎝ \frac{P_{H_{2}} [{F e}^{2 +}]}{{[H^{+}]}^{2}} ⎞ ⎠ = 0.44 - \frac{0.0591}{2} \times l o g \frac{1 \times 0.001}{1} E_{c e l l} = 0.35 V$