First slide

Nernst Equation

Question

The potential of the cell containing two hydrogen electrodes as represented below Pt, H_2(g) | H⁺ (10^-6M) || H⁺(10^-4M)|H_2(g), Pt at 298 K is

Moderate

A

– 0.118 V
B

– 0.0591 V
C

0.118 V
D

0.0591 V

Solution

cell representation $Pt, H_{2 (g)} / H^{+} (10^{- 6} M) / / H^{+} (10^{- 4} M) / H_{2 (g)}, Pt$
Cell reaction $H_{2 (1 atm)} + 2 H^{+} (10^{- 4} M) \to 2 H^{+} (10^{- 6} M) + H_{2 (1 atm)}$
Applying Nernst equation
$E_{cell} = E_{cell}^{o} - \frac{0.0591}{2} \log \frac{{(10^{- 6})}^{2}}{{(10^{- 4})}^{2}};$
$E_{cell} = 0 + 4 \times \frac{0.0591}{2} = + 0.1182 V$

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