Write the Nernst equation and emf of the following cells at 298 K: (i) $Mg ( s )\left| Mg ^{2+}(0.001 M ) \| Cu ^{2+}(0.0001 M )\right| Cu ( s )$ (ii) $Fe ( s )\left| Fe ^{2+}(0.001 M ) \| H ^{+}(1 M )\right| H 2( g )( bar ) \mid Pt ( s )$

Home » Write the Nernst equation and emf of the following cells at 298 K: (i) (ii)

Write the Nernst equation and emf of the following cells at 298 K: (i) $Mg ( s )\left| Mg ^{2+}(0.001 M ) \| Cu ^{2+}(0.0001 M )\right| Cu ( s )$
(ii) $Fe ( s )\left| Fe ^{2+}(0.001 M ) \| H ^{+}(1 M )\right| H 2( g )( bar ) \mid Pt ( s )$

CBSE | Chemistry | Class 12 | Electrochemistry | NCERT

Write the Nernst equation and emf of the following cells at 298 K: (i) $Mg ( s )\left| Mg ^{2+}(0.001 M ) \| Cu ^{2+}(0.0001 M )\right| Cu ( s )$
(ii) $Fe ( s )\left| Fe ^{2+}(0.001 M ) \| H ^{+}(1 M )\right| H 2( g )( bar ) \mid Pt ( s )$

Solution:

Nernst equation in electrochemistry is a relationship between the reduction potential of a reaction (either a half-cell or full-cell reaction) and various parameters such as the standard electrode potential, temperature, and activities (often approximated by concentrations) of the chemical species undergoing reduction and oxidation. It was given this name in honor of Walther Nernst, a German physical chemist who came up with the equation.

(i) For the given reaction, the Nernst equation can be given as:
$E_{c e l l}=E_{c e l l}^{0}-\frac{0.591}{n} \log \frac{\left[ Mg ^{2+}\right]}{\left[ Cu ^{2+}\right]}=0.34-(-2.36)-\frac{0.0591}{2} \log \frac{0.001}{0.0001} 2.7-\frac{0.0591}{2} \log 10$
$=2.7-0.02955$
$=2.67 V { approximately }$

(ii) For the given reaction, the Nernst equation can be given as:
$E_{c e l l}=E_{c e l l}^{0}-\frac{0.591}{n} \log \frac{\left[F e^{2+}\right]}{\left[H^{+}\right]^{2}}$
$=0-(-0.14)-\frac{0.0591}{n} \log \frac{0.050}{(0.020)^{2}}$
$=0.52865 V$
$=0.53 V { (approximately) }$