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Nernst Equation

Write the Ner...

Question

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

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

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Solution

Nernst equation: It is a mathematical equation that is used to calculate the emf of a cell.
Cell representation:
The left-hand side of a cell represents the anode where the oxidation takes place.
The right-hand side of a cell represents the cathode where reduction takes place.
The middle part of a cell representation denotes the salt bridge.
(i)
$C e l l e q u a t i o n : M g (s) + C u^{2 +} (a q) \to M g^{2 +} (a q) + C u (s)$

$(n = 2)$

Nernst equation : $E_{c e l l} = E_{c e l l}^{\circ} - \frac{0.0591}{2} l o g \frac{[M g^{2 +}]}{C u^{2 +}}$

EMF of the cell:
$E_{c e l l} = [0.34 - (- 2.37)] - \frac{0.0591}{2} l o g \frac{[10^{- 3}]}{10^{- 4}}$

$= 2.71 - 0.02955 = 2.68 V$

$E M F = 2.68 V$

(ii)
Cell equation : $F e (s) + 2 H^{+} (a q) \to F e^{2 +} (a q) + H_{2} (g)$

$(n = 2)$

Nernst equation : $E_{c e l l} = E_{c e l l}^{\circ} - \frac{0.0591}{2} l o g \frac{[F e^{2 +}]}{[H^{+}]^{2}}$

EMFof the cell,
$E_{c e l l} = [0 - (- 0.44)] \frac{0.0591}{2} l o g \frac{[10^{- 3}]}{[1]^{2}}$

$= 0.44 - \frac{0.591}{2} \times (- 3)$

$= 0.44 + 0.0887$

$= 0.5287 V \approx 0.53 V$

$E M F = 0.53 V$

(iii)
$C e l l E q u a t i o n : S n (s) + 2 H^{+} (a q) \to S n^{2 +} (a q) + H_{2} (g)$

$(n = 2)$

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

EMF of the cell,
$E_{c e l l} = [0 - (- 0.14) - \frac{0.0591}{2} l o g \frac{[0.05]}{[0.02]^{2}}]$

$= 0.14 - \frac{0.0591}{2} \times (2.097)$

= 0.14 - 0.0620 = 0.08V

$E M F = 0.08 V$

(iv)
Cell equation : $2 B r^{-} (l) + 2 H^{+} (a q) \to B r_{2} (l) + H_{2} (g)$

$(n = 2)$

Nernst equation : $E_{c e l l} = E_{c e l l}^{\circ} - \frac{0.0591}{2} l o g \frac{1}{[B r^{-}]^{2} [H^{+}]^{2}}$

EMF of the cell,

$E_{c e l l} = [0 - 1.08] - \frac{0.0591}{2} l o g \frac{1}{(0.01)^{2} \times (0.03)^{2}}$

$= - 1.08 - \frac{0.0591}{2} l o g (1.11.1 \times 10^{7})$

$= - 1.08 - \frac{0.0591}{2} (7.0457)$

$= - 1.08 - 0.208 = - 1.288 V$

$E M F = - 1.288 V$

( $∴$ Oxidation will occur at hydrogen electrode and reduction at $B r_{2}$ electrode)

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Similar questions

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).

298 K

M g (s) | M g^{+ 2} (0.001 M) ∥ C u^{+ 2} (0.0001 M) | C u (s)

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

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

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)

298 K

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

E_{c e l l}^{\circ} = + 0.44 V

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)

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

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