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