Question

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).

Answer 1

(i) For the given reaction, the Nernst equation can be given as:

$E_{cell}$ = $E^{\ominus}_{cell}$ - $\frac{0.0591}{n}$log$\frac{[{Mg}^{2+}]}{[Cu^{2+}]}$

= {0.34-(-2.36)}-$\frac{0.0591}{2}$log $\frac{.001}{.0001}$

= 2.7- $\frac{0.0591}{2}$ log10

= 2.7 - 0.02955

= 2.67 V (approximately)

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(ii) For the given reaction, the Nernst equation can be given as:

$E_{cell}$ =$E^{\ominus}_{cell}$ - $\frac{0.0591}{n}$ log $\frac{[Fe^{2+}]}{[H^+]^2}$

= {0-(-0.44)}- $\frac{0.0591}{2}$ log $\frac{0.0001}{1^2}$

= 0.44-0.02955(-3)

= 0.52865 V

= 0.53 V (approximately)

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(iii) For the given reaction, the Nernst equation can be given as:

$E_{cell}$ =$E^{\ominus}_{cell}$- $\frac{0.0591}{n}$ log$\frac{[Sn^{2+}]}{[H^+]^2}$

= {0-(-0.14)}- $\frac{0.0591}{2}$log$\frac{0.050}{(0.020)^2}$

= 0.14-0.0295 $\times$ log125

=0.14-0.62

=0.78 V

= 0.08 V (approximately)

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(iv) For the given reaction, the Nernst equation can be given as:

$E_{cell}$=$E^{\ominus}_{cell}$-$\frac{0.0591}{n}$ log $\frac{1}{[Br^-]^2[H^+]^2}$

= (0-1.09)- $\frac{0.0591}{2}$log $\frac{1}{(0.010)^2(0.030)^2}$

= -1.09-0.02955 x log $\frac{1}{0.000000009}$

= -1.09-0.02955 x log $\frac{1}{9 \times 10^{-8}}$

= -1.09-0.02955 x log(1.11 $\times$ 107)

= - 1.09 - 0.02955(0.0453+7)

= 1.09-0.208

=-1.298 V