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Gibb's Energy and Nernst Equation

Calculate the...

Question

Calculate the EMF $(inV)$ of the following electrochemical cell at $25^{o} C$ :
$P t (s) | H_{2} (g, 1 bar) | H^{+} (a q, 0.01 M) | | C u^{2 +} (0.1 M) | C u$
Give your answer upto two decimals only.
(Given: $E^{o} for C u^{2 +} | C u = 0.34 V$ , $\frac{2.303 R T}{F} = 0.059 V$ )

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Solution

The half-cell reactions are:
$H_{2} (g) ⇌ 2 H^{+} (a q) + 2 e^{-} . . . Oxidation$
$C u^{2 +} + 2 e^{-} ⇌ C u (s) . . . Reduction$
Adding both the equations,
$C u^{2 +} (a q) + H_{2} (g) ⇌ C u (s) + 2 H^{+} (a q)$
We know,
$E_{c e l l}^{o} = E_{r i g h t}^{o} - E_{l e f t}^{o}$
$\Rightarrow E_{c e l l}^{o} = 0.34 - 0 = + 0.34 V$
Using Nernst equation,
$E_{c e l l} = E_{c e l l}^{o} - \frac{0.059}{n} log \frac{[H^{+}]^{2}}{[C u^{2 +}]}$
$\Rightarrow E_{c e l l} = + 0.34 - \frac{0.059}{2} log \frac{[0.01]^{2}}{[0.1]}$
$\Rightarrow E_{c e l l} = 0.34 - \frac{0.059}{2} log 10^{- 3} \Rightarrow E_{c e l l} = 0.34 + \frac{0.059}{2} \times 3 \Rightarrow E_{c e l l} = 0.34 + 0.088 \Rightarrow E_{c e l l} = 0.428 V$

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Similar questions

The cell potential for the electrochemical reaction shown below depends upon the

C l^{-}

C u^{2 +}

concentrations. Calculate the cell potential (in V) at

25^{0} C

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C l^{-}

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C l^{-}

\Leftrightarrow

Cu(s) + 2AgCl(s) ; E = 0.12 V

Q. Calculate the emf of the cell

C u / C u^{2 +} (1 M) / / A g^{+} (0.01 M) / A g

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E^{o} A g = 0.8

E^{o} C u = 0.34

Z n | Z n^{2 +} (a q) | | C u^{2 +} (a q) | C u

E^{o}

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Z n

0.35

- 0.76

Find EMF of cell (nearest integer) if

Z n^{2 +}

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C u^{2 +} + e^{-} \to C u^{+}

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?

Q. For the following electrochemical cell at 298K,

P t (s) | H_{2} (g, 1 b a r) | H^{+} (a q, 1 M) | | M^{4 +} (a q), M^{2 +} (a q) | P t (s)

E_{c e l l} = 0.092

\frac{[M^{2 +} (a q)]}{[M^{4 +} (a q)]} = 10^{x}

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The value of x is :

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