# Write the Nernst equation and emf of the following cells at 298 K:

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).

Solution:

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

$E_{\text {cell }}=E_{\text {cell }}^{\ominus}-\frac{0.0591}{n} \log \frac{\left[\mathrm{Mg}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}$

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

$=2.7-\frac{0.0591}{2} \log 10$

= 2.7 − 0.02955

= 2.67 V (approximately)

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

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

$=\{0-(-0.44)\}-\frac{0.0591}{2} \log \frac{0.001}{1^{2}}$

$=0.44-0.02955(-3)$

= 0.52865 V

= 0.53 V (approximately)

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

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

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

= 0.14 − 0.0295 × log125

= 0.14 − 0.062

= 0.078 V

= 0.08 V (approximately)

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

$E_{\text {cell }}=E_{\text {cell }}^{\ominus}-\frac{0.0591}{n} \log \frac{1}{\left[\mathrm{Br}^{-}\right]^{2}\left[\mathrm{H}^{+}\right]^{2}}$

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

$=-1.09-0.02955 \times \log \frac{1}{0.00000009}$

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

$=-1.09-0.02955 \times \log \left(1.11 \times 10^{7}\right)$

$=-1.09-0.02955(0.0453+7)$

$=-1.09-0.208$

$=-1.298 \mathrm{~V}$

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