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Nernst Equation

The EMF of th...

Question

The EMF of the cell
$M | M^{n +} (0.02 M) | | H^{+} (1 M) | H_{2} (g) (1 atm), P t$ at $25^{\circ} C$ is $0.81 V$ . Calculate the valency of the metal if the standard oxidation potential of the metal is $0.76 V$

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Solution

$M (s) \to M^{n +} (a q) + n e^{-}$ ----1
$n H^{+} (1 M) + n e^{-} \to \frac{n}{2} H_{2} (g)$ ---2
Adding 1 and 2 we get,
$n H^{+} (1 M) + M (s) \to M^{n +} (0.02 M) + \frac{n}{2} H_{2} (1 a t m)$
$E = 0.81$
$E^{o} = E_{R}^{o} - E_{L}^{o}$
= 0-(-0.76)
= 0.76
$E = E^{o} - \frac{0.059}{n} l o g (1 \times 0.02)$
$0.81 = 0.76 - \frac{0.059}{n} l o g (2 \times 10^{- 2})$
$0.05 = \frac{- 0.059}{n} \times (- 2 + l o g 2)$
$n = 2$

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Similar questions

M | M^{n +} (0.2 M) | | H^{+} (1 M) | H_{2} (g) (1 a t m)

25^{\circ} C

M | M^{n +} (0.01 M) | | H^{+} (1 M) | H_{2} (g) (1 a t m),

25^{o}

0.82

0.76

\frac{2.303 R T}{F} = 0.06

25^{o}

M | M^{n +} (0.02 M) | | H^{+} (1 M) | H_{2} (g) (1 a t m), P t

25^{o} C

0.81 V

0.76 V

log 0.02 = - 1.7

298 K

0.074 V

M | M^{n +} (a q, 0.0083 M) | | M^{n +} (a q, 0.148 M) | M

l o g (0.056) = - 1.25

P t, H_{2 (g)} (1 atm) / H^{+}, p H = x | |

0.67 V

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p H

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